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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.

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

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  • $\begingroup$ 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
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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 $rank(A-2I)=2$

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  • $\begingroup$ 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
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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.

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

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