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Introduction
Suppose we have a set \(A\). Any map of the form \(A \times A \mapsto A\) is called a (binary) operation on the set \(A\). It is important that the range of the operation not leave \(A\), i.e., that closure with respect to the operation hold. For example, the usual subtraction operation on the natural numbers \(\mathbb {N}\) is not closed: for instance, \(6 - 9 = -3 \not\in \mathbb {N}\). On the integers, however, subtraction is clearly closed.
For simplicity, let us denote the operation by a dot \(\bullet\). A set with a binary operation (closed with respect to the operation) \((A, \bullet )\) is called a magma or groupoid.
A magma \((A, \bullet )\) is called a semigroup if
- \((a \bullet b) \bullet c = a \bullet (b \bullet c), \; \forall a,b,c \in A\), i.e., it does not matter in what order the operation is performed. This property is called associativity of the operation.
A semigroup \((A, \bullet )\) is called a monoid if
- there exists a special element \(e \in A\) such that \(a \bullet e = e \bullet a = a \; \forall a \in A\). This element is called the identity element with respect to the operation \(\bullet\). When \(\bullet\) is addition, \(e\) is denoted by \(0\). When \(\bullet\) is multiplication, \(e\) is denoted by \(1\).
A monoid \((A, \bullet )\) is called a group if
- every element is invertible, i.e., \(\forall a \in A \; \exists b \in A: \; a \bullet b= b \bullet a = e\). When \(\bullet\) is addition, \(b\) is denoted by \((-a)\). When \(\bullet\) is multiplication, \(b\) is denoted by \(a^{-1}\).
A group \((A, \bullet )\) is called an abelian group if
- the operation \(\bullet\) on \(A\) is commutative: \(a \bullet b = b \bullet a , \; \forall a,b \in A\).
A set \(\mathbb {F}\) with two operations, usually denoted \(+, \cdot\) and called addition and multiplication, is called a ring if
\((\mathbb {F}, +)\) is an abelian group under addition;
addition and multiplication are compatible, namely, expansion of parentheses is compatible:
\[
\begin{align} \alpha \cdot (\beta + \gamma ) & = \alpha \cdot \beta + \alpha \cdot \gamma \\ (\beta + \gamma ) \cdot \alpha & = \beta \cdot \alpha + \gamma \cdot \alpha \end{align}
\]
\(\forall \alpha , \beta , \gamma \in \mathbb {F}\). This property is also called distributivity.
A ring is called commutative if its multiplication is commutative. The same terminology applies to an associative ring. A commutative associative ring with a unit (an identity element for multiplication) is called a field if every one of its nonzero elements is invertible (under multiplication). That is, a field is a ring that is almost an abelian group under multiplication. The difference is that in an abelian group all elements must be invertible, while in a field only the nonzero elements are invertible.
Put another way, a field is a set with addition and multiplication such that under addition it is an abelian group, under multiplication it is almost an abelian group (\(0\) is not invertible, but otherwise all the properties hold), and addition and multiplication are linked by distributivity.
Let a field \(\mathbb {F}\) be given (typically \(\mathbb {R}\) or \(\mathbb {C}\)). We denote the identity elements of this field with respect to addition and multiplication by \(0,1\). We call the elements of this field scalars. A quadruple \((V, \mathbb {F},+, \cdot )\), consisting of a set \(V\), a field of scalars \(\mathbb {F}\), an addition operation \(+: V \times V \to V\), and a scalar multiplication operation \(\cdot \; : \; \mathbb {F} \times V \to V\), is called a vector space if the following conditions hold:
Under addition \((V, +)\) is an abelian group;
For the scalar multiplication operation the following properties hold:
\(\alpha \cdot (\beta \cdot \overrightarrow {x}) = (\alpha \cdot \beta ) \cdot \overrightarrow {x}, \quad \forall \alpha , \beta \in \mathbb {F}, \; \forall \overrightarrow {x} \in V\) (associativity of scalar multiplication). Note that the first and second multiplications are multiplications of a scalar by an element of \(V\), but the 3rd multiplication is multiplication of a scalar by a scalar. These operations are fundamentally different in nature (\(\mathbb {F} \times V \mapsto V\) and \(\mathbb {F} \times \mathbb {F} \mapsto \mathbb {F}\)), even though this is usually not marked explicitly;
\(1 \cdot \overrightarrow {x} = \overrightarrow {x}, \; \forall \overrightarrow {x} \in V\) (unitarity: multiplying by the identity element of the field \(\mathbb {F}\) under multiplication preserves the vector);
\((\alpha + \beta ) \cdot \overrightarrow {x} = \alpha \cdot \overrightarrow {x} + \beta \cdot \overrightarrow {x}, \quad \forall \alpha , \beta \in \mathbb {F}, \; \forall \overrightarrow {x} \in V\) (distributivity of scalar multiplication with respect to addition of scalars);
\(\alpha \cdot (\overrightarrow {x} + \overrightarrow {y}) = \alpha \cdot \overrightarrow {x} + \alpha \cdot \overrightarrow {y}, \quad \forall \alpha \in \mathbb {F}, \; \forall \overrightarrow {x}, \overrightarrow {y} \in V\) (distributivity of scalar multiplication with respect to addition of vectors).
From these properties it is easy to derive other natural properties.
Prove that for any \(\overrightarrow {x} \in V\) we have \(0 \cdot \overrightarrow {x} = \overrightarrow {0}\), where \(0 \in \mathbb {F}\) is the additive identity element in \(\mathbb {F}\), and \(\overrightarrow {0} \in V\) is the additive identity element in \(V\).
\(0 \cdot \overrightarrow {x} = (0 + 0) \cdot \overrightarrow {x} = 0 \cdot \overrightarrow {x} + 0 \cdot \overrightarrow {x}\), hence \[
\overrightarrow {0} = 0 \cdot \overrightarrow {x} + \left(-\left(0 \cdot \overrightarrow {x} \right)\right) = 0 \cdot \overrightarrow {x} + 0 \cdot \overrightarrow {x} + \left(-\left(0 \cdot \overrightarrow {x} \right)\right) = 0 \cdot \overrightarrow {x}
\]
Prove that for any \(\overrightarrow {x} \in V\) we have \((-1) \cdot \overrightarrow {x} = -\overrightarrow {x}\), where \((-1) \in \mathbb {F}\) is the additive inverse of the multiplicative identity element \(1\) of the field \(\mathbb {F}\), and \(-\overrightarrow {x} \in V\) is the additive inverse of \(\overrightarrow {x} \in V\).
