3
$\begingroup$

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

enter image description here


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 .

$\endgroup$
2
$\begingroup$

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 $3$.

$\endgroup$
  • $\begingroup$ can you give me best link to read eigenvectors by inspection . thanks for hints. $\endgroup$ – Mithlesh Upadhyay Oct 12 '15 at 7:05
  • $\begingroup$ 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 $2v$. $\endgroup$ – André 3000 Oct 12 '15 at 16:09
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ thanks for hints , please can you hint , how I guess these vectors , thank you. $\endgroup$ – Mithlesh Upadhyay Oct 12 '15 at 7:07
  • $\begingroup$ 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. $\endgroup$ – mfl Oct 12 '15 at 15:11
1
$\begingroup$

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

enter image description here enter image description here

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.

$\endgroup$
  • 1
    $\begingroup$ @Silence Your welcome $\endgroup$ – M.Badaoui Oct 12 '15 at 8:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.