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I have a $5\times 5$ matrix and I need to find the Jordan form and its inverse. I know how to find the inverse. But for the Jordan form I am screwed.

The matrix is $$\begin{bmatrix}3 & 0 & 0 & 0 & 0\\2 & 3 & 0 & 0 & 0\\1 & 0 & 2 & 2 & 0\\0 & 0 & 0 & 3 & 2\\0 & 0 & 0 & 0 & 2\end{bmatrix}.$$

Please help me to find the eigenvalues and number of Jordan blocks. Is there any easy method to know the eigenvalues other than solving $\det(A-\lambda I)=0$?

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2 Answers

Note that $$ M=\left[\begin{array}{cc|ccc} 3 & 0 & 0 & 0 & 0\\ 2 & 3 & 0 & 0 & 0\\ \hline 1 & 0 & 2 & 2 & 0\\ 0 & 0 & 0 & 3 & 2\\ 0 & 0 & 0 & 0 & 2 \end{array}\right] =\begin{pmatrix}A&0\\C&D\end{pmatrix} $$ is lower block triangular. Therefore the eigenvalues of $M$ are precisely the eigenvalues of $A$ and $D$. Yet $A$ and $D$ are triangular matrices. So you can read off their eigenvalues $3,3,2,3,2$ immediately from their diagonals.

Having the eigenvalues, you may find the eigenvectors and the dimensions of the eigenspaces the "old" way (i.e. by inspecting the system $(M-\lambda I)x=0$. Alternatively, note that $$ F=\begin{pmatrix}1&0\\0&0\\0&0\end{pmatrix} \ \Rightarrow\ \begin{pmatrix}I&0\\-F&I\end{pmatrix} \begin{pmatrix}A&0\\C&D\end{pmatrix} \begin{pmatrix}I&0\\F&I\end{pmatrix} =\begin{pmatrix}A&0\\0&D\end{pmatrix}. $$ Therefore $M$ is similar to $A\oplus D$ and its Jordan form is just the direct sum of the Jordan forms of $A$ and $D$. Now the Jordan form of $A$ is clearly $J_2(3)$, the $2\times2$ Jordan block for the eigenvalue $3$. And the Jordan form of $D$ is $3\oplus J_2(2)$, because the nullity of $D-2I$ is $1$. Putting the pieces together, the Jordan form of $M$ is $J_2(3)\oplus 3\oplus J_2(2)$. You may confirm this using some online computer algebra system.

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+1, nice and clean. –  Andreas Caranti Feb 26 '13 at 10:18
    
@user1551 If you dont mind I want to ask some more doubts, how you got the eigen values of A as 2,3. If the diagonal elements are the eigen value then A must have 3,3. But its 2,3.Can you please tell me how –  Iya Feb 26 '13 at 10:21
    
@Iya Thanks. It's just a typo and I have corrected it. –  user1551 Feb 26 '13 at 10:23
    
@user1551 thanks a lot –  Iya Feb 26 '13 at 10:40
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When you do cofactor expansion on $A-\lambda I$ you'll find that it's always the case that either the first row or the first column has only one nonzero entry. So taking the determinant is very very quick and I know of no other way to get eigenvalues other than guessing them.

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can you please tell me how many jordan block are here. –  Iya Feb 26 '13 at 8:31
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