We have \[
\overrightarrow {0} = \left(1 + (- 1)\right) \overrightarrow {x} = 1 \cdot \overrightarrow {x} + (-1)\cdot \overrightarrow {x} = \overrightarrow {x} + (-1)\cdot \overrightarrow {x}.
\]
Typically, when one speaks of a vector space \((V, \mathbb {F},+, \cdot )\), it is denoted simply by \(V\), since it is usually clear from context over which field it is defined and how its operations are introduced.
In this chapter, unless stated otherwise, we use the field of real numbers.
Examples of spaces. Basis and dimension
Let \(\mathbb {R}^{n}\) be the set of columns of \(n\) real numbers: \[
\mathbb {R}^{n} := \left\{ \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \quad : \quad x_i \in \mathbb {R}, \; \forall i \in \left\{ 1, \ldots , n\right\} \right\} .
\] Addition and scalar multiplication by elements of \(\mathbb {R}\) are introduced componentwise. Show that \((\mathbb {R}^{n}, \mathbb {R}, + , \cdot )\) is a vector space.
Both operations are closed: the sum of two columns of \(n\) real numbers and the product of a column by a real number are again columns of \(n\) real numbers.
All the axioms are checked componentwise and reduce to the corresponding properties of the field \(\mathbb {R}\). For example, commutativity of addition: the \(i\)-th component of \(\overrightarrow {x} + \overrightarrow {y}\) equals \(x_i + y_i = y_i + x_i\), i.e., the \(i\)-th component of \(\overrightarrow {y} + \overrightarrow {x}\), and this holds for every \(i\). Associativity of addition, both distributivity laws, associativity of scalar multiplication, and unitarity (\(1 \cdot \overrightarrow {x} = \overrightarrow {x}\), since \(1 \cdot x_i = x_i\)) are verified analogously.
The additive identity element is the zero column \(\overrightarrow {0} = \left[\begin{smallmatrix} 0 & \cdots & 0 \end{smallmatrix}\right]^T\), and the additive inverse of \(\overrightarrow {x}\) is the column \(-\overrightarrow {x}\) of numbers \(-x_i\). Altogether \((\mathbb {R}^{n}, +)\) is an abelian group, and all the properties of scalar multiplication hold.
Let \(C[0,1]\) be the set of continuous real-valued functions on \([0,1]\). Addition and scalar multiplication by elements of \(\mathbb {R}\) are introduced in the obvious way. Show that \((C[0,1], \mathbb {R}, + , \cdot )\) is a vector space.
The operations are introduced pointwise: \((f + g)(t) := f(t) + g(t)\), \((\alpha f)(t) := \alpha f(t)\). The key point is closure: from calculus it is known that the sum of two continuous functions and the product of a continuous function by a number are continuous, i.e., the operations do not take us outside \(C[0,1]\).
All the axioms are checked pointwise and reduce to the properties of \(\mathbb {R}\) β at each point \(t \in [0,1]\) the reasoning is the same as for the components in the example with \(\mathbb {R}^{n}\). The identity element is the identically zero function (it is continuous), and the inverse of \(f\) is the function \(-f\) (also continuous).
Show that if \(\mathbb {R}\) is replaced by \(\mathbb {C}\) in the previous examples, we again obtain vector spaces.
In the checks above we nowhere used any specific properties of the real numbers β only the fact that \(\mathbb {R}\) is a field: associativity and commutativity of addition and multiplication, distributivity, and the existence of \(0\), \(1\), and inverse elements. \(\mathbb {C}\) is also a field, so all the reasoning carries over verbatim: \(\mathbb {C}^{n}\) (columns of \(n\) complex numbers) and the space of continuous complex-valued functions on \([0,1]\) are vector spaces over \(\mathbb {C}\).
In this example we consider various sets of sequences of real numbers. For each of them we must check whether it is a vector space. Addition and scalar multiplication are defined componentwise in each example.
\(\mathbb {R}^{\infty }:= \left\{ (x_1, x_2, \ldots ) \; : \; x_i \in \mathbb {R}, \; \forall i \in \mathbb {N}\right\}\) β the space of all sequences of real numbers;
\(\left\{ (x_1, x_2, \ldots ) \; : \; x_i \in \mathbb {R}, \; \forall i \in \mathbb {N}; \quad \sup_{n}x_n < \infty \right\}\) β the space of sequences of real numbers bounded above;
\(l^{\infty }:= \left\{ (x_1, x_2, \ldots ) \; : \; x_i \in \mathbb {R}, \; \forall i \in \mathbb {N}; \quad \sup_{n}\left|x_n\right| < \infty \right\}\) β the space of bounded sequences of real numbers;
\(c:= \left\{ (x_1, x_2, \ldots ) \; : \; x_i \in \mathbb {R}, \; \forall i \in \mathbb {N}; \quad \exists \lim_{n \to \infty }x_n\right\}\) β the space of convergent sequences of real numbers;
\(\left\{ (x_1, x_2, \ldots ) \; : \; x_i \in \mathbb {R}, \; \forall i \in \mathbb {N}; \quad \lim_{n \to \infty }x_n = 1\right\}\) β the space of sequences of real numbers converging to one;
\(c_0:= \left\{ (x_1, x_2, \ldots ) \; : \; x_i \in \mathbb {R}, \; \forall i \in \mathbb {N}; \quad \lim_{n \to \infty }x_n = 0\right\}\) β the space of sequences of real numbers converging to zero;
\(c_{00}:= \left\{ (x_1, x_2, \ldots ) \; : \; x_i \in \mathbb {R}, \; \forall i \in \mathbb {N}; \quad \exists N \in \mathbb {N}: \forall n > N \; x_n = 0 \right\}\) β the space of sequences of real numbers that are equal to zero from some index onward.
