Exercise 1197Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

The canonical basis and matrix representation

#canonical basis#matrix representation#linear map

Statement

Let $f:\mathbb{R}^2\to\mathbb{R}^2$ be defined by $f(x,y)=(2x+y, x-y)$ . Write the matrix of $f$ in the canonical basis, and in the basis $\mathcal{B}=((1,1),(1,-1))$ .

Hint

The columns of the matrix in the canonical basis are the images of the canonical basis vectors. For the new basis, compute $P^{-1}AP$ .

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Solution

Canonical matrix.
$A = \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}.$

Transition matrix.
$P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \qquad P^{-1} = \frac12\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.$

New matrix.
$P^{-1}AP = \frac12\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.$
Compute:
$\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}.$
Then
$\frac12\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix} = \frac12\begin{pmatrix} 3 & 3 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix} 3/2 & 3/2 \\ 3/2 & -1/2 \end{pmatrix}.$

Conclusion. In $\mathcal{B}$ , the matrix of $f$ is $\begin{pmatrix} 3/2 & 3/2 \\ 3/2 & -1/2 \end{pmatrix}$ .

The canonical basis and matrix representation

Statement

Solution

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