Answer to Question #106054 in Linear Algebra for Rutuja

Question

Define T:R^3-->R^3 by T(x,y,z)=(-x,x-y,3x+2y+z). Check if T satisfies the polynomial (x-1)(x+1)^2.

Answer 1

First we have to show that ,$T:R^3 \rightarrow R^3$

Define by $T(x,y,z)=(-x,x-y,3x+2y+z)$ is a linear operator.

Let $(x,y,z),(x',y',z') \in R^3$ and $a,b \in R$

Where $R$ is the underlying field.

Now ,$T\{ a(x,y,z)+b(x',y',z') \}=$ $T\{ (ax,ay,az)+(bx',by',bz') \}=$

$T$ $=\{ (ax+bx',ay+by',az+bz') \}$

$=( - ax-bx,ax-ay+bx'-by',3ax+2ay+az+3bx'+2by'+bz')$

$=(-ax,ax-ay,3ax+2ay+az)$ +$(-bx',bx'-by',3bx'+2by'+bz')$

$=a(-x,x-y,3x+2y+z)+b(-x',x'-y',3x'+2y'+z')$

$=aT(x,y,z)+bT(x',y',z')$

Hence $T$ is a linear transformation.

Now, the matrix representation of $T$ with respect to the usual basis

$\{ (1,0,0),(0,1,0),(0,0,1) \}$is the following:

$\begin{pmatrix} -1&0&0\\ 1&-1&0\\ 3&2&1 \end{pmatrix}$

Since it is a Lower triangular matrix ,its eigen values are $-1,-1,\, and \, 1$ .

So its characteristics polynomial is $(x+1)^2(x-1)$ .

Again by Cayley Hamilton theorem ,

every square matrix satisfies its characteristic polynomial.

Therefore $T$ satisfies $(x-1)(x+1)^2$ .