(Spaces of sequences)
In every item the vector space axioms (associativity, distributivity, etc.) hold automatically, since the operations are componentwise β the only question is closure with respect to addition and scalar multiplication (and the existence of a zero element, which follows from closure under multiplication by \(0\)).
Yes: the sum of any two sequences and the product of a sequence by a number are again sequences, with no additional conditions.
No: there is no closure under multiplication by a negative scalar. For example, \(\overrightarrow {x} = (-1, -2, -3, \ldots )\) is bounded above, but \((-1) \cdot \overrightarrow {x} = (1, 2, 3, \ldots )\) is not.
Yes: if \(\sup_n \left|x_n\right| = M_x < \infty\) and \(\sup_n \left|y_n\right| = M_y < \infty\), then \(\left|x_n + y_n\right| \leq M_x + M_y\) and \(\left|\alpha x_n\right| \leq \left|\alpha \right| M_x\) for all \(n\).
Yes: by the arithmetic of limits, if \(x_n \to a\) and \(y_n \to b\), then \(x_n + y_n \to a + b\) and \(\alpha x_n \to \alpha a\), i.e., sums and scalar multiples of convergent sequences converge.
No: if \(x_n \to 1\) and \(y_n \to 1\), then \(x_n + y_n \to 2 \neq 1\). (Also, this set has no zero element: the zero sequence converges to \(0\), not to \(1\).)
Yes: the same reasoning as in item 4, with \(a = b = 0\): \(x_n + y_n \to 0\), \(\alpha x_n \to 0\).
Yes: if \(x_n = 0\) for \(n > N_x\) and \(y_n = 0\) for \(n > N_y\), then \(x_n + y_n = 0\) for \(n > \max \left\{ N_x, N_y\right\}\), and \(\alpha x_n = 0\) for \(n > N_x\).
A nonempty subset \(U \subseteq V\) of a vector space \(V\) is called a subspace if it is closed with respect to the operations:
\(\overrightarrow {x} + \overrightarrow {y} \in U, \; \forall \overrightarrow {x}, \overrightarrow {y} \in U\);
\(\alpha \cdot \overrightarrow {x} \in U, \; \forall \alpha \in \mathbb {F}, \forall \overrightarrow {x} \in U\).
Prove that a subspace is closed with respect to taking the inverse element, i.e., for any \(\overrightarrow {x}\in U\) we have \((-\overrightarrow {x}) \in U\).
We have \(-\overrightarrow {x} = (-1) \cdot \overrightarrow {x}\).
Consider the spaces \(\mathbb {R}^{\infty }, c, c_0, c_{00}, l^{\infty }\). Which is a subspace of which?
We have \(c_{00} \subset c_{0} \subset c \subset l^{\infty } \subset \mathbb {R}^{\infty }\).
Definition 1 Let \(V\) be some space over a field \(\mathbb {F}\). A collection of vectors \(B \subset V\) (finite or infinite) is called linearly independent if every (finite) linear combination of its elements that equals the zero vector is trivial. That is, \(\forall n \in \mathbb {N}, \; \forall \left\{ \overrightarrow {v}_i\right\}_{i=1}^n \subset B\) and for any collection of scalars \(\left\{ \alpha_i\right\}_{i=1}^n\) such that \(\alpha_1\overrightarrow {v}_1 + \ldots + \alpha_n \overrightarrow {v}_n = \overrightarrow {0}\), we have \(\alpha_1 = \cdots = \alpha_n = 0\).
Find some linearly independent collections of vectors in \(\mathbb {R}^{n}\).
For example, \(\left\{ \overrightarrow {e}_i\right\}_{i=1}^n\), where \(\overrightarrow {e}_i\) is the column with \(1\) in position \(i\) and zeros elsewhere.
Any subset of \(\left\{ \overrightarrow {e}_i\right\}_{i = 1}^n\) also works.
Consider the spaces \(\mathbb {R}^{\infty }, c, c_0, c_{00}, l^{\infty }\). Find linearly independent collections there.
For example, \(\left\{ \overrightarrow {e}_i\right\}_{i \in \mathbb {N}}\), where \(\overrightarrow {e}_i\) is the sequence with \(1\) in position \(i\) and zeros elsewhere.
Any subset of \(\left\{ \overrightarrow {e}_i\right\}_{i \in \mathbb {N}}\) also works.
Definition 2 Recall that for a collection \(B\) of vectors, the linear span is the set of all possible finite linear combinations of them. Notation: \(\left\langle B\right\rangle\) or \(\operatorname {span}(B)\). We also say that \(B\) spans \(\left\langle B\right\rangle\).
It is easy to show that the linear span of a set of vectors is the smallest (under inclusion) vector space containing those vectors.
For a vector space, a basis is any linearly independent system of vectors that spans the space.
If a space \(V\) has a finite basis, the space is called finite-dimensional. If this basis has \(n\) vectors, we say that the dimension of \(V\) equals \(n\) and write \(\operatorname {dim}(V) = n\).
It is easy to prove that if a space has at least one finite basis, then all its other bases are also finite, and the number of vectors in all of them is the same. Thus, the dimension of a finite-dimensional space is well defined.
Let \(\mathbb {R}^{m \times n}\) be the space of \(m \times n\) matrices. Find a finite basis for this space. What is the dimension of this space? What will the coefficients of the expansion of an arbitrary matrix \(A \in \mathbb {R}^{m \times n}\) in the found basis be equal to?
Consider the matrix units \(E^{(i,j)} \in \mathbb {R}^{m \times n}\), \(i \in \left\{ 1, \ldots , m\right\}\), \(j \in \left\{ 1, \ldots , n\right\}\): the matrix \(E^{(i,j)}\) has \(1\) in position \((i,j)\) and zeros everywhere else. Any matrix expands uniquely in terms of them: \[
A = \sum _{i=1}^{m} \sum _{j=1}^{n} a_{i,j} \cdot E^{(i,j)},
\] i.e., the collection \(\left\{ E^{(i,j)}\right\}\) spans \(\mathbb {R}^{m \times n}\), and the expansion coefficients are exactly the entries of the matrix \(a_{i,j}\). Linear independence: if \(\sum_{i,j} \alpha_{i,j} E^{(i,j)} = O\), then, looking at position \((i,j)\) of the left-hand side, we get \(\alpha_{i,j} = 0\) for all \(i, j\). Altogether this is a basis, and \[
\dim \mathbb {R}^{m \times n} = mn
\]
The dimension equals \(mn\); the basis is given by the matrix units \(E^{(i,j)}\), and the expansion coefficients are the entries of the matrix \(a_{i,j}\).
