# find the eigenvalues of $5×5$ matrix

The product of the non-zero eigenvalues of the matrix is ____.

using characteristic equation , it is too lengthy to find eigenvalues of $5×5$ matrix , I'm looking for any short trick to solve this question .

Hint: Note that $(1,0,0,0,1)^T$ and $(0,1,1,1,0)^T$ are eigenvectors by inspection. What are their corresponding eigenvalues? (You just need to compute $Av$ for each of the above.) Now observe that the matrix has only $2$ linearly independent columns, so its kernel has dimension $3$. Thus $0$ is also an eigenvalue with multiplicity $3$.

• can you give me best link to read eigenvectors by inspection . thanks for hints. – Mithlesh Upadhyay Oct 12 '15 at 7:05
• I'm afraid there's no such link. By "by inspection," I just meant that this matrix is of a particular form so that we figure out the eigenvectors without computation. Since the first and last columns are the same, multiplying by the column vector $v = (1,0,0,0,1)^T$ yields the sum of these two columns, which is $2v$. – André 3000 Oct 12 '15 at 16:09

Hint

Note that \left(\begin{align}1 \\ 0 \\ 0\\ 0\\ 1\end{align}\right) and \left(\begin{align}1 \\ 0 \\ 0\\ 0\\ -1\end{align}\right) are two linearly independent eigenvectors.

Also note that

\left(\begin{align}0 \\ 1 \\ 1\\ 1\\ 0\end{align}\right), \left(\begin{align}0 \\ -2 \\ 1\\ 1\\ 0\end{align}\right) and \left(\begin{align}0 \\ 1 \\ 1\\ -2\\ 0\end{align}\right) are three linearly independent eigenvectors.

• thanks for hints , please can you hint , how I guess these vectors , thank you. – Mithlesh Upadhyay Oct 12 '15 at 7:07
• Note that $(x,0,0,0,y)$ and $(0,x,y,z,0)$ are invariant subspaces. Thus you can study the eigenvalues of $\left(\begin{matrix}1 & 1\\ 1 & 1\end{matrix}\right)$ and $\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1\\ 1 & 1 & 1\end{matrix}\right)$ separately. – mfl Oct 12 '15 at 15:11

Let $A$ and $B$ be repectively the following matrices:

Then $A= B +I_5$ where $I_5$ is the identity matrix.

Since B and I commutes then the $eigenvalues(A)=eigenvalues(B) +eigenvalues(I_5)$ (with the same order), that is $eigenvalue(A)=eigenvalue(B) +1$ (for every eigenvalue of B).

Now commuting the eigenvalues of $B$ is easy since in the first row we have only one nonzero entry.