# How to find Jordan normal form and basis?

I have a matrix:

$$A = \begin{pmatrix}-1&5&4\\ 4&-6&-6\\ -8&16&14 \end{pmatrix}$$

simplifying:

$$A = \begin{pmatrix}-1&5&4\\ 2&-3&-3\\ -4&8&7 \end{pmatrix}$$

What steps should I reproduce to find Jordan form and Jordan basis, I am really really confuswd without good step by step example!

What I really understand that I have to find eigenvalues, so I did:

$$\det|A-I\lambda| = \begin{vmatrix}-1 - \lambda&5&4\\ 2&-3-\lambda&-3\\ -4&8&7-\lambda \end{vmatrix}$$

so I found cubic equation which is:

$$-\lambda^3+3\lambda^2-5\lambda+3 = 0$$ $$-(\lambda-1)^3-2(\lambda-1)$$

So as I can judge Jordan normal form will have three eigenvalues that equal to $\lambda = 1$ on its main diagonal, below main diagonal all the values should be "$0$". But I bet it is not all. I do not know how to proceed (If I am right at all), and do not know how to find Jordan's basis after :(

• I don't see how you get that characteristic equation from that matrix. And I don't see how you get that factorization from that characteristic equation. – Doug M Sep 14 '17 at 21:08
• @DougM, I missed the power, fixed it now – user477211 Sep 14 '17 at 21:17

I have characteristic equation

$-\lambda^3 + 7\lambda^2 -16\lambda + 12 = -(\lambda - 2)^2(\lambda-3)$

lets get the easy ones out of the way.

$\lambda = 3$

$\pmatrix{-4&5&4\\4&-9&-6\\-8&16&11}\mathbf v_1 = 0\\ \mathbf v_1 = \pmatrix{3\\-4\\8}$

$\lambda = 2$

$\pmatrix{-5&5&4\\4&-8&-6\\-8&16&12}\mathbf v_2 = 0\\ \mathbf v_2 = \pmatrix{4\\-4\\8}$

Curious.

When you have $Av = 0$ you can simplify $A$ with elementary row operations to find $v.$

Now for the odd man...

$A \mathbf v_3 = 2v_3 + v_2\\ (A - 2I)v_3 =v_2$

If you perform any row operations, that is the equivalent to multiply on the left some matrix, you must also apply the same operations to $\mathbf v_2$

$\pmatrix{-3&5&4\\4&-8&-6\\-8&16&12}v_3 =\pmatrix{4\\-4\\8}\\ \pmatrix{-3&5&4\\0&-4&-2\\0&0&0}v_3 =\pmatrix{4\\4\\0}\\ \mathbf v_3 = \pmatrix{-3\\-1\\0}$

$\pmatrix{-1&5&4\\4&-6&-6\\-8&16&14}\pmatrix{3&4&-3\\-4&-4&-1\\8&8&0} = \pmatrix{3&4&-3\\-4&-4&-1\\8&8&0}\pmatrix{3\\&2&1\\&&2}$

Update:

Since I had a doubled eigenvalue and only found one vector for that eigenvalue, I took a bit of a short cut. I knew that I had this Jordan block, and

$A[v_2,v_3] = [v_2,v_3]\begin{bmatrix} 2&1\\&2\end{bmatrix} = [2v_2,v_2+v_3]$ and solved for that.

There is another way to do it, and it might line up with your notes better.

If you have a defective eigenvalue. You can find the generalized eigenvector this way.

(It starts out the same)

We have found $\bf v_2$ as an eigenvector.

$(A-\lambda I)\mathbf v_3 = \mathbf v_2\\ (A-\lambda I)^2\mathbf v_3 = (A-\lambda I) \mathbf v_2$

Since $bf v_2$ is an eigenvector $(A-\lambda I) \mathbf v_2 = \mathbf 0$

$(A-\lambda I)^2\mathbf v_3 = \mathbf 0$

And in this problem

$(A-\lambda I)^2 = \pmatrix{-3&9&6\\4&-12&-8\\-8&24&16}$

$\mathbf v_3 = \pmatrix{-3\\-1\\0}$ is in the kernel of $(A-\lambda I)^2$

• If I understood correctly we found two eigenvalues and then two eigenvectors. As it is very well seen $\lambda = 2$ has power $2$ and we have only 2 vectors found, so we need one more which should be $v_3$. If above is correct then I am confused how you got $A\mathbf{v}_3 = 2v_3+v_2$ – user477211 Sep 15 '17 at 7:12
• @d.sci I have added some additional information. – Doug M Sep 15 '17 at 16:46
• yes, thats what I was thinking about, "if you do not have enough eigenvectors, find generalized eigenvectors", but how did you get equation with matrices? (The last one befrore your update) I can not follow this. – user477211 Sep 15 '17 at 18:33