Answer to Question #162110 in Linear Algebra for Sourav Mondal

Question

Let T : R2-+ R2be a linear operator with 

matrix [ (w.r.t. the standard basis).

[ 7 1

-1 -1]

Use Cayley-Hamilton theorem to 

check whether T is invertible or not. If T is 

invertible, obtain T-1(x, y) for 

(x,y) E R2 . If T is not invertible, obtain 

the minimum polynomial of T.

Answer 1

Let the associated matrix be denoted by A. So we have that, $A = \begin{bmatrix} 7 & 1 \\ -1& -1 \end{bmatrix}$

$|A- \lambda I| = \begin{vmatrix} 7- \lambda & 1 \\ -1 & -1- \lambda \end{vmatrix} = (7- \lambda) (-1-\lambda)- 1(-1)$

= $\lambda^2 - 6\lambda - 6$

So we have that the characteristics polynomial is $\lambda^2 - 6\lambda - 6$ . But Cayley Hamilton's theorem states that every square matrix satisfies it characteristics equation. So we check whether $A^2 - 6 A -6I = 0$

$A^2 = \begin{bmatrix} 48 &6\\ -6&0 \end{bmatrix}\\ -6A = \begin{bmatrix} -42&-6\\ 6&6 \end{bmatrix}\\ -6I= \begin{bmatrix} -6&0\\ 0&-6 \end{bmatrix}$

So we have that $A^2 -6A -6I = \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix}$

as desired. So we have that A is invertible and hence T is invertible.

Since $A^2-6A-6I =0$

Multiplying through by $A^{-1}$ , we have

$A-6I-6A^{-1}= 0\\ 6A^{-1}= A-6I\\ A^{-1}= \frac{1}{6}(A-6I)$

$= \frac{1}{6}(\begin{bmatrix} 7&1\\-1&-1\end{bmatrix} - \begin{bmatrix} 6&0\\ 0&6 \end{bmatrix})$

$= \begin{bmatrix} \frac{1}{6}&\frac{1}{6}\\ \frac{-1}{6}&\frac{-7}{6} \end{bmatrix}$

Hence $T^{-1} =( \frac{1}{6}x+ \frac{1}{6}y, \frac{-1}{6}x -\frac{7}{6}y)$