Is A diagonalizable? If yes find a matrix such that is a diagonal matrix, if not, find a matrix such that is Jordan

Question

Let A = $\begin{bmatrix} 9&4&5 \\ -4&0&-3 \\ -6&-4&-2 \end{bmatrix}$ $\in$ $M_{3x3}$$(\mathbb{R})$.

Is A diagonalizable?

Justify your answer.

If yes, find a matrix $Q$ such that $Q^{-1}AQ$ is a diagonal matrix.

If not, find a matrix $Q$ such that $Q^{-1}AQ$ is the Jordan canonical form for A. Write out the diagonal matrix or the Jordan canonical form.

Okay update, so I found out the matrix is not diagonalizable. However, I am having trouble understanding the theory behind representing the matrix in Jordan Form.

I have that the eigenvalues are $\lambda_{1,2}$ = 2 and $\lambda_{3}$ = 3.

Could someone please provide a nice explanation as to how I get $Q^{-1}AQ$ as the Jordan canonical form for A.

Answer 1

Test for Diagonalization:

Let $T$ be a linear operator on an n-dimensional vector space $V$. Then $T$ is diagonalizable if and only if both of the following conditions hold.

1. The characteristic polynomial of $T$ splits.
2. For each eigenvalue $\lambda$ of $T$, the multiplicity of $\lambda$ equals $n - rank(T — \lambda I).$

Here, $2,2,3$ are the eigenvalues.

But multiplicity of $2$ = $2 \neq 3-rank(A-2I)$, since $rank(A-2I)=2$

Answer 2

A is diagonalizable iff A is nondefective, i.e., the dimension of A's eigenspace is the same with its generalized eigenspace. Or you can directly calculate its Jordan canonical form. For the diagonalizable matrix, it automatically turns to be a diagonal matrix.

Answer 3

I should check and find its eigenvalues. if you found its eigenvalues and they are not independent so A is not diagonalizable. because the columns of Q(3 by 3 matrix) are linear independent eigenvectors of A. This is a theorem.