Consider the spaces \(\mathbb {R}^{\infty }, c, c_0, c_{00}, l^{\infty }\). For which of them is the collection \(\left\{ \overrightarrow {e}_i\right\}_{i=1}^{\infty }\) a basis?
Typically, the vectors in a basis are ordered. That is, a basis is spoken of as an ordered collection.
Determine whether the given set of square matrices of order \(n\) is a linear subspace of the space of all square matrices of order \(n\), and if it is, find its dimension:
the set of matrices with zero first row
the set of diagonal matrices
the set of upper triangular matrices
the set of skew-symmetric matrices
the set of nonsingular matrices
Yes; the dimension equals \(n(n - 1)\)
Yes; the dimension equals \(n\)
Yes; the dimension equals \(n(n+1) / 2\)
Yes; the dimension equals \(n(n - 1) / 2\)
No, since
\[
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}
\]
Prove that for any natural number \(n\) the given set of functions forms a finite-dimensional vector space; find the dimension and give a basis for this space:
the set of polynomials of degree at most \(n\) (denoted \(\left. \mathcal{P}^{(n)}\right)\)
the set of even polynomials of degree at most \(n\)
the set of odd polynomials of degree at most \(n\)
In all three cases closure is obvious: the sum of polynomials of degree at most \(n\) and a scalar multiple of such a polynomial are again a polynomial of degree at most \(n\), and evenness/oddness (i.e., having only even/only odd powers of \(t\)) is preserved under these operations.
The monomials \(1, t, \ldots , t^n\) span \(\mathcal{P}^{(n)}\) by the definition of a polynomial. Linear independence: if the polynomial \(\alpha_0 + \alpha_1 t + \ldots + \alpha_n t^n\) is identically zero (as a function), then all its coefficients are zero (a nonzero polynomial of degree \(\leq n\) has at most \(n\) roots). Altogether this is a basis, of dimension \(n + 1\).
A polynomial is an even function if and only if all its nonzero coefficients stand at even powers (compare the coefficients of \(f(t)\) and \(f(-t)\)). Hence the even polynomials of degree at most \(n\) are exactly the linear combinations of the monomials \(1, t^2, \ldots , t^{2k}\), \(k = [n/2]\), and they are linearly independent as a subset of the basis from item 1. The dimension is \(k + 1 = [n/2] + 1\).
Similarly: oddness is equivalent to the nonzero coefficients standing only at odd powers. Basis \(t, t^3, \ldots , t^{2k-1}\), \(k = [(n+1)/2]\); dimension \([(n+1)/2]\).
The dimension equals \(n + 1\); basis: \(1, t, t^{2}, \ldots , t^{n}\)
The dimension equals \([n / 2]+1\); basis: \(1, t^{2}, \ldots , t^{2 k}(k = [n / 2])\)
The dimension equals \([(n+1) / 2]\); basis: \(t\), \(t^{3}, \ldots , t^{2 k - 1}(k = [(n+1) / 2])\)
Let \(x_0 \in \mathbb {R}\) be an arbitrary fixed number. Prove that the polynomials of the form \[
1, x - x_0, (x-x_0)^2, (x-x_0)^3, \ldots , (x-x_0)^n
\] form a basis in the space of polynomials of degree at most \(n\), and find the coordinate column of the polynomial \[
f(x) = a_0 + a_1x + \ldots + a_nx^n
\] in this basis.
If \[
f(x) = a_0 + a_1x + \ldots + a_nx^n = b_0 + b_1(x-x_0) + b_2(x-x_0)^2 + \ldots + b_n(x-x_0)^n,
\] then \(b_k = \frac{f^{(k)}(x_0)}{k!}, \; k \in \left\{ 0, \ldots , n\right\}\).
Let \(V, V'\) be two vector spaces over the same field \(\mathbb {F}\). A map \(f: V \to V'\) is called an isomorphism if it is bijective and preserves the operations:
\(f\) is one-to-one;
\(f(\overrightarrow {x} + \overrightarrow {y}) = f(\overrightarrow {x}) + f(\overrightarrow {y}), \; \forall \overrightarrow {x}, \overrightarrow {y} \in V\);
\(f(\lambda \overrightarrow {x}) = \lambda f(\overrightarrow {x}), \; \forall \lambda \in \mathbb {F}, \; \forall \overrightarrow {x} \in V\).
Any finite-dimensional space \(V\) over \(\mathbb {F}\) of dimension \(n\) is isomorphic to \(\mathbb {F}^n\). Indeed, choose a basis \(\overrightarrow {e}_1, \ldots , \overrightarrow {e_n}\) in \(V\). Let \(\overrightarrow {x}\) be an arbitrary vector from \(V\). It turns out that \(\overrightarrow {x}\) is uniquely represented in the form
\[
\overrightarrow {x} = \alpha _1\overrightarrow {e}_1 + \ldots + \alpha _n \overrightarrow {e}_n, \quad \alpha _1, \ldots , \alpha _n \in \mathbb {F}
\]
Then define the map \(f\) as follows: to \(\overrightarrow {x}\) assign the column of its coordinates in the basis \(\left\{ \overrightarrow {e}_i\right\}\):
\[
f(\overrightarrow {x}) := \left[\begin{smallmatrix} \alpha _1 \\ \vdots \\ \alpha _n \end{smallmatrix}\right]
\]
It is easy to check that \(f\) is indeed an isomorphism from \(V\) to \(\mathbb {F}^n\). Such an isomorphism for a finite-dimensional space is called canonical.
Note that the isomorphism above depends on the basis chosen in \(V\). For example, consider the example with polynomials. There, besides the obvious basis \(1, x, \ldots , x^n\), another basis is considered. It is clear that for some fixed polynomial \(P_k\) of degree \(k \leq n\) the coordinates in one basis and in the other will differ.
