Ch05. Eigenvalues and Eigenvectors

5.4 Eigenvectors and Linear Transformations

The Matrix of a Linear Transformation

• Given any $\textbf{x}$ in $\text{V}$, the coordinate vector $[\textbf{x}]_\mathcal{B}$ is in $\mathbb{R}^n$ and the coordinate vector of its image, $[T(\textbf{x})]_\mathcal{C}$, is in $\mathbb{R}^m$, as shown in Fig. 1. below.

• The connection between $[\textbf{x}]_\mathcal{B}$ and $[T(\textbf{x})]_\mathcal{C}$ is easy to find.

• Let $\left\{ \textbf{b}_1, \cdots , \textbf{b}_n \right\}$ be the basis $\mathcal{B}$ for$V$. If $\textbf{x}=r_1\textbf{b}_1+\cdots+r_n\textbf{b}_n$ then, $$[\textbf{x}]_\mathcal{B} = \begin{bmatrix}r_1 \\ \vdots \\ r_n \end{bmatrix}$$

• And $$T(\textbf{x}) = T(r_1\textbf{b}_1+\cdots+r_n\textbf{b}_n)=r_1T(\textbf{b}_1)+\cdots+r_nT(\textbf{b}_n) \tag{1}$$ because $T$ is linear.

• Now, since the coordinate mapping from $W$ to $\mathbb{R}^m$ is linear, equation (1) leads to $$[T(\textbf{x})]_\mathcal{C} = r_1[T(\textbf{b}_1)]_\mathcal{C}+\cdots+r_n[T(\textbf{b}_n)]_\mathcal{C} \tag{2}$$

• Since $\mathcal{C}$-coordinate vectors are in $\mathbb{R}^m$, the vector equation (2) can be written as a matrix equation, namely, $$[T(\textbf{x})]_\mathcal{C} = M [\textbf{x}]_\mathcal{B} \tag{3}$$ where $$M = \begin{bmatrix} [T(\textbf{b}_1)]_\mathcal{C} & [T(\textbf{b}_2)]_\mathcal{C} & \cdots & [T(\textbf{b}_n)]_\mathcal{C} \end{bmatrix} \tag{4}$$
• The matrix $M$ is a matrix representation of $T$, called the matrix for $T$ relative to the bases $\mathcal{B}$ and $C$. See Fig. 2 below:

$M$ 이 실제로 coordinate vector를 linear transform하여 목표로하는 basis의 coordinate vector로 mapping

Example 1

Suppose $\mathcal{B}=\left\{\textbf{b}_1, \textbf{b}_2 \right\}$ is a basis for $V$ and $\mathcal{C}=\left\{\textbf{c}_1,\textbf{c}_2,\textbf{c}_3\right\}$ is a basis for $W$. Let $T: V \rightarrow W$ be a linear transformation with the property that

$$T(\textbf{b}_1)=3\textbf{c}_1-2\textbf{c}_2+5\textbf{c}_3 \text{ and } T(\textbf{b}_2)=4\textbf{c}_1+ 7\textbf{c}_2-\textbf{c}_3$$

Find the matrix $M$ for $T$ relative to $\mathcal{B}$ and $\mathcal{C}$.

Solution

The $C$-coordinate vectors of the images of $\textbf{b}_1$ and $\textbf{b}_2$ are $$[T(\textbf{b}_1)]_\mathcal{C} = \begin{bmatrix} 3 \\ -2 \\ 5 \end{bmatrix} \text{ and } [T(\textbf{b}_2)]_\mathcal{C} = \begin{bmatrix} 4 \\ 7 \\ -1 \end{bmatrix}$$

• Hence $$M=\begin{bmatrix} 3 & 4 \\ -2 & 7 \\ 5 & -1 \end{bmatrix}$$

If $\mathcal{B}$ and $\mathcal{C}$ are bases for the same space $V$ and if $T$ is the identity transformation $T(\textbf{x})=\textbf{x}$ for $\textbf{x} \in V$, then Matrix $M$ in (4) is just a change-of-coordinates matrix.

4.7절 참고

Linear Transformations from V into V

• In the common case where $W$ is the same as $V$ and the basis $\mathcal{C}$ is the same as $\mathcal{B}$, the matrix $M$ in (4) is called the matrix for $T$ relative to $\mathcal{B}$, or simply the $\mathcal{B}$-matrix for $T$, and is denoted by $[T]_\mathcal{B}$.

Example2

The mapping $T: \mathbb{P}_2 \rightarrow \mathbb{P}_2$ defined by $$T(a_0+a_1t +a_2t^2) = a_1 +2a_2 t$$ is a linear transformation.

