Answer to Question #121536 in Linear Algebra for Brain

Question

Define T in L.F3/ by
T.z1; z2; z3/ D .2z2; 0; 5z3/:
Find all eigenvalues and eigenvectors of T.

Answer 1

We know that $(1,0,0),(0,1,0),(0,0,1)$ is the standard basis of $R^3$ .

Given $T(z_1,z_2,z_3) = ( 2z_2, 0, 5z_3)$ .

Now, $T(1,0,0) =(0,0,0)$ , $T(0,1,0) =(2,0,0)$ and $T(0,0,1)=(0,0,5)$ .

So, Matrix $T = \begin{bmatrix} 0 & 2 &0\\ 0 & 0 & 0 \\ 0 & 0 & 5 \end{bmatrix}$ .

Since, T is upper-triangular Matrix, so eigen-values of T are $0,0,5$ .

Eigenvector of T with respect to eigen-value 0 :

$[T -0I]X = O \implies \begin{bmatrix} 0 & 2 &0\\ 0 & 0 & 0 \\ 0 & 0 & 5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$\implies x_2 = 0, x_3 = 0$ and let $x_1 = k$

So, eigenvector is $[ 1 \ 0 \ 0 ]^t$ .

Eigenvalue 0 has arithmetic multiplicity 2, but rank of T is 2, so geometric multiplicity corresponding to 0 is $= n - \rho(T ) = 3- 2 = 1$. So, corresponding to eigenvalue 0, only one eigenvector exists and calculated above.

Eigenvector of T with respect to eigen-value 5 :

$[T -5I]X = O \implies \begin{bmatrix} -5 & 2 &0\\ 0 & -5 & 0 \\ 0 & 0 & 0\end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$\implies x_1 = 0, x_2 = 0$ and let $x_3 = k$

So, eigenvector is $[ 0 \ 0 \ 1 ]^t$ .