References: edX online course MIT 8.05 Week 6.
Suppose we have an operator that has a matrix representation in some basis . In some cases (not all!) it is possible to transform to a different basis in which is a diagonal matrix. If the operator transforms from the basis to , then we’ve seen that transforms according to
The diagonalization problem is therefore to find the matrix (if it exists).
Assuming does exist, we can look at the situation in two ways.
- The matrix representation of is diagonal in the basis.
- The operator is diagonal in the original basis.
To prove (2), we start with the fact that is diagonal in the basis, so that (no implied sums in what follows):
From the last line, we see that is an eigenvector of the operator with the same eigenvalue as the eigenvector of the operator .
Thus we can write, in the original basis , the equation
where is the diagonalized version of the matrix . This is known as a similarity transformation.
All this is fine, but we still haven’t seen how to find . It turns out that the columns of are the eigenvectors of . We can see this from the following argument.
In the basis , each has the form of a column vector with a 1 in the th position and zeroes everywhere else. Thus the transformation to the basis can be written as
The 1 is in the th position in the column vector representing and picks out the elements in column of in the product . Since is the th eigenvector of , the columns of are the eigenvectors of .
Example Suppose we have
The eigenvalues are found in the usual way, from the characteristic determinant, which is
We can find the eigenvectors by solving for each eigenvalue, where is a 2-element column vector. (I won’t go through this, since it’s just algebra.) Placing the eigenvectors as the columns in a matrix , we have
The inverse of a matrix is
So for our matrix, we have
Doing the matrix products (just a lot of arithmetic) we get
The resulting matrix is diagonal, and the diagonal entries are the eigenvalues of . It’s worth noticing that the traces of and are the same (both = 2), as are the determinants (both ).