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$$\left( \begin{array}{ccc} 0 & 1 & 2 \\ -5 &-3 & -7 \\ 1 & 0 & 0 \end{array} \right) $$

I figured out the eigenvalues are all -1 from the characteristic polynomial, but I'm not sure how to find the 1's on the subdiagonal. I know they're 1, but I'm not sure how that's determined. I also know that the eigenspace is 0 on the matrix where we subtract -1 on the diagonal. Thank you.

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  • $\begingroup$ Do you need to know only the Jordan normal form or also the basis with respect to which the matrix is in JNF? $\endgroup$ – Bman72 Dec 15 '14 at 21:05
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Here's a way to find the minimal polynomial, in this case:

We know that the characteristic polynomial is going to be $(x+1)^3$. The minimal polynomial must divide that.

We find $$ A + I = \pmatrix{1&1&2\\-5&-2&-7\\1&0&1} $$

Method 1: what is $(A+I)^2$? If it's zero, the min polynomial is $(x+1)^2$. Otherwise, it's $(x+1)^3$

Method 2: It is pretty clear that rank$(A + I) \geq 2$. Of the possible Jordan forms $J$, only one satisfies rank$(J+I) = 2$.

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  • $\begingroup$ Thanks, I used method 1 on my exam today. $\endgroup$ – Joe Dec 17 '14 at 1:08
  • $\begingroup$ It should be $-2$, not $-4$, on term $a_{22}$. $\endgroup$ – Aaron Maroja Nov 3 '15 at 13:06
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I will map it out and you can fill in the details.

Approach 1:

We are given:

$$A = \begin{pmatrix} 0 & 1 & 2 \\ -5 &-3 & -7 \\ 1 & 0 & 0 \end{pmatrix} $$

Setting up and solving

$$|A - \lambda I| = 0 \implies -(\lambda + 1)^3 = 0 \implies \lambda_{1,2,3} = -1$$

To find the eigenvectors, we would take $[A - \lambda_i I] = 0$, assuming we have unique eigenvalues. Here we do not. Do you know how to determine the algebraic versus geometric multiplicity?

We attempt to the find the first eigenvector using $[A + I]v_1 = 0$ and the row-reduced-echelon-form (RREF) produces:

$$v_1 = \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix} $$

Unfortunately, we cannot generate anymore linearly independent eigenvectors ($\dim ker(A)=1$), so have to resort to generalized eigenvectors.

To find the second (generalized) eigenvectors, we set up and solve $[A + I]v_2 = v_1$, and after finding the RREF yields:

$$v_2 = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} $$

To find the third (generalized) eigenvectors, we set up and solve $[A + I]v_3 = v_2$, and after finding the RREF yields:

$$v_3 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} $$

We can use these eigenvectors to form $P$ as:

$$P = \begin{pmatrix} -1 & 1 & 0 \\ -1 & -2 & 1 \\ 1 & 0 & 0 \end{pmatrix} $$

We can now use this to find our Jordan matrix $J$ as:

$$J = P^{-1} A P = \begin{pmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{pmatrix}$$

Approach 2:

Have you learned about the characteristic polynomial and minimal polynomials and ways to infer the Jordan form?

You can find discussions in previous MSE posts like How can one find Jordan forms, given the characteristic and minimal polynomials?, Jordan normal form for a characteristic polynomial $(x-a)^5$, and these lecture notes.

Try stepping through these whimsical notes and see if you can work through approach 2.

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  • $\begingroup$ Thanks, I used the minimum polynomial to show it wasn't annilihated on my exam today. $\endgroup$ – Joe Dec 17 '14 at 1:08

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