6.10.1: Examples and Elementary Properties

Definition: Linear Transformations of Vector Spaces

If \(V\) and \(W\) are two vector spaces, a function \(T : V \to W\) is called a linear transformation if it satisfies the following axioms.

\[\begin{array}{lll} \mbox{T1. } \quad T(\mathbf{v} + \mathbf{v}_1) = T(\mathbf{v}) + T(\mathbf{v}_1) & & \mbox{for all } \mathbf{v} \mbox{ and } \mathbf{v}_1 \mbox{ in } V. \\ \mbox{T2. } \quad T(\mathbf{rv}) = rT(\mathbf{v}) & & \mbox{for all } \mathbf{v} \mbox{ in } V \mbox{ and } r \mbox{ in } \mathbb{R}. \end{array} \nonumber \]

Figure \(\PageIndex{1}\)

A linear transformation \(T : V \to V\) is called a linear operator on \(V\). The situation can be visualized as in the diagram.

Axiom T1 is just the requirement that \(T\) preserves vector addition. It asserts that the result \(T(\mathbf{v} + \mathbf{v}_{1})\) of adding \(\mathbf{v}\) and \(\mathbf{v}_{1}\) first and then applying \(T\) is the same as applying \(T\) first to get \(T(\mathbf{v})\) and \(T(\mathbf{v}_{1})\) and then adding. Similarly, axiom T2 means that \(T\) preserves scalar multiplication. Note that, even though the additions in axiom T1 are both denoted by the same symbol \(+\), the addition on the left forming \(\mathbf{v} + \mathbf{v}_{1}\) is carried out in \(V\), whereas the addition \(T(\mathbf{v}) + T(\mathbf{v}_{1})\) is done in \(W\). Similarly, the scalar multiplications \(r\mathbf{v}\) and \(rT(\mathbf{v})\) in axiom T2 refer to the spaces \(V\) and \(W\), respectively.

We have already seen many examples of linear transformations \(T : \mathbb{R}^n \to \mathbb{R}^m\). In fact, writing vectors in \(\mathbb{R}^n\) as columns, Theorem2.6.2 shows that, for each such \(T\), there is an \(m \times n\) matrix \(A\) such that \(T(\mathbf{x}) = A\mathbf{x}\) for every \(\mathbf{x}\) in \(\mathbb{R}^n\). Moreover, the matrix \(A\) is given by \(A = \left[ \begin{array}{cccc} T(\mathbf{e}_{1}) & T(\mathbf{e}_{2}) & \cdots & T(\mathbf{e}_{n}) \end{array} \right]\) where \(\{\mathbf{e}_{1}, \mathbf{e}_{2}, \dots, \mathbf{e}_{n}\}\) is the standard basis of \(\mathbb{R}^n\). We denote this transformation by \(T_{A} : \mathbb{R}^n \to \mathbb{R}^m\), defined by

\[T_A(\mathbf{x}) = A\mathbf{x} \quad \mbox{for all } \mathbf{x} \mbox{ in } \mathbb{R}^n \nonumber \]

Example \(\PageIndex{1}\) lists three important linear transformations that will be referred to later. The verification of axioms T1 and T2 is left to the reader.

Example \(\PageIndex{1}\)

If \(V\) and \(W\) are vector spaces, the following are linear transformations:

Identify operator \(V \rightarrow V \quad 1_V: V \rightarrow V \quad\) where \(1_V(\mathbf{v})=\mathbf{v}\) for all \(\mathbf{v}\) in \(V\) Zero transformation \(V \rightarrow W \quad 0: V \rightarrow W \quad\) where \(0(\mathbf{v})=\mathbf{0}\) for all \(\mathbf{v}\) in \(V\) Scalar operator \(V \rightarrow V \quad a: V \rightarrow V \quad\) where \(a(\mathbf{v})=a \mathbf{v}\) for all \(\mathbf{v}\) in \(V\) (Here \(a\) is any real number.)

The symbol \(0\) will be used to denote the zero transformation from \(V\) to \(W\) for any spaces \(V\) and \(W\). It was also used earlier to denote the zero function \(\left[a, b\right] \to \mathbb{R}\).

The next example gives two important transformations of matrices. Recall that the trace \(tr \mathbf{A}\) of an \(n \times n\) matrix \(A\) is the sum of the entries on the main diagonal.

Example \(\PageIndex{2}\)

Show that the transposition and trace are linear transformations. More precisely,

\[\begin{array}{lll} R: \mathbf{M}_{mn} \to \mathbf{M}_{nm} & & \mbox{where } R(A) = A^T \mbox{ for all } A \mbox{ in } \mathbf{M}_{mn} \\ S: \mathbf{M}_{mn} \to \mathbb{R} & & \mbox{where } S(A) = tr \mathbf{A} \mbox{ for all } A \mbox{ in } \mathbf{M}_{nn} \end{array} \nonumber \]

are both linear transformations.