Find the dimension and a basis of the linear span of the given system of columns: \(\overrightarrow {c}_{166}, \overrightarrow {c}_{198}, \overrightarrow {c}_{199}, \overrightarrow {c}_{201}\). Matrices and vectors: \(\overrightarrow {c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \overrightarrow {c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right], \overrightarrow {c}_{199} = \left[\begin{smallmatrix} 3 \\ - 5 \\ 7 \\ 2 \end{smallmatrix}\right], \overrightarrow {c}_{201} = \left[\begin{smallmatrix} 1 \\ - 7 \\ 5 \\ - 2 \end{smallmatrix}\right]\).
Recall that the linear span of columns does not change under elementary transformations of the columns. We have
\[
\begin{aligned} \begin{bmatrix} \overrightarrow {c}_{166} & \overrightarrow {c}_{198} & \overrightarrow {c}_{199} & \overrightarrow {c}_{201} \end{bmatrix} & = \begin{bmatrix} 1 & 1 & 3 & 1 \\ 1 & 1 & -5 & -7 \\ 1 & 1 & 7 & 5 \\ 1 & 3 & 2 & -2 \end{bmatrix} \\ & \mathrel {\leadsto } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 0 & -8 & -8 \\ 1 & 0 & 4 & 4 \\ 1 & 2 & -1 & -3 \end{bmatrix} \\ & \mathrel {\leadsto } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & -8 & -8 & 0 \\ 1 & 4 & 4 & 0 \\ 1 & -3 & -1 & 2 \end{bmatrix} \\ & \mathrel {\leadsto } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & -8 & 0 & 0 \\ 1 & 4 & 0 & 0 \\ 1 & -3 & 2 & 2 \end{bmatrix} \\ & \mathrel {\leadsto } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & -8 & 0 & 0 \\ 1 & 4 & 0 & 0 \\ 1 & -3 & 2 & 0 \end{bmatrix} \end{aligned}
\]
We brought the matrix to column echelon form. The rank of the matrix equals \(3\), and this is the dimension of the space. As a basis one may take the nonzero columns of the echelon form, since there are exactly \(3\) of them and they are linearly independent.
The dimension equals \(3\); basis: \[
\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}, \qquad \begin{bmatrix} 0 \\ -8 \\ 4 \\ -3 \end{bmatrix}, \qquad \begin{bmatrix} 0 \\ 0 \\ 0 \\ 2 \end{bmatrix}.
\]
Set up a system of equations defining the linear span of the given system of columns: \(\overrightarrow {c}_{166}, \overrightarrow {c}_{198}, \overrightarrow {c}_{199}, \overrightarrow {c}_{201}\). Matrices and vectors: \(\overrightarrow {c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \overrightarrow {c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right], \overrightarrow {c}_{199} = \left[\begin{smallmatrix} 3 \\ - 5 \\ 7 \\ 2 \end{smallmatrix}\right], \overrightarrow {c}_{201} = \left[\begin{smallmatrix} 1 \\ - 7 \\ 5 \\ - 2 \end{smallmatrix}\right]\).
The linear span is always a vector space, i.e., it contains the zero vector \(\overrightarrow {0}\). Hence \(\overrightarrow {0}\) is one of the solutions of the desired system, and so the system must be homogeneous: \(A \overrightarrow {x} = \overrightarrow {0}\). We need to find the matrix \(A\).
Let us check the rank of the given system. Write the columns as rows and bring the matrix to echelon form.
\[
\begin{bmatrix} \overrightarrow {c}^T_{166} \\ \overrightarrow {c}^T_{198} \\ \overrightarrow {c}^T_{199} \\ \overrightarrow {c}^T_{201} \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 3 \\ 3 & -5 & 7 & 2 \\ 1 & -7 & 5 & -2 \end{bmatrix} \mathrel {\leadsto } \cdots \mathrel {\leadsto } \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & -8 & 4 & -3 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix}
\]
The dimension of the linear span equals 3, and a basis may be given by the following vectors:
\[
\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}, \qquad \begin{bmatrix} 0 \\ -8 \\ 4 \\ -3 \end{bmatrix}, \qquad \begin{bmatrix} 0 \\ 0 \\ 0 \\ 2 \end{bmatrix}.
\]
For details, see the worked problem 20.14(5).
So, for the desired system we have a fundamental matrix:
\[
\Phi = \begin{bmatrix} 1 & 0 & 0 \\ 1 & -8 & 0 \\ 1 & 4 & 0 \\ 1 & -3 & 2 \end{bmatrix}
\]
We need to obtain the coefficient matrix \(A\).
Recall that if \(\Phi\) is a fundamental matrix of the HSLE \(A\overrightarrow {x} = \overrightarrow {0}\), then the system \(\Phi^T \overrightarrow {y} = \overrightarrow {0}\) defines the linear span of the rows of \(A\). That is, the fundamental matrix for the system \(\Phi^T \overrightarrow {y} = \overrightarrow {0}\), when transposed, is one of the possible choices of the matrix \(A\). In our case we have
\[
\Phi ^T \overrightarrow {y} = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & -8 & 4 & -3 \\ 0 & 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}
\]
Let us bring \(\Phi^T\) to reduced form:
\[
\begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & -8 & 4 & -3 \\ 0 & 0 & 0 & 2 \end{bmatrix} \mathrel {\leadsto } \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & -1/2 & 3/8 \\ 0 & 0 & 0 & 1 \end{bmatrix} \mathrel {\leadsto } \begin{bmatrix} 1 & 0 & 3/2 & 0 \\ 0 & 1 & -1/2 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}
\]
The fundamental matrix for the system \(\Phi^T \overrightarrow {y} = \overrightarrow {0}\):
\[
\begin{bmatrix} -3/2 \\ 1/2 \\ 1 \\ 0 \end{bmatrix} =: A^T
\]
The desired system (one of the possible choices):
\[
0 = \overrightarrow {0} = A \overrightarrow {x} = -\frac{3}{2}x_1 + \frac{1}{2}x_2 + x_3 \iff 0 = 3x_1 -x_2 - 2x_3
\]
\(3 x_{1} - x_{2} - 2 x_{3} = 0\).