1. Find the $\mathcal{B}$-matrix for $T$, when $\mathcal{B}$ is the basis $\left\{ 1, t, t^2 \right\}$
2. Verify that $[T(\textbf{p})]_\mathcal{B} = [T]_\mathcal{B}[\textbf{p}]_\mathcal{B}$

Solution

1. Compute the images of the basis vectors: $$T(1) = 0 \\ T(t) = 1 \\ T(t^2) = 2t$$

2. Then write the $\mathcal{B}$-coordinate vector of $T(1), T(t)$, and $T(t^2)$ and place them together as the $\mathcal{B}$-matrix for $T$: $$[T(1)]_\mathcal{B} = \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}, [T(t)]_\mathcal{B} = \begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}, [T(t^2)]_\mathcal{B} = \begin{bmatrix}0 \\ 2 \\ 0 \end{bmatrix} \\ [T]_\mathcal{B} = \begin{bmatrix}0 & 1 & 0\\ 0& 0 &2 \\ 0& 0& 0 \end{bmatrix}$$

3. For a general $\textbf{p}(t)= a_0+a_1t+a_2t^2$, $$\begin{align} [T(\textbf{p})]_\mathcal{B}&=[a_1 + 2a_2 t]_\mathcal{B} = \begin{bmatrix} a_1 \\ 2a_2 \\ 0 \end{bmatrix}\\ &=\begin{bmatrix}0 & 1 & 0\\ 0& 0 &2 \\ 0& 0& 0 \end{bmatrix}\begin{bmatrix} a_0 \\ a_1 \\ a_2 \end{bmatrix} = [T]_\mathcal{B}[\textbf{p}]_\mathcal{B} \end{align}$$

4. SeeFig. 4 below

Linear Transformations on R to the n Power

A가 대각화가 가능한 경우, $A$의 eigenvector들로 구성된 basis $\mathcal{B}$이 존재하며, Linear Transform $T$에 대한 $\mathcal{B}$-matrix for $T$는 diagonal matrix임. 아래의 Theorem 8은 이 사실을 보여줌.

Theorem 8:

Suppose $A=PDP^{-1}$ where $D$ is a diagonal $n \times n$ matrix. If $\mathcal{B}$ is the basis for $\mathbb{R}^n$ formed from the columns of $P$, then $D$ is the $\mathcal{B}$-matrix for the transformation $\textbf{x} \mapsto A\textbf{x}$.

Proof

Denote the columns of $P$ by $\textbf{b}_1,\cdots,\textbf{b}_n$, so that $\mathcal{B}=\left\{\textbf{b}_1,\cdots,\textbf{b}_n\right\}$ and $P=\begin{bmatrix}\textbf{b}_1 \cdots \textbf{b}_n\end{bmatrix}$. In this case, $P$ is the change-of-coordinates matrix $P_\mathcal{B}$ discussed in Section 4.4, where

$$P[\textbf{x}]_\mathcal{B}=\textbf{x} \text{ and } [\textbf{x}]_\mathcal{B} = P^{-1} \textbf{x}$$

If $T(\textbf{x})=A\textbf{x}$ for $\textbf{x} \in \mathbb{R}^n$, then

Since $A=PDP^{-1}$, we have $[T]_\mathcal{B} = P^{-1}AP = D$.

Example3

Define $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ by $T(\textbf{x})=A\textbf{x}$, where $A=\begin{bmatrix} 7 & 2 \\ -4 & 1 \end{bmatrix}$. Find a basis $\mathcal{B}$ for $\mathbb{R}^2$ with the property that the $\mathcal{B}$-matrix for $T$ is a diagonal matrix.

Solution

From Example 2 in section 5.3 we know that $A=PDP^{-1}$, where $$P=\begin{bmatrix}1 & 1 \\ -1 & -2 \end{bmatrix} \text{ and } D=\begin{bmatrix}5 & 0 \\ 0 & 3 \end{bmatrix}$$

• The columns of $P$, call them $\textbf{b}_1$ and $\textbf{b}_2$, are eigenvectors of $A$. By Theorem 8, $D$ is the $\mathcal{B}$-matrix for $T$ when $\mathcal{B}=\left\{ \textbf{b}_1, \textbf{b}_2 \right\}$. The mappings $\textbf{x} \mapsto A\textbf{x}$ and $\textbf{u} \mapsto D\textbf{u}$ describe the same linear transformation, relative to different bases. $\blacksquare$