3
$\begingroup$

I don't believe this matrix is diagonalizable, but I wanted to make sure.

Here's the matrix:

$A=\begin{bmatrix} 4 & 0 & 0 \\2 & 2 & 0 \\ 0 & 2 & 2 \end{bmatrix}$

A nice thing about this matrix is that it is diagonal, so the eigenvectors are $\lambda_1 = 4$, and $\lambda_2 = 2$ with algebraic multiplicity $2$.

The associated eigenvector for $\lambda_1$ is $x_1 = (1,1,1)$. When trying to solve for the eigenvector of $\lambda_2$, one gets the reduced matrix

$C = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Here, there is only the trivial solution, and that can't be an eigenvector due to definition. Based on that, can I rightly say that this matrix is not diagonalizable?

  • 1
    I bet you meant "A nice thing...it is **triangular**..." , and not "diagonal", am I right?2012-12-02
  • 0
    Ha, I didn't catch that! Yes, I meant triangular.2012-12-02
  • 0
    @Tim: I am not sure what exactly do you mean by a "reduced matrix", but $A$ does have an eignvector for $\lambda_2$: $v=(0,0,1)^T$.2012-12-02
  • 0
    By reduced matrix, I mean a row equivalent matrix under GJE.2012-12-02

1 Answers 1

3

Yes, your argument works. You can also remark that since the minimal polynomial of the matrix is the same as its characteristic one, i.e. $(x-4)(x-2)^2\,$ , then as the min. pol. is not a product of different linear factors the matrix isn't diagonalizable.

  • 0
    I didn't think of that, thanks.2012-12-02