In quantum physics, if you’re given an operator in matrix form, you can find its eigenvectors and eigenvalues. For example, say you need to solve the following equation:

First, you can rewrite this equation as the following:

I represents the identity matrix, with 1s along its diagonal and 0s otherwise:

Remember that the solution to

exists only if the determinant of the matrix A – *a*I is 0:

det(A – *a*I) = 0

## How to find the eigenvalues

Any values of *a* that satisfy the equation det(A – *a*I) = 0 are eigenvalues of the original equation. Try to find the eigenvalues and eigenvectors of the following matrix:

First, convert the matrix into the form A – *a*I:

Next, find the determinant:

And this can be factored as follows:

You know that det(A – *a*I) = 0, so the eigenvalues of A are the roots of this equation; namely, *a*_{1} = –2 and *a*_{2} = –3.

## How to find the eigenvectors

How about finding the eigenvectors? To find the eigenvector corresponding to *a*_{1}, substitute *a*_{1} — the first eigenvalue, –2 — into the matrix in the form A – *a*I:

So you have

Because every row of this matrix equation must be true, you know that

And that means that, up to an arbitrary constant, the eigenvector corresponding to *a*_{1} is the following:

Drop the arbitrary constant, and just write this as a matrix:

How about the eigenvector corresponding to *a*_{2}? Plugging *a*_{2}, –3, into the matrix in A –*a*I form, you get the following:

Then you have

And that means that, up to an arbitrary constant, the eigenvector corresponding to *a*_{2} is

Drop the arbitrary constant:

So the eigenvalues of this matrix operator

are *a*_{1} = –2 and *a*_{2} = –3. And the eigenvector corresponding to *a*_{1} is

The eigenvector corresponding to *a*_{2} is