Math Help - Finding the eigenvalues of a 4x4 matrix?

1. Finding the eigenvalues of a 4x4 matrix?

Hi!

I need help finding the eigenvalues of a 4x4 matrix. Been working with the same problem for 2 days now. I just can't solve it. Help me pls!
The matrix (A) is:

$

\left( \begin{array}{cccc} 1 & -3 & 2 & -1 \\ -3 & 9 & -6 & 3 \\ 2 & -6 & 4 & -2 \\ -1 & 3 & -2 & 1\end{array} \right)

$

(Notice that the matrix is symmetric, if that is to any help)

I have come this far (If I'm on the right track).

det $

\left( \begin{array}{cccc} 1-\lambda & -3 & 2 & -1 \\ -3 &
9-\lambda & -6 & 3 \\ 2 & -6 & 4-\lambda & -2 \\ -1 & 3 & -2 & 1-\lambda \end{array} \right)

$
= 0

How do I solve for L from this? I would appreciate any help.

Thanks!
/Gordina

$\begin{bmatrix}{\lambda}-1&3&-2&1\\3&{\lambda}-9&6&-3\\-2&6&{\lambda}-4&2\\1&-3&2&{\lambda}-1\end{bmatrix}$

This gives a characteristic polynomial of:

${\lambda}^{4}-15{\lambda}^{3}={\lambda}^{3}({\lambda}-15)=0$

Now, solve this quartic and you have your eigenvalues.

3. Hi,

Another approach :
Originally Posted by Gordina
I need help finding the eigenvalues of a 4x4 matrix. Been working with the same problem for 2 days now. I just can't solve it. Help me pls!
The matrix (A) is:
$

\left( \begin{array}{cccc} 1 & -3 & 2 & -1 \\ -3 & 9 & -6 & 3 \\ 2 & -6 & 4 & -2 \\ -1 & 3 & -2 & 1\end{array} \right)

$

(Notice that the matrix is symmetric, if that is to any help)
Let $v=\begin{pmatrix}1\\-3\\2\\-1 \end{pmatrix}$. Note that $A=\begin{pmatrix}v&-3v&2v&-v \end{pmatrix}$ so $\mathrm{Im}(A)=\mathrm{Span}(\{v\})$ which implies $\dim\left(\mathrm{Im}(A)\right)=\dim\left(\mathrm{ Span}(\{v\})\right)=1$. As $\dim\left(\mathrm{Im}(A)\right)+\dim\left(\ker(A)\ right)=\dim(\mathbb{R}^4)=4$, one has $\dim\left(\ker(A)\right)=3>0$ hence 0 is an eigenvalue of $A$.

The spectral theorem tells us that because $A$ is symmetric, it is diagonalizable hence :

(1) _ the multiplicity of an eigenvalue of $A$ equals the dimension of the corresponding eigenspace ;
(2) _ the sum of the dimension(s) of the eigenspace(s) of $A$ equals $\dim(\mathbb{R}^4)=4$.

Thanks to (1), we get that the multiplicity of the eigenvalue 0 is $\dim\left(\ker(A-0\cdot\mathrm{I}_3)\right)=\dim\left(\ker(A)\right )=3$. As $\dim\left(\ker(A)\right)=3\neq4$, (2) implies that one eigenvalue $\alpha$ (of multiplicity 1) remains to be found. This can easily be done using the equality $\mathrm{tr}(A)=\sum_{\lambda\in\mathrm{Sp(A)}}\lam bda$ where $\mathrm{Sp}(A)=\{0,\alpha\}$ is the spectrum of $A$ and $\mathrm{tr}(A)$ is the trace of $A$.

If this proof is correct it could probably be generalized to any matrix $A=\begin{pmatrix} v_1 v& v_2v&\cdots& v_n v\end{pmatrix}$ where $v=\begin{pmatrix} v_1 \\ v_2\\ \vdots \\v_n\end{pmatrix}\in\mathbb{R}^n$. It would show that the two eigenvalues of $A$ are 0 (multiplicity = $n-1$) and $\mathrm{tr}(A)$ (multiplicity = 1).