Solution

Axioms T1 and T2 for transposition are \((A + B)^{T} = A^{T} + B^{T}\) and \((rA)^{T} = r(A^{T})\), respectively (using Theorem [thm:002240]). The verifications for the trace are left to the reader.

Example \(\PageIndex{3}\)

If \(a\) is a scalar, define \(E_{a} : \mathbf{P}_{n} \to \mathbb{R}\) by \(E_{a}(p) = p(a)\) for each polynomial \(p\) in \(\mathbf{P}_{n}\). Show that \(E_{a}\) is a linear transformation (called evaluation at \(a\)).

Solution

If \(p\) and \(q\) are polynomials and \(r\) is in \(\mathbb{R}\), we use the fact that the sum \(p + q\) and scalar product \(rp\) are defined as for functions:

\[(p + q)(x) = p(x) + q(x) \quad \mbox{ and } \quad (rp)(x) = rp(x) \nonumber \]

for all \(x\). Hence, for all \(p\) and \(q\) in \(\mathbf{P}_{n}\) and all \(r\) in \(\mathbb{R}\):

\[\begin{aligned} E_a(p + q) &= (p + q)(a) = p(a) + q(a) = E_a(p) + E_a(q), \quad \mbox{ and } \\ E_a(rp) &= (rp)(a) = rp(a) = rE_a(p).\end{aligned} \nonumber \]

Hence \(E_{a}\) is a linear transformation.

The next example involves some calculus.

Example \(\PageIndex{4}\)

Show that the differentiation and integration operations on \(\mathbf{P}_{n}\) are linear transformations. More precisely,

\[\begin{aligned} D&: \mathbf{P}_n \to \mathbf{P}_{n-1} \quad \mbox{where } D\left[p(x)\right] = p^\prime(x) \mbox{ for all } p(x) \mbox{ in } \mathbf{P}_n \\ I&: \mathbf{P}_n \to \mathbf{P}_{n+1} \quad \mbox{where } I\left[p(x)\right] = \int_{0}^{x}p(t)dt \mbox{ for all } p(x) \mbox{ in } \mathbf{P}_n\end{aligned} \nonumber \]

are linear transformations.

Solution

These restate the following fundamental properties of differentiation and integration.

\[\begin{array}{l} \left[p(x) + q(x)\right]^\prime = p^\prime(x) + q^\prime(x) \quad \mbox{ and } \quad \left[rp(x)\right]^\prime = (rp)^\prime(x) \\ \\ \int_{0}^{x}\left[p(t) + q(t)\right]dt = \int_{0}^{x}p(t)dt + \int_{0}^{x}q(t)dt \quad \mbox{ and } \quad \int_{0}^{x}rp(t)dt = r\int_{0}^{x}p(t)dt \end{array} \nonumber \]

The next theorem collects three useful properties of all linear transformations. They can be described by saying that, in addition to preserving addition and scalar multiplication (these are the axioms), linear transformations preserve the zero vector, negatives, and linear combinations.

Theorem \(\PageIndex{1}\)

Let \(T : V \to W\) be a linear transformation.

1. \(T(\mathbf{0}) = \mathbf{0}\).
2. \(T(-\mathbf{v}) = -T(\mathbf{v})\) for all \(\mathbf{v}\) in \(V\).
3. \(T(r_{1}\mathbf{v}_{1} + r_{2}\mathbf{v}_{2} + \cdots + r_{k}\mathbf{v}_{k}) = r_{1}T(\mathbf{v}_{1}) + r_{2}T(\mathbf{v}_{2}) + \cdots + r_{k}T(\mathbf{v}_{k})\) for all \(\mathbf{v}_{i}\) in \(V\) and all \(r_{i}\) in \(\mathbb{R}\).

Proof.

1. \(T(\mathbf{0}) = T(0\mathbf{v}) = 0T(\mathbf{v}) = \mathbf{0}\) for any \(\mathbf{v}\) in \(V\).
2. \(T(-\mathbf{v}) = T \left[ (-1)\mathbf{v} \right] = (-1)T(\mathbf{v}) = -T(\mathbf{v})\) for any \(\mathbf{v}\) in \(V\).
3. The proof of Theorem 2.6.1 goes through.

The ability to use the last part of Theorem \(\PageIndex{1}\) effectively is vital to obtaining the benefits of linear transformations. Example \(\PageIndex{5}\) and Theorem \(\PageIndex{2}\) provide illustrations.

