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It is first time I try normally to figure out how it works.

So there's the matrix:

$$A=\begin{pmatrix} 0 & 6 & 4 & 0 \\ 0 & 4 & 0 & 2 \\ 0 & -1 & 2 & -1 \\1 & -3 & -2 & 2 \end{pmatrix}$$

1.) First of all we have to find eignevalue:

$$\mathrm{det}|A-\lambda{I}|= \begin{vmatrix} 0 - \lambda & 6 & 4 & 0 \\ 0 & 4-\lambda & 0 & 2 \\ 0 & -1 & 2-\lambda & -1 \\ 1 & -3 & -2 & 2-\lambda \end{vmatrix} = (\lambda-2)^4$$

So eigenvalue $\lambda = 2$

2.) Substitute $\lambda$ value into the matrix, got:

$$A_{\varphi}=\begin{pmatrix} -2 & 6 & 4 & 0 \\ 0 & 2 & 0 & 2 \\ 0 & -1 & 0 & -1 \\ 1 & -3 & -2 & 0 \end{pmatrix}$$

3.) As far as I understand I have to find rank of $A_{\varphi}$. Therefore $\mathrm{rank}A_{\varphi}=2$

Because $$A_{\varphi}=\begin{pmatrix} 0 & -1 & 0 & -1 \\ 1 & -3 & -2 & 0 \end{pmatrix}$$

I do not know how to procced from here...

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  • $\begingroup$ The answers to the 'related' questions appearing in the side bar are pretty good and should be useful for this question! $\endgroup$ – Vincent Jun 13 '17 at 8:49
  • $\begingroup$ $$J=\begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 0 \\0 & 0 & 0 & 2 \end{pmatrix}$$ $\endgroup$ – Vladimir Vaschenko Jun 13 '17 at 8:58
  • $\begingroup$ @Vincent, well, I would not ask if I could handle it using exist eamples. That's why I wrote so much latex code step by step $\endgroup$ – M.Mass Jun 13 '17 at 8:58
  • $\begingroup$ You have to find vector(s ?) $\begin{pmatrix}a\\b\\c\\d\\\end{pmatrix}$ such that $\begin{pmatrix} 0 & -1 & 0 & -1 \\ 1 & -3 & -2 & 0 \end{pmatrix}\begin{pmatrix}a\\b\\c\\d\\\end{pmatrix}$. $\endgroup$ – Jean Marie Jun 13 '17 at 9:19
  • $\begingroup$ @JeanMarie, so should I solve system of linear equations, I guess? (It just reminds me problem of finding eignevectors) $\endgroup$ – M.Mass Jun 13 '17 at 10:04
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You can easily, though annoyingly carefully, check that $\;(A-2I)^3=0\;$ and that means the minimal polynomial of your matrix is $\;(x-2)^3\;$ , which means its Jordan form must have at least (and, in this case, obviously also at most) a Jordan block of size three, so

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

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  • $\begingroup$ Doesn't that follow just from the fact that rank(A-2I)=2? $\endgroup$ – Vladimir Vaschenko Jun 14 '17 at 5:29
  • $\begingroup$ @VladimirVaschenko Just that fact leaves still two possibilities: either 1 block of size 3 and one block of size 1 as depicted above or two blocks of size 2. You need some more care (as DonAntonio put it) to rule out that last case. $\endgroup$ – Vincent Jun 14 '17 at 12:46
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I'm gonna go ahead and assume DonAntonio is right about the Jordan normal form itself. However I guess you also want to write down a basis with respect to which $A$ takes that form.

