Find Jordan form of the following matrix: $$\left(\begin{matrix}4&-5&2 \\ 5&-7&3\\ 6&-9&4 \end{matrix}\right)$$

So I got stuck pretty much trying to find the eigenvalues.

Related question: Is the characteristic polynomial of the characteristic matrix, equals to the characteristic polynomial of the transpose of the characteristic matrix? Since their determinants are equal...


  • $\begingroup$ See here : en.wikipedia.org/wiki/Matrix_similarity $\endgroup$ – Isomorphic Dec 9 '13 at 15:55
  • $\begingroup$ Where did you get stuck trying to find the eigenvalues? Have you found the characteristik polynomial? If not, why not? $\endgroup$ – fgp Dec 9 '13 at 16:02
  • $\begingroup$ I got to an expression of a 3rd degree. Isn't there a more simple way? Maybe some row operations I missed? $\endgroup$ – err Dec 9 '13 at 16:04
  • $\begingroup$ No.. No.. I would not think it would be a better way to go for short paths when you are just starter learning something.. if you have some problem with finding roots of characteristic polynomial we could help... $\endgroup$ – user87543 Dec 9 '13 at 16:08

The characteristic polynomial is given by $|A - \lambda I| = 0$, yielding:

$$\begin{vmatrix}4 - \lambda &-5 & 2 \\ 5 &-7 - \lambda & 3\\ 6 & -9 & 4 - \lambda \end{vmatrix} = 0$$ Using determinants, this reduces to the characteristic polynomial:

$$\lambda^2 - \lambda^3 = 0 \rightarrow -\lambda^2(\lambda - 1) \rightarrow \lambda_1 = 1, \lambda_{2,3} = 0, $$

To find the eigenvectors, we solve $[A-\lambda_i I]v_i = 0$

For the first eigenvalue, $\lambda_1 = 1$, we have:

$$v_1 = (1,1,1).$$

For the second eigenvalue, $\lambda_2 = 0$, we have a single eigenvector, which is:

$$v_2 = (1,2,3).$$

For the third, we need a generalized eigenvector and use $[A-\lambda I]v_3 = v_2$, yielding:

$$v_3 = (-1,-1,0).$$

I purposely left out the details so you can fill them in.

Also, you now have everything you need to find the Jordan Normal Form and there are other approaches to finding it.

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  • $\begingroup$ Can you explain how you got to that characteristic polynomial please? $\endgroup$ – err Dec 9 '13 at 19:12
  • $\begingroup$ Thank you very much, I succeeded in solving it! Thank you. Now another question if you may. If J is the jordan form of A, then there is a nonsingular matrix C such that $A=C^-1JC$. How do I calculate C if A isn't diaganozable? $\endgroup$ – err Dec 9 '13 at 20:39
  • $\begingroup$ Not all matrices are diagonalizable, like yours. We write $J=P{−1}AP$, where $P$ is made from the columns of the eigenvectors for the individual eigenvalues. Regards $\endgroup$ – Amzoti Dec 9 '13 at 21:56
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    $\begingroup$ Nice balance: enough of a push while leaving some work for the OP! $\endgroup$ – amWhy Dec 10 '13 at 0:04
  • $\begingroup$ Amzoti, but what's the order of the collumms of P? That is, if for instance the Jordan form of A is $$B=\left(\begin{matrix}0&1&0 \\ 0&0&0\\ 0&0&1 \end{matrix}\right)$$, and for two eigenvalues who are 0, there is one eigenvector, what's the order of the vectors I should write in P? Is it first $(0, 0, 0)$, then $(1, 2, 3)$ (eigenvector I found) or reversed? Thanks! $\endgroup$ – err Dec 10 '13 at 12:36

Answer to related question: $$|tI-A|=|(tI-A)^T|=|(tI)^T-A^T|=|tI-A^T|$$ So yes, $A$ and $A^T$ have the same characteristic polynomials.

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When you compute $\det(tI-A)$, you should find the characteristic polynomial to be $t^3 - t^2 = t^2(t-1)$. So, your eigenvalues are $0$ and $1$. Note that $0$ has one eigenvector and one generalized eigenvector, since $\dim \ker(A) = 1$.

Neat way to find your eigenvector for $1$: check the row sums.

As for your extra question: not only do $A$ and $A^T$ have the same characteristic polynomial, they actually have the same Jordan form as well!

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