Example \(\PageIndex{5}\)

Let \(T : V \to W\) be a linear transformation. If \(T(\mathbf{v} - 3\mathbf{v}_{1}) = \mathbf{w}\) and \(T(2\mathbf{v} - \mathbf{v}_{1}) = \mathbf{w}_{1}\), find \(T(\mathbf{v})\) and \(T(\mathbf{v}_{1})\) in terms of \(\mathbf{w}\) and \(\mathbf{w}_{1}\).

Solution

The given relations imply that

\[\begin{aligned} T(\mathbf{v}) - 3T(\mathbf{v}_1) &= \mathbf{w} \\ 2T(\mathbf{v}) - T(\mathbf{v}_1) &= \mathbf{w}_1\end{aligned} \nonumber \]

by Theorem [thm:020817]. Subtracting twice the first from the second gives \(T(\mathbf{v}_1) = \frac{1}{5}(\mathbf{w}_1 - 2\mathbf{w})\). Then substitution gives \(T(\mathbf{v}) = \frac{1}{5}(3\mathbf{w}_1 - \mathbf{w})\).

The full effect of property (3) in Theorem \(\PageIndex{1}\) is this: If \(T : V \to W\) is a linear transformation and \(T(\mathbf{v}_{1}), T(\mathbf{v}_{2}), \dots, T(\mathbf{v}_{n})\) are known, then \(T(\mathbf{v})\) can be computed for every vector \(\mathbf{v}\) in \(span \mathbf{\{\mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n}\}}\). In particular, if \(\{\mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n}\}\) spans \(V\), then \(T(\mathbf{v})\) is determined for all \(\mathbf{v}\) in \(V\) by the choice of \(T(\mathbf{v}_{1}), T(\mathbf{v}_{2}), \dots, T(\mathbf{v}_{n})\). The next theorem states this somewhat differently. As for functions in general, two linear transformations \(T : V \to W\) and \(S : V \to W\) are called equal (written \(T = S\)) if they have the same action; that is, if \(T(\mathbf{v}) = S(\mathbf{v})\) for all \(\mathbf{v}\) in \(V\).

Theorem \(\PageIndex{2}\)

Let \(T : V \to W\) and \(S : V \to W\) be two linear transformations. Suppose that \(V = span \mathbf{\{\mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n}\}}\). If T\((\mathbf{v}_{i}) = S(\mathbf{v}_{i})\) for each \(i\), then \(T = S\).

Proof. If \(\mathbf{v}\) is any vector in \(V = pan \mathbf{\{\mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n}\}}\), write \(\mathbf{v} = a_{1}\mathbf{v}_{1} + a_{2}\mathbf{v}_{2} + \cdots + a_{n}\mathbf{v}_{n}\) where each \(a_{i}\) is in \(\mathbb{R}\). Since \(T(\mathbf{v}_{i}) = S(\mathbf{v}_{i})\) for each \(i\), Theorem [thm:020817] gives

\[\begin{aligned} T(\mathbf{v}) &= T(a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \cdots + a_n\mathbf{v}_n) \\ &= a_1T(\mathbf{v}_1) + a_2T(\mathbf{v}_2) + \cdots + a_nT(\mathbf{v}_n) \\ &= a_1S(\mathbf{v}_1) + a_2S(\mathbf{v}_2) + \cdots + a_nS(\mathbf{v}_n) \\ &= S(a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \cdots + a_n\mathbf{v}_n) \\ &= S(\mathbf{v})\end{aligned} \nonumber \]

Since \(\mathbf{v}\) was arbitrary in \(V\), this shows that \(T = S\).

Example \(\PageIndex{6}\)

Let \(V = span \mathbf{\{\mathbf{v}_{1}, \dots, \mathbf{v}_{n}\}}\). Let \(T : V \to W\) be a linear transformation. If \(T(\mathbf{v}_{1}) = \cdots = T(\mathbf{v}_{n}) = \mathbf{0}\), show that \(T = 0\), the zero transformation from \(V\) to \(W\).

Solution

The zero transformation \(0 : V \to W\) is defined by \(0(\mathbf{v}) = \mathbf{0}\) for all \(\mathbf{v}\) in \(V\) (Example [exa:020771]), so \(T(\mathbf{v}_{i}) = 0(\mathbf{v}_{i})\) holds for each \(i\). Hence \(T = 0\) by Theorem \(\PageIndex{2}\).

Theorem \(\PageIndex{2}\) can be expressed as follows: If we know what a linear transformation \(T : V \to W\) does to each vector in a spanning set for \(V\), then we know what \(T\) does to every vector in \(V\). If the spanning set is a basis, we can say much more.