Sum and intersection of subspaces
Let \(U,W\) be two linear subspaces of the same ambient space \(V\): \(U,W \subset V\).
The intersection \(U \cap W\) consists of vectors belonging to both subspaces. Let us verify that the subspace conditions hold:
If \(x, y \in U \cap W\), then \(x, y \in U\) and \(x, y \in W\). Since \(U\) and \(W\) are subspaces, \(x+y \in U\) and \(x+y \in W\), hence \(x+y \in U \cap W\).
If \(x \in U \cap W\) and \(\alpha \in \mathbb {R}\), then \(\alpha x \in U\) and \(\alpha x \in W\), hence \(\alpha x \in U \cap W\).
Thus, \(U \cap W\) is a subspace.
The union \(U \cup W\) is a subspace if and only if it is closed with respect to addition and scalar multiplication. If \(U \not\subset W\) and \(W \not\subset U\), then there exist \(u \in U \setminus W\) and \(w \in W \setminus U\). Then their sum \(u + w\) belongs to neither \(U\) nor \(W\): if \(u+w \in U\) held, then \(w = (u+w) - u \in U\) (since \(u \in U\)), contradicting \(w \notin U\); similarly \(u + w \notin W\). Hence, \(u+w \notin U \cup W\), which violates closure. Therefore, \(U \cup W\) is a subspace only in the case where one of the subspaces is contained in the other.
The sum of the spaces \(U,W\) is the set consisting of all possible sums of vectors from \(U\) and \(W\):
\[
U + W := \left\{ \overrightarrow {u} + \overrightarrow {w} \; : \; \overrightarrow {u} \in U, \; \overrightarrow {w} \in W \right\} \subset V
\]
Prove that the sum of any two subspaces is itself a subspace.
Note that \[
U + W = \left\langle U \cup W\right\rangle
\] Indeed, \(U + W \subseteq \left\langle U \cup W\right\rangle\), since \(\overrightarrow {u} + \overrightarrow {w}\) is a linear combination of vectors from \(U \cup W\). Conversely, any finite linear combination of vectors from \(U \cup W\) splits into two parts: a combination of vectors from \(U\) (it lies in \(U\), since \(U\) is a subspace) plus a combination of vectors from \(W\) (lies in \(W\)); altogether this is an element of \(U + W\). And the linear span of any set is a subspace.
Grassmannβs formula: in a finite-dimensional \(V\) we have \(\dim (U + W) = \dim (U) + \dim (W) - \dim (U \cap W)\).
Prove Grassmannβs formula.
Consider bases of the subspaces \(U\) and \(W\):
Let \(\{ \overrightarrow {u}_1, \dots , \overrightarrow {u}_k\}\) be a basis of \(U\).
Let \(\{ \overrightarrow {w}_1, \dots , \overrightarrow {w}_m\}\) be a basis of \(W\).
Denote \(U \cap W = S\) and choose a basis \(\{ s_1, \dots , s_r\}\) for it. Extend the basis of \(S\) to a basis of \(U\) by adding vectors \(\overrightarrow {u}_{r+1}, \dots , \overrightarrow {u}_k\), and to a basis of \(W\) by adding vectors \(\overrightarrow {w}_{r+1}, \dots , \overrightarrow {w}_m\).
We claim that the set of vectors \(\{ s_1, \dots , s_r, \overrightarrow {u}_{r+1}, \dots , \overrightarrow {u}_k, \overrightarrow {w}_{r+1}, \dots , \overrightarrow {w}_m\}\) is a basis of \(U + W\). That it spans \(U + W\) is clear: any vector of the sum is (a vector from \(U\)) + (a vector from \(W\)), and each summand expands in its own part of the collection. Let us check linear independence. Suppose
\[
\sum _{i=1}^{r} \gamma _i s_i + \sum _{i=r+1}^{k} \lambda _i \overrightarrow {u}_i + \sum _{i=r+1}^{m} \mu _i \overrightarrow {w}_i = \overrightarrow {0}
\]
Denote \(\overrightarrow {v} := \sum_{i=r+1}^{m} \mu_i \overrightarrow {w}_i\). On the one hand, \(\overrightarrow {v} \in W\). On the other hand, from the equality above, \(\overrightarrow {v} = -\sum \gamma_i s_i - \sum \lambda_i \overrightarrow {u}_i \in U\). Hence, \(\overrightarrow {v} \in U \cap W = S\), i.e., \(\overrightarrow {v}\) expands in terms of the basis \(s_1, \ldots , s_r\). But then, equating the two expansions of \(\overrightarrow {v}\) in the linearly independent system \(\{ s_i\} \cup \{ \overrightarrow {w}_i\}\) (this is a basis of \(W\)), we get \(\mu_{r+1} = \ldots = \mu_m = 0\). What remains is \(\sum \gamma_i s_i + \sum \lambda_i \overrightarrow {u}_i = \overrightarrow {0}\), and this is a linear combination of the basis vectors of \(U\) β all coefficients are zero.
Therefore,
\[
\dim (U + W) = r + (k - r) + (m - r) = k + m - r = \dim U + \dim W - \dim (U \cap W).
