Determining the matrix of a linear map

Linear maps: Matrices of Linear Maps

Determining the matrix of a linear map

As a result of the theorem Linear map determined by the image of basis a linear map from $\mathbb{R}^n$ to $\mathbb{R}^m$ is uniquely determined by the images of a basis $\basis{\vec{a}_1,\ldots ,\vec{a}_n}$ for $\mathbb{R}^n$ . If this basis is the standard basis $\basis{\vec{e}_1,\ldots,\vec{e}_n}$ , then we can write down the corresponding matrix immediately using the theorem Linear maps in coordinate space defined by matrices. But if the basis differs from the standard basis, we have to perform some calculations first. These calculations are based on the following technique.

Let $L:\mathbb{R}^n\to\mathbb{R}^m$ be a linear map and $\alpha=\basis{\vec{a}_1,\ldots,\vec{a}_n}$ a basis for $\mathbb{R}^n$ . Form the $(n\times(n+m))$ -matrix

$\matrix{\vec{a}_1& L( \vec{a}_1)\\ \vec{a}_2& L( \vec{a}_2)\\ \vdots&\vdots\\ \vec{a}_n& L( \vec{a}_n)}$

If we bring this matrix to the reduced echelon form, then the identity matrix appears on the left while the images of the standard basis vectors under $L$ appear on the right (as row vectors). In other words, the $(n\times m)$ -submatrix on the right is the transposed of $L_{\varepsilon}$ .

Consider two rows in which we put a vector on the left and its corresponding image on the right:

$\matrix{\vec{a}& L(\vec{a})\\ \vec{b}& L( \vec{b})}$ Adding the rows yields

$\matrix{\vec{a}+\vec{b}& L(\vec{a})+ L(\vec{b})}$ Because $L$ is linear, the right hand side is equal to $L(\vec{a}+\vec{b})$ ; addition of rows therefore yields another row with a vector on the left and its image on the right. This also applies to scalar multiplication: multiplying the first row by $\alpha$ yields

$\matrix{\alpha\vec{a}& \alpha L(\vec{a})}$ which is, thanks to the linearity of $L$ , again a row with a vector $\alpha\vec{a}$ on the left and its image $L(\alpha \,\vec{a})$ on the right.

It follows that if we are going to perform row reduction operations in a system of this kind of rows, each row in this system keeps the characteristic that it contains a vector on the left and its corresponding image on the right.

Because $\basis{\vec{a}_1,\ldots,\vec{a}_n}$ is a basis, the reduced echelon form of the left part of the matrix will be the identity matrix. As a consequence, for $i=1,\ldots,n$ , the $i$ -th row vector of the $(n\times m)$ -submatrix on the right of the reduced echelon form is the image of the $i$ -th standard basis vector. We conclude that the matrix of $L$ consists of the corresponding column vectors, that is, is equal to the transposed matrix of the $(n\times m)$ -submatrix on the right of the reduced echelon form.

A linear map $L :\mathbb{R}^3\rightarrow\mathbb{R}^3$ is given by

$\begin{array}{rcl} L( \rv{ -1 , 1 , 1 } )&=&\rv{ 1 , -3 , -1 } \\ L( \rv{ -2 , 2 , 1 } )&=&\rv{ 2 , -4 , -1 } \\ L(\rv{ 2 , -1 , -1 } )&=&\rv{ -1 , 4 , 2 } \end{array}$
Determine the matrix $L_{\varepsilon}$ of $L$ with respect to the standard basis $\varepsilon=\basis{\vec{e}_1,\vec{e}_2,\vec{e}_3}$ .

$L_{\varepsilon} =$ $\matrix{0 & 1 & 0 \\ 1 & 0 & -2 \\ 1 & 1 & -1 \\ }$

In terms of matrices, the conditions on $L_{\varepsilon}$ can be written as

$\matrix{-1 & 1 & 1 \\ -2 & 2 & 1 \\ 2 & -1 & -1 \\ }\, L_{\varepsilon}^\top = \matrix{1 & -3 & -1 \\ 2 & -4 & -1 \\ -1 & 4 & 2 \\ }$ wherein $L_{\varepsilon}^\top$ is the transpose of $L_{\varepsilon}$ . We find $L_{\varepsilon}^\top$ by performing row reduction operations on the matrix

$\left(\left.\begin{array}{ccc}-1 & 1 & 1 \\ -2 & 2 & 1 \\ 2 & -1 & -1 \end{array}\ \right|\begin{array}{ccc}1 & -3 & -1 \\ 2 & -4 & -1 \\ -1 & 4 & 2 \end{array}\right)$ until it obtains the reduced echelon form:

$\begin{array}[t]{ll} \left(\left.\begin{array}{ccc}-1 & 1 & 1 \\ -2 & 2 & 1 \\ 2 & -1 & -1 \end{array}\ \right|\begin{array}{ccc}1 & -3 & -1 \\ 2 & -4 & -1 \\ -1 & 4 & 2 \end{array}\right) & \begin{array}[t]{ll} \sim\left(\begin{array}{rrr|rrr} -2 & 2 & 1 & 2 & -4 & -1 \\ -1 & 1 & 1 & 1 & -3 & -1 \\ 2 & -1 & -1 & -1 & 4 & 2 \end{array}\right) & \\\\ \sim\left(\begin{array}{rrr|rrr} 1 & -1 & -{{1}\over{2}} & -1 & 2 & {{1}\over{2}} \\ -1 & 1 & 1 & 1 & -3 & -1 \\ 2 & -1 & -1 & -1 & 4 & 2 \end{array}\right) & \\\\ \sim\left(\begin{array}{rrr|rrr} 1 & -1 & -{{1}\over{2}} & -1 & 2 & {{1 }\over{2}} \\ 0 & 0 & {{1}\over{2}} & 0 & -1 & -{{1}\over{2}} \\ 2 & -1 & -1 & -1 & 4 & 2 \end{array}\right) & \\\\ \sim\left(\begin{array}{rrr|rrr} 1 & -1 & -{{1}\over{2}} & -1 & 2 & {{1}\over{2}} \\ 0 & 0 & {{1}\over{2}} & 0 & -1 & -{{1}\over{ 2}} \\ 0 & 1 & 0 & 1 & 0 & 1 \end{array}\right) & \\\\ \sim\left(\begin{array}{rrr|rrr} 1 & -1 & -{{1}\over{2}} & -1 & 2 & {{1}\over{2}} \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & {{1 }\over{2}} & 0 & -1 & -{{1}\over{2}} \end{array}\right) & \\\\ \sim\left(\begin{array}{rrr|rrr} 1 & 0 & -{{1 }\over{2}} & 0 & 2 & {{3}\over{2}} \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & {{1}\over{2}} & 0 & -1 & -{{1}\over{2}} \end{array}\right) & \\\\ \sim\left(\begin{array}{rrr|rrr} 1 & 0 & - {{1}\over{2}} & 0 & 2 & {{3}\over{2}} \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & -2 & -1 \end{array}\right) & \\\\ \sim\left(\begin{array}{rrr|rrr} 1 & 0 & 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & -2 & -1 \end{array}\right) \end{array} \end{array}$ The $(3\times 3)$ -submatrix on the right is equal to

$L_{\varepsilon}^\top = \matrix{0 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & -2 & -1 \\ }$ so the answer is:

$L_{\varepsilon} = \matrix{0 & 1 & 0 \\ 1 & 0 & -2 \\ 1 & 1 & -1 \\ }$

New example