Determine the Jordan canonical form of the matrix: $\quad\begin{bmatrix} -7 & 9 \\ -4 & 5 \end{bmatrix}$

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    $\begingroup$ Welcome to MSE! Have you done any of the work and can share that? Is there a particular point you are getting stuck? Did you find the eigenvalues/eigenvectors, for example? $\endgroup$ – Amzoti Jun 20 '13 at 1:27

Hints, to find the Jordan Normal Form, we can use the following approach (and others are possible too):

  • (1) Find the eigenvalues (we have a multiplicity $2$ eigenvalue $\lambda_{1,2} = -1$)
  • (2) Find the eigenvectors (we find one eigenvector and a second generalized eigenvector)
  • (3) Write $A = P \cdot J \cdot P^{-1}$, where $J$ is the Jordan block (eigenvalues) and P is the corresponding eigenvectors.


To find the eigenvalues/eigenvectors, we set up and solve the characteristic polynomial $|A - \lambda I| = 0$, so we get ($v_i's$ are the eigenvectors):

  • $\lambda_1 = -1, ~v_1 = (3, 2)$
  • $\lambda_2 = -1, ~v_2 = (-\dfrac{1}{2}, 0)$

Jordan Form

The JF is given by:

$$A = PJP^{-1} = \begin{bmatrix}3 & -1/2\\2 & 0\end{bmatrix} \cdot \begin{bmatrix}-1 & 1\\0 & -1\end{bmatrix} \cdot \begin{bmatrix}0 & 1/2\\-2 & 3\end{bmatrix}$$

Note what makes up $J$ and the columns of $P$.

  • $\begingroup$ Nicely put + (xtra characters) $\endgroup$ – Namaste Jun 20 '13 at 1:32
  • $\begingroup$ Thanks! I'm teaching myself Jordan form for an exam, so I am rather shaky with it at the moment. I found an eigenvector of $\lambda=-1$ with multiplicity 2, with corresponding eigenvectors $[1\ 2/3]^{T}$ and $[3/2\ 1]^{T}$ (which I don't think is correct...). $\endgroup$ – Kirk Fogg Jun 20 '13 at 1:43
  • $\begingroup$ Thanks, this helps a lot. $\endgroup$ – Kirk Fogg Jun 20 '13 at 2:09
  • $\begingroup$ @KirkFogg: You are very welcome. Please note that the eigenvectors are not unique, but using RREF (Gaussian Elimination), you should be able to derive the ones I showed. Also, for the second eigenvector, we setup and solve $[A]\lambda I]v_2 = v_1$, to find a generalized eigenvector. Give a whirl! $\endgroup$ – Amzoti Jun 20 '13 at 2:16
  • $\begingroup$ Oh, I bet! There is so many ways kids (and anyone) can be taken advantage of, or make grievous choices...with all their online social networking. $\endgroup$ – Namaste Jun 22 '13 at 2:01

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