Question

Let f:R²→R² be defined by f(x,y)=(-y,-x)

i) show that f is linear

ii)Determine a basis for the kernel of f and the nullity of f

iii) Determine the basis for the range of f and the rank of f

iv) Determine whether f is invertible or not

Answer 1

i) Let f:R²→R² be defined by f(x,y)=(-y,-x)

Let $\beta =\{ (1,0)= e_1, (0,1)= e_2 \}$ be a standard ordered pair of R²

Then f(0,1) = (1,0)= 1(1,0)+0(0,1)

$[F]_{\beta}=\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}= A(ay)$

$A(x)= |A-xI| = \begin{vmatrix} -x & 1 \\ 1 & -x \end{vmatrix}= x^2-1$ Hence linear

ii)

$F(x)= Ax\\ A= \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$

$dom = R^2, co-dom = R^2\\ dim \space dom = 2, dim \space co-dom = 2\\ Ax= 0 \implies \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ \implies \begin{pmatrix} x+y\\ x-y \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ x+y=0\\ x-y=0\\ \implies x=0; y=0\\ \therefore Kernel \space F = \{(0,0)\} \\ Nullity =0\\$

iii)

$dom = R^2, co-dom = R^2\\ dim \space dom = 2, dim \space co-dom = 2\\ Ax= 0 \implies \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ \implies \begin{pmatrix} x+y\\ x-y \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ x+y=0\\ x-y=0\\ \implies x=0; y=0\\ \therefore Range \space F = R^2\\ \implies Range = \{(1,0), (0,1)\}$

iv)

$For \space y \epsilon R$ to find $x \epsilon R$ such that

$f(x)=y\\ f(x,y)=(-y,-x)$

This shows that f is invertible