# finding the dimension of eigenspace with characteristic and minimal polynomial

I have a problem in understanding how author find the dimension of eigenspace of 3 and 1 in following example. How can I calculate it?

For example, suppose you're given a $6 \times 6$ matrix and you calculate that its characteristic polynomial is $(t-3)^4(t-i)^2$, that its minimal polynomial is $(t-3)^2 (t-i)^2$, that the 3-eigenspace is 3-dimensional, and that the i-eigenspace is 1-dimensional.

Thanks for your help

Let us take for example the eigenvalue $\;\lambda=3\;$: that the exponent of its linear factor in the minimal polynomial is two means that two is the largest block in the Jordan Block (=JB) corresponding to $\;3\;$, which means that the JB, which clearly has size $\;4\times4\;$ , has exactly three blocks: one of size two, two of size one. Something like the following, with the different colors marking the different block in $\;JB_3\;$ :

$$JB_{\lambda=3}=\begin{pmatrix} \color{red}3&\color{red}1&0&0\\ \color{red}0&\color{red}3&0&0\\ 0&0&\color{green}3&0\\ 0&0&0&\color{blue}3\end{pmatrix}$$

Since the number of blocks in $\;JB_\lambda\;$ gives us $\;\dim V_\lambda\;$ ,we're done.

• Isn't two Jordan blocks of size 2 a possibility? – Rab Jun 13 '17 at 10:05
• Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni. – Rab Jun 13 '17 at 10:10
• @RabMakh You're completely correct, but in this case we're given that $\;\dim V_3=3\;$ , so there can't be only two blocks but three. – DonAntonio Jun 13 '17 at 11:20
• Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $\';2\times2\;$ blocks, I believe. – DonAntonio Jun 13 '17 at 11:22
• @DonAntonio "3-eigenspace is 3-dimensional", it is given ... – Peter Melech Jun 16 '17 at 7:52

That implies its Jordan normal form is:

$\begin{pmatrix}3&1&0&0&0&0\\ 0&3&0&0&0&0\\ 0&0&3&0&0&0\\ 0&0&0&3&0&0\\ 0&0&0&0&i&1\\ 0&0&0&0&0&i\end{pmatrix}$

• why we don't have two 2*2 blocks for eigenvalue 3? – fateme jl Jun 15 '17 at 12:05
• Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1 – Peter Melech Jun 16 '17 at 7:48

The Jordan form of your matrix must be of the type$$\begin{pmatrix}3&*&0&0&0&0\\0&3&*&0&0&0\\0&0&3&*&0&0\\0&0&0&3&0&0\\0&0&0&0&i&*\\0&0&0&0&0&i\end{pmatrix},$$where each $*$ is either $0$ or $1$. But if, for instance, the $*$ from the fifth row is equal to $0$, then the exponent of $t-i$ in the minimal polynomial of the matrix must be $1$. Since it is $2$, that $*$ must be equal to $1$. Now, see what the other three $*$ must be so that the exponent of $t-3$ in the minimal polynomial of the matrix is $2$.