Question #162110

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.


1
Expert's answer
2021-02-19T15:15:21-0500

Let the associated matrix be denoted by A. So we have that, A=[7111]A = \begin{bmatrix} 7 & 1 \\ -1& -1 \end{bmatrix}

AλI=7λ111λ=(7λ)(1λ)1(1)|A- \lambda I| = \begin{vmatrix} 7- \lambda & 1 \\ -1 & -1- \lambda \end{vmatrix} = (7- \lambda) (-1-\lambda)- 1(-1)

= λ26λ6\lambda^2 - 6\lambda - 6

So we have that the characteristics polynomial is λ26λ6\lambda^2 - 6\lambda - 6 . But Cayley Hamilton's theorem states that every square matrix satisfies it characteristics equation. So we check whether A26A6I=0A^2 - 6 A -6I = 0

A2=[48660]6A=[42666]6I=[6006]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 A26A6I=[0000]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 A26A6I=0A^2-6A-6I =0

Multiplying through by A1A^{-1} , we have

A6I6A1=06A1=A6IA1=16(A6I)A-6I-6A^{-1}= 0\\ 6A^{-1}= A-6I\\ A^{-1}= \frac{1}{6}(A-6I)

=16([7111][6006])= \frac{1}{6}(\begin{bmatrix} 7&1\\-1&-1\end{bmatrix} - \begin{bmatrix} 6&0\\ 0&6 \end{bmatrix})

=[16161676]= \begin{bmatrix} \frac{1}{6}&\frac{1}{6}\\ \frac{-1}{6}&\frac{-7}{6} \end{bmatrix}

Hence T1=(16x+16y,16x76y)T^{-1} =( \frac{1}{6}x+ \frac{1}{6}y, \frac{-1}{6}x -\frac{7}{6}y)


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