# finding eigenvalues of a linear transformation and determine if its diagonalizable

Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be a linear transformation such that $T(a_1,a_2,a_3) = (-3a_3, a_1+5a_3,a_2-a_3)$.

(i) Find all the eigenvalues of T

(ii) For each eigenvalue $\alpha$, write $E_\alpha$ as a span of some basis.

(iii) Is T diagnolizable?

This is my solution. Is it correct? First, we find the standard matrix representation.

The standard basis $e_1=(1,0,0)$, $e_2=(0,1,0)$, $e_3=(0,0,1)$

Then the images of these vectors are $T(e_1) = (0,1,0)$, $T(e_2) = (0,0,1)$, $T(e_3) = (-3,5,-1)$

Hence the standard matrix representation is given by:

$[ \begin{matrix} 0 & 0 & -3\\ 1 & 0 & 5\\ 0 & 1 & -1 \end{matrix}]$

now we need to find the eigenvalues:

Solve $|M-\alpha I|=0$ for $\alpha$, you get $\alpha^3 + \alpha^2-5\alpha=-3$. the roots are $\alpha_1=1, \alpha_2=-3,\alpha_3=1$ which are the eigenvectors.

(ii) To find $E_1$ and $E_{-3}$, we compute $E_1=Null(M-\alpha_1 I)$ and $E_3=Null(M-alpha_3 I)$ which turns to be: $E_1 = span \{(-1,2,1)\}$ and $E_3 = span \{(1,-2,1)\}$

(iii) To determine if the matrix is diagnolizable, we compute roots of $det(\alpha I-M)=0$, which turns to be the polynomial $\alpha^3 + \alpha^2 -5\alpha + 3=0$ and this has the roots 1, -3, 1. Since they are not distinct then the matrix is not diagonlizable.

• Your roots do not match the polynomial you wrote. – Tobias Kildetoft Apr 22 '15 at 11:15
• thanks. Is it correct now? – user233500 Apr 22 '15 at 11:20
• identity matrix has repeating eigenvalues. what you need for diagonalizablity is to have an eigenbasis. that the is sum of the dimensions of the null spaces add up to the dimension of the whole sapce. – abel Apr 22 '15 at 11:46
• I see. so the dimension of an eigenspace is its multiplicity. hence, $dim(E_1)+dim(E_{-3})=2+1=3=dim(\mathbb{R}^3)$. @abel – user233500 Apr 22 '15 at 12:21
• there are two types of multiplicity:(i) algebraic multiplicity, (ii) geometric multiplicity. algebraic multiplicity is the number of times the eiegenvalue repeats. geometric multiplicity is the dimension of the null space associated with that eigenvalue. in the case of eigenvalue $-1,$ you have algebraic multiplicity $2$ and geometric multiplicity $1$ because rank of $(A-I)$ is two. – abel Apr 22 '15 at 14:00

The criterion for a matrix $T$ to be diagonalisable is either:

• The dimension of each eigenspace $E_\alpha$ is equal to the multiplicity $m_1$ of $\alpha$ as a root of the characteristic polynomial $\chi_{T}(x)$, or:

• Each eigenvalue $\alpha$ is a simple root of the minimal polynomial of $T$.

First method:

Using row-reduction, we obtain: $$I-T=\begin{bmatrix}1&0&3\\-1&1&-5\\0&-1&2\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&0&3\\0 &1&-2\\0&0&0\end{bmatrix}$$ This matrix has rank 2, hence $\dim E_1=1< m_1=2$. So $T$ is *not * diagonisable.

Moreover, solving the system, we find a basis for $E_1$ is the vector: $$\begin{bmatrix}-3\\2\\1\end{bmatrix}.$$

Second method: If the matrix is diagonalisable, its minimal polynomial must be $(x-1)(x+3)=x^2+2x-3=(x+1)^2-4$, hence $T$ must satisfy $(T+I)^2=4I$. However $$(T+I)^2=\begin{bmatrix}1&0&-3\\1&1&5\\0&1&0\end{bmatrix}^2=\begin{bmatrix}1&-3&-3\\2&6&2\\1&1&5\end{bmatrix}.$$