Answer to Question #167037 in Linear Algebra for Nikhil Rawat

Question

Let T: R^2→R^2 be a linear opeator with matrix

[ 7 1]

[-1 1] (w.r.t. the standard basis). Use Cayley haMilton theorem to check whether T is invertible or not. If T is invertible, obtain T^-1(x,y) for (x,y)∈ R^2. If T is not invertible, obtain the minimal polynomial of T.

Answer 1

Solution:

Given matrix, $A=\begin{bmatrix} 7 \ \ 1\\-1\ 1\end{bmatrix}$

First we find characteristic polynomial, $p(t)=\begin{vmatrix} A-tI\end{vmatrix}$

$=\begin{vmatrix} 7-t \ \ \ \ 1\\-1\ \ \ 1-t\end{vmatrix}$

$=(7-t)(1-t)-1(-1) \\ =7-t-7t+t^2+1 \\ =t^2-8t+8$

Now, applying Cayley hamilton theorem,

$O=p(A)=A^2-8A+8I$

where $O$ is zero matrix and $I$ is identity matrix of order 2.

So, $A^2-8A=-8I$

$\Rightarrow-\dfrac18A^2+A=I$

$\Rightarrow A(-\dfrac18A+I)=I$ ...(i)

Similarly, we can write this as: $(-\dfrac18A+I)A=I$ ...(ii)

Thus, from (i) and (ii), matrix $(-\dfrac18A+I)$ is the inverse of matrix $A$ .

$\therefore A^{-1}=(-\dfrac18A+I)$

$=-\frac18\begin{bmatrix} 7 \ \ 1\\-1\ 1\end{bmatrix}+\begin{bmatrix} 1 \ \ 0\\0\ 1\end{bmatrix}$

$=\begin{bmatrix} -\frac78 \ \ -\frac18\\\frac18\ -\frac18\end{bmatrix}+\begin{bmatrix} 1 \ \ 0\\0\ 1\end{bmatrix}$

$=\begin{bmatrix} \frac18 \frac{-1}8\\\frac18\ \frac78\end{bmatrix}$

This is required answer.