Exercise 131Linear algebra › Linear mapsL1Added 3 Nov 2026

An isomorphism

#isomorphism#inverse#linear map

Statement

Let $f:\mathbb{R}^2\to\mathbb{R}^2$ be defined by $f(x,y)=(2x+y,\ x-y)$ . Show that $f$ is an isomorphism and compute $f^{-1}$ .

Hint

Show that $\ker f=\{0\}$ , or compute the inverse of the matrix of $f$ .

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Solution

Matrix.
$A = \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}, \qquad \det A = -2-1=-3\neq0.$
So $A$ is invertible and $f$ is an isomorphism.

Inverse.
$A^{-1} = -\frac13\begin{pmatrix} -1 & -1 \\ -1 & 2 \end{pmatrix} = \frac13\begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix},$
so $f^{-1}(u,v)=\tfrac13\,(u+v,\ u-2v)$ .

Verification.
$f\left(\frac{u+v}{3}, \frac{u-2v}{3}\right) = \left(\frac{2(u+v)+(u-2v)}{3},\ \frac{(u+v)-(u-2v)}{3}\right)=(u,v).$
Since $A$ is square, this already gives $f^{-1}\circ f=\mathrm{id}$ as well.

An isomorphism

Statement

Solution

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