Let's see what that means. You are looking for a basis $\mathbf{v}_1, \mathbf{v}_2 ,\mathbf{v}_3, \mathbf{v}_4$ such that with respect to that basis, the matrix $A$ takes the form $J_A$ written in DonAntonio's answer. Since w.r.t. to their own basis the $\mathbf{v}_i$ look like $(1, 0, 0, 0)^T$, $(0, 1, 0, 0)^T$ etc we see that they should have the following properties:

  • Under the action of $A$, vector $\mathbf{v}_1$ is sent to 2 times itself - equivalently it is in the kernel of the matrix $A_\varphi$ you wrote.
  • Under the action of $A$ vector $\mathbf{v}_2$ is sent to 2 times itself plus $\mathbf{v}_1$. Equalently $A_\varphi$ sends $\mathbf{v}_2$ to vector $\mathbf{v}_1$.
  • Under the action of $A$ vector $\mathbf{v}_3$ is sent to 2 times itself plus $\mathbf{v}_2$. Equalently $A_\varphi$ sends $\mathbf{v}_3$ to vector $\mathbf{v}_2$.
  • Under the action of $A$, vector $\mathbf{v}_4$ is sent to 2 times itself - equivalently it is in the kernel of the matrix $A_\varphi$ you wrote.

Now to find vectors $\mathbf{v}_i$ that fit above description it seems fair to first find all possibilities for $\mathbf{v}_1$ and $\mathbf{v}_4$, in other words to find $\ker A_\varphi$.

You got quite far. The matrix $A_\varphi$ has rank 2, which means that it also has a 4 - 2 = 2 dimensional kernel. This kernel is the same as the eigenspace of the original matrix $A$ at eigenvalue 2. Using your technique we can find all vectors in this space. You already found one: $\mathbf{a} = \begin{pmatrix} 6 \\ 0 \\ 3 \\ 0 \end{pmatrix}$. Here is another (just from staring at the equations you wrote down in the comments): $\mathbf{b} = \begin{pmatrix} 1 \\ 1 \\ -1 \\-1 \end{pmatrix}$.

It is important to realize that any vector of the form $\lambda \mathbf{a} + \mu \mathbf{b}$ is in the kernel of $A_\varphi$ and it would be a bit premature to just pick $\mathbf{v}_1 = \mathbf{a}$ and $\mathbf{v}_4 = \mathbf{b}$. The trouble being of course that we need to choose $\mathbf{v}_1$ in such a way that we can solve $A_\varphi \mathbf{v}_2 = \mathbf{v}_1$. But at least we can be happy that we have some concrete vectors we can work with.

In order to find $\mathbf{v}_2$ we can either solve $A_\varphi \mathbf{v}_2 = \lambda \mathbf{a} + \mu \mathbf{b}$ such that $\lambda, \mu$ are not both zero or solve $A_\varphi^2 \mathbf{v}_2 = 0$ and from all the solutions pick one that is not already a solution of $A_\varphi \mathbf{v} = 0$, i.e. that is not of the form $\lambda \mathbf{a} + \mu \mathbf{b}$. There are multiple choices for $\mathbf{v}_2$ possible but once you pick one it automatically gives us the corresponding choice of $\mathbf{v}_1$.

Now at that point we need to use all the work we did on finding $\mathbf{v}_2$ to see if there are two linearly independent choices for $\mathbf{v}_2$ we could have taken. In that case the Jordan form $J_A$ in DonAntonio's answer was wrong and the true Jordan form is $J_A' = \begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{pmatrix}$. However I trust DonAntonio and don't expect that to happen.

So that leaves us with the easy task of picking a $\mathbf{v}_4$ (any element in $\ker A_\varphi$ that is not a scalar multiple of $\mathbf{v}_1$ will do) and the harder task of finding $\mathbf{v}_3$. But this is again solving a linear equation since we know that $A_\varphi \mathbf{v}_3 = \mathbf{v}_2$.

One more remark: I short-cutted the answer a little by already knowing the shape of $J_A$, but if we would not know it then from the fact that the $\ker A_\varphi$ is 2-dimensional we could already say that the Jordan form was either DonAntonio's $J_A$ or the matrix $J_A'$ I wrote above and so finding which one and finding the corresponding vectors $\mathbf{v}_i$ would be almost the same as what I wrote in this answer.

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