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.


closed as off-topic by user370967, José Carlos Santos, André 3000, Krish, Ove Ahlman Nov 29 '17 at 8:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, José Carlos Santos, André 3000, Krish, Ove Ahlman
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Kindly include your attempt and describe where do you get stuck. $\endgroup$ – Siong Thye Goh Nov 29 '17 at 6:19
  • $\begingroup$ where should i begin? $\endgroup$ – Tiny Nov 29 '17 at 6:20
  • $\begingroup$ what do you know about diagonalization? $\endgroup$ – Siong Thye Goh Nov 29 '17 at 6:21
  • $\begingroup$ in this case the matrix is not diagonalizable since its eigenvectors are not linearly independent $\endgroup$ – Tiny Nov 29 '17 at 7:33
  • $\begingroup$ $A$ is not diagonalizable, because it fails the diagonalizability test.or use the fact "an nxn matrix is diagonalizable iff it has n linearly independent eigenvectors" $\endgroup$ – Chinnapparaj R Nov 29 '17 at 7:37

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 $rank(A-2I)=2$

  • $\begingroup$ ok perfect! so now how do i find the matrix Q such that Q-1QA is the jordan canonical form for A $\endgroup$ – Tiny Nov 29 '17 at 8:03
  • $\begingroup$ There are totally two linearly independent eigenvectors, hence there are two blocks in JCF, namely a 2x2 block corresponding to 2 and one single 1x1 block corresponding to 3. $\endgroup$ – Chinnapparaj R Nov 29 '17 at 13:09
  • $\begingroup$ can you please give symbolics? or a link to a good example $\endgroup$ – Tiny Nov 29 '17 at 16:46
  • $\begingroup$ could you please expand on the jordan canonical form $\endgroup$ – Tiny Nov 30 '17 at 13:36

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.

  • $\begingroup$ in this case the matrix is not diagonalizable since its eigenvectors are not linearly independent $\endgroup$ – Tiny Nov 29 '17 at 7:29

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.


Not the answer you're looking for? Browse other questions tagged or ask your own question.