4. Looks promising!

To galactus:
Are you sure lambda should be to the left in the matrix? My book says different. I guess you'r right, since you surely know better than me

I also wan't to know how you got the characteristic polynomial of the matrix. Did you use cofactor expansion? Or is there an easier way? (Please say there's an easier way.)

Anyway, the two answers upove seems intressting, since both characteristic polynomials and diagonalization is a part of my course. I will look at this tomorrow, and get back to you. Now I need some sleep...

Thanks a lot for the help!

Edit:
galactus, I now noticed that you also have the rows in different orders than me. What am I doing wrong here?

5. Finding eigenvalues of a 4x4 matrix.

I asked this question in the homework forum, but desided to move it here, since I'm not in a hurry.

My problem:
How do I find the eigenvalues of this matrix:

$
$
$\begin{bmatrix}1&-3&2&-1\\-3&9&-6&3\\2&-6&4&-2\\-1&3&-2&1\end{bmatrix}
$

(Notice that it is symmetric, if that helps)

As I understand I need to solve
${\lambda}$ from:

det $
\begin{bmatrix}1-{\lambda}&-3&2&-1\\-3&9-{\lambda}&-6&3\\2&-6&4-{\lambda}&-2\\-1&3&-2&1-{\lambda}\end{bmatrix}
$
= 0

The only way I know how to do this is by cofactor expansion. But that would take all day for me. There's got to be a faster way.

galactus said:

$\begin{bmatrix}{\lambda}-1&3&-2&1\\3&{\lambda}-9&6&-3\\-2&6&{\lambda}-4&2\\1&-3&2&{\lambda}-1\end{bmatrix}$

This gives a characteristic polynomial of:

${\lambda}^{4}-15{\lambda}^{3}={\lambda}^{3}({\lambda}-15)=0$ "

Why is galactus using a different matrix, and how did he get it? And how did he get the characteristic polynomial of that matrix?

I've been working on this problem for several days now. I just can't solve it. I would be so happy if anyone could help me.

/Gordina

6. Originally Posted by Gordina
I asked this question in the homework forum, but desided to move it here, since I'm not in a hurry.

My problem:
How do I find the eigenvalues of this matrix:

$
$
$\begin{bmatrix}1&-3&2&-1\\-3&9&-6&3\\2&-6&4&-2\\-1&3&-2&1\end{bmatrix}
$

(Notice that it is symmetric, if that helps)

As I understand I need to solve
${\lambda}$ from:

det $
\begin{bmatrix}1-{\lambda}&-3&2&-1\\-3&9-{\lambda}&-6&3\\2&-6&4-{\lambda}&-2\\-1&3&-2&1-{\lambda}\end{bmatrix}
$
= 0

The only way I know how to do this is by cofactor expansion. But that would take all day for me. There's got to be a faster way.

galactus said:

$\begin{bmatrix}{\lambda}-1&3&-2&1\\3&{\lambda}-9&6&-3\\-2&6&{\lambda}-4&2\\1&-3&2&{\lambda}-1\end{bmatrix}$

This gives a characteristic polynomial of:

${\lambda}^{4}-15{\lambda}^{3}={\lambda}^{3}({\lambda}-15)=0$ "

Why is galactus using a different matrix, and how did he get it? And how did he get the characteristic polynomial of that matrix?

I've been working on this problem for several days now. I just can't solve it. I would be so happy if anyone could help me.

/Gordina

You are using $det(A- \lambda I)= 0$ which is certainly what I would tend to do do. Galactus is using $det(\lambda I- A)= 0$, just multiplying the entire characteristice equation by -1. Of course, those two equations have the same solutions. Perhaps Galactus just doesn't like writing " $-\lambda$"! I suspect he got the characteristic equation by doing exactly what you said would "take all day"- expanding the determinant.