\]
Find the dimension and a basis of the sum and intersection of the linear subspaces \(U\), \(W\) of the \(4\)-dimensional arithmetic space, spanned by the systems of vectors \(\overrightarrow {a}_{1}, \overrightarrow {a}_{2}\) and \(\overrightarrow {b}_{1}, \overrightarrow {b}_{2}\), respectively, where \[
\overrightarrow {a}_1 = \left[\begin{smallmatrix} 1 \\ 0 \\ 0 \\ 0 \end{smallmatrix}\right], \quad \overrightarrow {a}_2 = \left[\begin{smallmatrix} 1 \\ 0 \\ 1 \\ 0 \end{smallmatrix}\right], \quad \overrightarrow {b}_1 = \left[\begin{smallmatrix} 0 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \quad \overrightarrow {b}_2 = \left[\begin{smallmatrix} 2 \\ 1 \\ 0 \\ 1 \end{smallmatrix}\right]
\]
Let the first subspace be \(U\) and the second \(W\). Let us simplify the condition. Note that the linear span of rows does not change under elementary row transformations. We have
\[
\begin{align} \begin{bmatrix} \overrightarrow {a}_1^T \\ \overrightarrow {a}_2^T \end{bmatrix} & = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 \end{bmatrix} \mathrel {\leadsto } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} \\ \begin{bmatrix} \overrightarrow {b}_1^T \\ \overrightarrow {b}_2^T \end{bmatrix} & = \begin{bmatrix} 0 & 1 & 1 & 1 \\ 2 & 1 & 0 & 1 \end{bmatrix} \mathrel {\leadsto } \begin{bmatrix} 1 & 1/2 & 0 & 1/2 \\ 0 & 1 & 1 & 1 \\ \end{bmatrix} \mathrel {\leadsto } \begin{bmatrix} 1 & 0 & -1/2 & 0 \\ 0 & 1 & 1 & 1 \\ \end{bmatrix} \end{align}
\]
Let
\[
\overrightarrow {a}'_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \quad \overrightarrow {a}'_2 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \quad \overrightarrow {b}'_1 = \begin{bmatrix} 1 \\ 0 \\ -1/2 \\ 0 \end{bmatrix}, \quad \overrightarrow {b}'_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \end{bmatrix}
\]
We have \(U = \left\langle \overrightarrow {a}'_1, \overrightarrow {a}'_2\right\rangle , \; W = \left\langle \overrightarrow {b}'_1, \overrightarrow {b}'_2\right\rangle\).
Let us find the dimension and basis of the sum. The sum is spanned by the union of the spanning systems: \(U + W = \left\langle \overrightarrow {a}'_1, \overrightarrow {a}'_2, \overrightarrow {b}'_1, \overrightarrow {b}'_2\right\rangle\). Write the vectors as rows and bring the matrix to echelon form:
\[
\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & -1/2 & 0 \\ 0 & 1 & 1 & 1 \end{bmatrix} \mathrel {\leadsto } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -1/2 & 0 \\ 0 & 1 & 1 & 1 \end{bmatrix} \mathrel {\leadsto } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
\]
(we subtracted row 1 from row 3; added row 2 with coefficient \(1/2\) to row 3 and swapped rows.) Altogether \(\dim (U + W) = 3\), and as a basis of the sum we can take the nonzero rows of the echelon form (transposed):
\[
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \qquad \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \end{bmatrix}, \qquad \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}
\]
Let us find the dimension and basis of the intersection. Let \(\overrightarrow {x} \in U \cap W\) be an arbitrary vector from the intersection. Then for some scalars \(\alpha_1, \alpha_2, \beta_1, \beta_2\) we have
\[
\overrightarrow {x} = \alpha _1\overrightarrow {a}'_1 + \alpha _2\overrightarrow {a}'_2 = \beta _1\overrightarrow {b}'_1 + \beta _2\overrightarrow {b}'_2 \quad \iff \quad \begin{bmatrix} \alpha _1 \\ 0 \\ \alpha _2 \\ 0 \end{bmatrix} = \begin{bmatrix} \beta _1 \\ \beta _2 \\ -\frac{1}{2}\beta _1 + \beta _2 \\ \beta _2 \end{bmatrix}
\]
From the 2nd (or 4th) coordinate, \(\beta_2 = 0\); then from the 1st coordinate, \(\beta_1 = \alpha_1\), and from the 3rd, \(\alpha_2 = -\frac{1}{2}\alpha_1\). Altogether
\[
\overrightarrow {x} = \alpha _1 \overrightarrow {a}'_1 - \frac{\alpha _1}{2} \overrightarrow {a}'_2 = \frac{\alpha _1}{2} \cdot \left[\begin{smallmatrix} 2 \\ 0 \\ -1 \\ 0 \end{smallmatrix}\right], \quad \alpha _1 \in \mathbb {R} \text{ -- arbitrary}
\]
Thus, \(U \cap W = \left\langle \left[\begin{smallmatrix} 2 \\ 0 \\ -1 \\ 0 \end{smallmatrix}\right]\right\rangle\) is a one-dimensional subspace (this is exactly \(\left\langle \overrightarrow {b}'_1\right\rangle\): the vector \(\overrightarrow {b}'_1 = \overrightarrow {a}'_1 - \frac{1}{2}\overrightarrow {a}'_2\) lies in \(U\)).
Let us check ourselves with Grassmannβs formula: \(\dim (U + W) = \dim U + \dim W - \dim (U \cap W) = 2 + 2 - 1 = 3\). It checks out.
\(\dim (U + W) = 3\) with basis \(\left[\begin{smallmatrix} 1 \\ 0 \\ 0 \\ 0 \end{smallmatrix}\right], \left[\begin{smallmatrix} 0 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \left[\begin{smallmatrix} 0 \\ 0 \\ 1 \\ 0 \end{smallmatrix}\right]\); \(\; U \cap W = \left\langle \left[\begin{smallmatrix} 2 \\ 0 \\ -1 \\ 0 \end{smallmatrix}\right]\right\rangle\) of dimension \(1\).
The sum \(U + W\) of subspaces \(U,W \subset V\) is called direct if for every \(\overrightarrow {x} \in U + W\) its representation \(\overrightarrow {u} + \overrightarrow {w}\) as a sum of \(\overrightarrow {u} \in U\), \(\overrightarrow {w} \in W\) is unique. Notation: \(U \oplus W\).
The sum \(U + W\) is direct \(\iff\) \(U \cap W = \left\{ \overrightarrow {0}\right\}\).
(\(\Rightarrow\)) Suppose the sum is direct, and let \(\overrightarrow {x} \in U \cap W\). Then the vector \(\overrightarrow {x} \in U + W\) has two representations:
\[
\overrightarrow {x} = \underbrace{\overrightarrow {x}}_{\in U} + \underbrace{\overrightarrow {0}}_{\in W} = \underbrace{\overrightarrow {0}}_{\in U} + \underbrace{\overrightarrow {x}}_{\in W}
\]
By uniqueness of the representation, \(\overrightarrow {x} = \overrightarrow {0}\). That is, \(U \cap W = \left\{ \overrightarrow {0}\right\}\).
