# Math Help - Find Eigenvalues

1. ## Find Eigenvalues

Given a matrix with components

$
(A^{\pm})_{ij} = \delta_{ij} \pm a_i a_j
$

where a is a constant real vector, find the eigenvalues of the matrix.

I can't really see where to start with this - any hints to see the obvious that I may have missed?

2. Originally Posted by thelostchild
Given a matrix with components

$
(A^{\pm})_{ij} = \delta_{ij} \pm a_i a_j
$

where a is a constant real vector, find the eigenvalues of the matrix.

I can't really see where to start with this - any hints to see the obvious that I may have missed?

What is $A^{\pm}\,,\,a_i\,,\,a_j$? I suppose $\delta_{ij}$ is Kronecker's delta, but for that...

Tonio

3. Originally Posted by thelostchild
Given a matrix with components

$
(A^{\pm})_{ij} = \delta_{ij} \pm a_i a_j
$

where a is a constant real vector, find the eigenvalues of the matrix.

I can't really see where to start with this - any hints to see the obvious that I may have missed?
$\delta_{ij} = 0$ for all $i \neq j$

So this is a diagonal matrix, that is if I have understood your notation correctly.

4. Sorry for the misunderstanding with notation
$A^{\pm}$ are two nxn matrices

and $a_i$ are the components of an n component real vector.

Just looking at it again I can see it can be written as

$A^{\pm} = I \pm a a^T$

If that helps, although I still cant see a way of doing this

5. The formula uses the Kronecker delta:

$\delta_{ij} =\left \{\begin{matrix} 1, & \mbox{if } i=j \\ 0, & \mbox{if } i \ne j \end{matrix}\right.$

That is a short way of writing

$(A^{\pm})_{ij}= \begin{bmatrix}a_{11}=1 & a_{12}=0 & ...& a_{1n}=0\\a_{21}=0 & a_{22}=1 & ... &a_{2n}=0 \\ ...&...&...&...\\a_{n1}=0 & a_{n2}=0 & ... & a_{nn}=1\end{bmatrix} \pm a_i a_j$

But it doesn't make a whole lot of sense.

Originally Posted by thelostchild
Sorry for the misunderstanding with notation
$A^{\pm}$ are two nxn matrices

and $a_i$ are the components of an n component real vector.

Just looking at it again I can see it can be written as

$A^{\pm} = I \pm a a^T$

If that helps, although I still cant see a way of doing this
Yes, if "a" is just an nxn matrix then you have to find the two resulting matrices here: $A^{\pm} = I \pm a a^T = B_1, B_2$ and then you solve the characteristic equation for this matrix ; i.e., $det (\lambda I - B_i)$ to find its eigenvalues.

6. I would recommend looking at "easy" cases to start with (always a good strategy).

For n= 1, this operator is just "1+ a" and its operation is just multiplying the number x: $(1+a)x= \lambda x$ for $\lambda= 1+ a$, of course.

For n= 2, this operator can be represented by the matrix $\begin{bmatrix}1+ a_1^2 & a_1a_2 \\ a_1a_2 & 1+ a_2^2\end{bmatrix}$. What are its eigenvalues?

For n= 3, this operator can be represented by the matrix $\begin{bmatrix}1+ a_1^2 & a_1a_2 & a_1a_3 \\ a_1a_2 & 1+ a_2^2 & a_2a_3 \\ a_1a_3 & a_2a_3 & 1+ a_3^2\end{bmatrix}$.
What are its eigenvalues?

7. Ok so looking at the results for those two cases I assume the general result is a repeated eigenvalue of n=1 for (n-1) times and then

$
\lambda=1 + \sum_{i=1}^{n} a_{i}^{2}
$

Thanks for the help

8. Originally Posted by thelostchild
Ok so looking at the results for those two cases I assume the general result is a repeated eigenvalue of n=1 for (n-1) times and then

$
\lambda=1 + \sum_{i=1}^{n} a_{i}^{2}
$

Thanks for the help
Well, I'm impressed! I suggested looking at simple cases as a general method of getting ideas. I hadn't actually solved the problem myself,

9. Originally Posted by thelostchild
Ok so looking at the results for those two cases I assume the general result is a repeated eigenvalue of n=1 for (n-1) times and then

$
\lambda=1 + \sum_{i=1}^{n} a_{i}^{2}
$

Thanks for the help

This doesn't seem to make sense: what are these $a_i's$ ?? The entries of a general n x n matrix are usually double indexed, so what are those elements whose squares' sum are you taking?

Tonio

10. Originally Posted by tonio
This doesn't seem to make sense: what are these $a_i's$ ?? The entries of a general n x n matrix are usually double indexed, so what are those elements whose squares' sum are you taking?

Tonio
Sorry if I wasn't clear above, a is an n component vector therefore it only has one index.