Example \(\PageIndex{3}\)

Let \(V\) and \(W\) be vector spaces and let \(\{\mathbf{b}_{1}, \mathbf{b}_{2}, \dots, \mathbf{b}_{n}\}\) be a basis of \(V\). Given any vectors \(\mathbf{w}_{1}, \mathbf{w}_{2}, \dots, \mathbf{w}_{n}\) in \(W\) (they need not be distinct), there exists a unique linear transformation \(T : V \to W\) satisfying \(T(\mathbf{b}_{i}) = \mathbf{w}_{i}\) for each \(i = 1, 2, \dots, n\). In fact, the action of \(T\) is as follows:

Given \(\mathbf{v} = v_{1}\mathbf{b}_{1} + v_{2}\mathbf{b}_{2} + \cdots + v_{n}\mathbf{b}_{n}\) in \(V\), \(v_{i}\) in \(\mathbb{R}\), then

\[T(\mathbf{v}) = T(v_1\mathbf{b}_1 + v_2\mathbf{b}_2 + \cdots + v_n\mathbf{b}_n) = v_1\mathbf{w}_1 + v_2\mathbf{w}_2 + \cdots + v_n\mathbf{w}_n. \nonumber \]

Proof. If a transformation \(T\) does exist with \(T(\mathbf{b}_{i}) = \mathbf{w}_{i}\) for each \(i\), and if \(S\) is any other such transformation, then \(T(\mathbf{b}_{i}) = \mathbf{w}_{i} = S(\mathbf{b}_{i})\) holds for each \(i\), so \(S = T\) by Theorem [thm:020878]. Hence \(T\) is unique if it exists, and it remains to show that there really is such a linear transformation. Given \(\mathbf{v}\) in \(V\), we must specify \(T(\mathbf{v})\) in \(W\). Because \(\{\mathbf{b}_{1}, \dots, \mathbf{b}_{n}\}\) is a basis of \(V\), we have \(\mathbf{v} = v_{1}\mathbf{b}_{1} + \dots + v_{n}\mathbf{b}_{n}\), where \(v_{1}, \dots, v_{n}\) are uniquely determined by \(\mathbf{v}\) (this is Theorem [thm:018721]). Hence we may define \(T : V \to W\) by

\[T(\mathbf{v}) = T(v_1\mathbf{b}_1 + v_2\mathbf{b}_2 + \cdots + v_n\mathbf{b}_n) = v_1\mathbf{w}_1 + v_2\mathbf{w}_2 + \cdots + v_n\mathbf{w}_n \nonumber \]

for all \(\mathbf{v} = v_{1}\mathbf{b}_{1} + \dots + v_{n}\mathbf{b}_{n}\) in \(V\). This satisfies \(T(\mathbf{b}_{i}) = \mathbf{w}_{i}\) for each \(i\); the verification that \(T\) is linear is left to the reader.

This theorem shows that linear transformations can be defined almost at will: Simply specify where the basis vectors go, and the rest of the action is dictated by the linearity. Moreover, Theorem [thm:020878] shows that deciding whether two linear transformations are equal comes down to determining whether they have the same effect on the basis vectors. So, given a basis \(\{\mathbf{b}_{1}, \dots, \mathbf{b}_{n}\}\) of a vector space \(V\), there is a different linear transformation \(V \to W\) for every ordered selection \(\mathbf{w}_{1}, \mathbf{w}_{2}, \dots, \mathbf{w}_{n}\) of vectors in \(W\) (not necessarily distinct).

Example \(\PageIndex{7}\)

Find a linear transformation \(T : \mathbf{P}_{2} \to \mathbf{M}_{22}\) such that

\[T(1 + x) = \left[ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right], \quad T(x + x^2) = \left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right], \quad \mbox{ and } \quad T(1 + x^2) = \left[ \begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array} \right]. \nonumber \]

Solution

The set \(\{1 + x, x + x^{2}, 1 + x^{2}\}\) is a basis of \(\mathbf{P}_{2}\), so every vector \(p = a + bx + cx^{2}\) in \(\mathbf{P}_{2}\) is a linear combination of these vectors. In fact

\[p(x) = \frac{1}{2}(a + b - c)(1 + x) + \frac{1}{2}(-a + b + c)(x + x^2) + \frac{1}{2}(a - b + c)(1 + x^2) \nonumber \]

Hence Theorem [thm:020916] gives

\[\begin{aligned} T\left[p(x)\right] &= \frac{1}{2}(a + b - c)\left[ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right] + \frac{1}{2}(-a + b + c)\left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right] + \frac{1}{2}(a - b + c)\left[ \begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array} \right] \\ &= \frac{1}{2} \left[ \begin{array}{rr} a + b - c & -a + b + c \\ -a + b + c & a - b + c \end{array} \right]\end{aligned} \nonumber \]