(\(\Leftarrow\)) Suppose \(U \cap W = \left\{ \overrightarrow {0}\right\}\), and suppose some \(\overrightarrow {x}\) has two representations: \(\overrightarrow {x} = \overrightarrow {u}_1 + \overrightarrow {w}_1 = \overrightarrow {u}_2 + \overrightarrow {w}_2\), where \(\overrightarrow {u}_i \in U, \overrightarrow {w}_i \in W\). Then
\[
\underbrace{\overrightarrow {u}_1 - \overrightarrow {u}_2}_{\in U} = \underbrace{\overrightarrow {w}_2 - \overrightarrow {w}_1}_{\in W} \in U \cap W = \left\{ \overrightarrow {0}\right\}
\]
That is, \(\overrightarrow {u}_1 = \overrightarrow {u}_2\) and \(\overrightarrow {w}_1 = \overrightarrow {w}_2\) β the representation is unique, the sum is direct.
Prove that the space \(V\) of all real-valued functions defined on all of \(\mathbb {R}\) is the direct sum of the subspace of even functions \(U\) and the subspace of odd functions \(W\). Find the projection of the function \(e^x\) onto the indicated subspaces.
Let \(f: \mathbb {R} \to \mathbb {R}\) be an arbitrary function, and suppose \(f(x) = g(x) + h(x)\), where \(g\) is even and \(h\) is odd. Then
\[
f(-x) = g(-x) + h(-x) = g(x) - h(x)
\]
Hence, \(g(x) = \frac{1}{2}(f(x) + f(-x))\), \(h(x) = \frac{1}{2}(f(x) - f(-x))\). That is, any function \(\mathbb {R} \mapsto \mathbb {R}\) can indeed be decomposed into a sum of an even and an odd function.
Moreover, it is clear that the intersection of the set of even functions and the set of odd functions is the set consisting only of the function identically equal to zero. Hence the sum is direct.
For the exponential we have
\[
\operatorname {Pr}_{U}^{\parallel W}(e^x) = \frac{e^x + e^{-x}}{2} = \operatorname {ch}(x), \qquad \operatorname {Pr}_{W}^{\parallel U}(e^x) = \frac{e^x - e^{-x}}{2} = \operatorname {sh}(x).
\]
Let \(U \subset V\) be a subspace. A subspace \(W \subset V\) is called a direct complement to \(U\) in \(V\) if \(V = U \oplus W\). In this case, every element \(\overrightarrow {x} \in V\) can be represented uniquely as a sum of elements from \(U\) and from \(W\). The first summand of this sum is called the projection of the vector \(\overrightarrow {x}\) onto the subspace \(U\) along the subspace \(W\) and is denoted \(\operatorname {Pr}_U^{\parallel W}(\overrightarrow {x})\). The second summand is, respectively, the projection onto \(W\) along \(U\). Thus, we have
\[
\overrightarrow {x} = \operatorname {Pr}_U^{\parallel W}\hspace{-0.5em}(\overrightarrow {x}) + \operatorname {Pr}_W^{\parallel U}(\overrightarrow {x})
\]
Every subspace of a finite-dimensional space has a direct complement.
Let \((\overrightarrow {e}_1, \ldots , \overrightarrow {e}_m)\) be a basis of \(U\). It is known that it can be extended to a basis \((\overrightarrow {e}_1, \ldots , \overrightarrow {e}_m, \overrightarrow {e}_{m+1}, \ldots , \overrightarrow {e}_n)\) of the whole space \(V\). Set \[
W := \left\langle \overrightarrow {e}_{m+1}, \ldots , \overrightarrow {e}_n\right\rangle
\] Then \(V = U + W\): any vector expands in the basis of \(V\), and it suffices to group the first \(m\) terms (this gives a vector from \(U\)) and the last \(n - m\) (a vector from \(W\)). Moreover, \(U \cap W = \left\{ \overrightarrow {0}\right\}\): if \(\overrightarrow {x}\) lies in the intersection, then it expands both in terms of \(\overrightarrow {e}_1, \ldots , \overrightarrow {e}_m\) and in terms of \(\overrightarrow {e}_{m+1}, \ldots , \overrightarrow {e}_n\); subtracting one expansion from the other, we obtain a nontrivial (if \(\overrightarrow {x} \neq \overrightarrow {0}\)) zero combination of the basis vectors of \(V\) β a contradiction. By the criterion from the example above, the sum is direct: \(V = U \oplus W\).
Give an example showing that the projection onto a subspace depends not only on the subspace itself, but also on the chosen complement.
Let \(V = \mathbb {R}^{2}\), \(U = \left\langle \overrightarrow {e}_1\right\rangle\) be the horizontal axis. Consider two different direct complements:
\[
W_1 = \left\langle \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]\right\rangle , \qquad W_2 = \left\langle \left[\begin{smallmatrix} 1 \\ 1 \end{smallmatrix}\right]\right\rangle
\]
(in both cases \(U \cap W_i = \left\{ \overrightarrow {0}\right\}\) and \(U + W_i = \mathbb {R}^{2}\)). Let us project the vector \(\overrightarrow {x} = \left[\begin{smallmatrix} 1 \\ 1 \end{smallmatrix}\right]\) onto \(U\):
\[
\overrightarrow {x} = \underbrace{\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]}_{\in U} + \underbrace{\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]}_{\in W_1} \; \Rightarrow \; \operatorname {Pr}_U^{\parallel W_1}(\overrightarrow {x}) = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right], \qquad \overrightarrow {x} = \underbrace{\left[\begin{smallmatrix} 0 \\ 0 \end{smallmatrix}\right]}_{\in U} + \underbrace{\left[\begin{smallmatrix} 1 \\ 1 \end{smallmatrix}\right]}_{\in W_2} \; \Rightarrow \; \operatorname {Pr}_U^{\parallel W_2}(\overrightarrow {x}) = \left[\begin{smallmatrix} 0 \\ 0 \end{smallmatrix}\right]
\]
The subspace \(U\) is the same, but the projections are different.