Answer to Question #337152 in Linear Algebra for devil

Question

Let T be the linear operator on R

2 defined by

T(x, y) = (−y, x)

i. What is the matrix of T in the standard ordered basis for R2 ?

ii. What is the matrix of T in the ordered basis B = {α1, α2}, where α1 = (1, 2) and α2 = (1, −1)?

iii. Prove that for every real number c the operator (T − cI) is invertible.

Answer 1

i. The operator $T(x,y)$ can be written in the form: $\left(\begin{array}{ll}-y\\ x\end{array}\right)=\left(\begin{array}{ll}0&-1\\1&0\end{array}\right)\left(\begin{array}{ll}x\\ y\end{array}\right)$. $\left(\begin{array}{ll}0&-1\\1&0\end{array}\right)$ is the matrix of the operator in the standard ordered basis. The matrix can be easily found by considering equality: $\left(\begin{array}{ll}-y\\ x\end{array}\right)=\left(\begin{array}{ll}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right)\left(\begin{array}{ll}x\\ y\end{array}\right)$. From the latter we get: $-y=a_{11}x+a_{12}y$ and $x=a_{21}x+a_{22}y$. We immediately receive: $a_{11}=0,a_{12}=-1$ and $a_{21}=1,a_{22}=0.$

ii. In order to find the matrix in the ordered basis $B=\{ \alpha_1,\alpha_2\}$, we consider the vector $\left(\begin{array}{ll}x\\ y\end{array}\right)$ in the basis $B$. We receive: $\left(\begin{array}{ll}x\\ y\end{array}\right)=\left(\begin{array}{ll}1\\ 2\end{array}\right)t_1+\left(\begin{array}{ll}1\\ -1\end{array}\right)t_2$. We receive: $x=t_1+t_2$, $y=2t_1-t_2$. Thus, $t_1=\frac{x+y}{3}$, $t_2=\frac{2x-y}3$. In the same way we consider the vector $\left(\begin{array}{ll}-y\\ x\end{array}\right)=\left(\begin{array}{ll}1\\ 2\end{array}\right)s_1+\left(\begin{array}{ll}1\\ -1\end{array}\right)s_2$. We receive: $s_1=\frac{-y+x}{3}$, $s_2=\frac{-2y-x}{3}$. The aim is to find the matrix satisfying: $\left(\begin{array}{ll}\frac{-y+x}{3}\\ \frac{-2y-x}{3}\end{array}\right)=\left(\begin{array}{ll}b_{11}&b_{12}\\b_{21}&b_{22}\end{array}\right)\left(\begin{array}{ll}\frac{x+y}{3}\\ \frac{2x-y}3\end{array}\right)$. We receive: $\frac{-y+x}{3}=b_{11}\frac{x+y}{3}+b_{12}\frac{2x-y}3$, $\frac{-2y-x}{3}=b_{21}\frac{x+y}{3}+b_{22}\frac{2x-y}3$. We set $y=-x$ in the first equation and receive: $b_{12}=\frac23$. Setting $y=2x$ yields: $b_{11}=-\frac{1}3$. We set $y=-x$ in the second equation and get: $b_{22}=\frac13$. Setting $y=2x$ provides: $b_{21}=-\frac53$. Thus, matrix $T$ in the ordered bases $B$ has the form: $\left(\begin{array}{ll}-\frac13&\frac23\\-\frac53&\frac13\end{array}\right)$.

iii. Matrix $T-cI$ has the form: $T-cI=\left(\begin{array}{ll}-c&-1\\1&-c\end{array}\right)$. $det(T-cI)=c^2+1>0$. Thus, the matrix is invertible for all real $c$. It means, that the operator is invertible.

Answer: i. $\left(\begin{array}{ll}0&-1\\1&0\end{array}\right)$; ii. $\left(\begin{array}{ll}-\frac13&\frac23\\-\frac53&\frac13\end{array}\right)$; iii. $det(T-cI)=c^2+1>0$. Therefore, the operator is invertible.