A similarity transformation of a matrix to a matrix is expressed as , where

 is an invertible matrix called the change of basis matrix.
 is an linear transformation matrix with respect to the basis .
 is the transformed representation of , such that performs the same linear transformation as but with respect to another basis .
Let a vector be with respect to the basis and with respect to the basis . Let another vector be with respect to the basis and with respect to the basis . Consider the transformation of these vectors as follows:
where the first two equations describe change of basis transformations and the last equation is a linear transformation of to in the same basis .
Combining the three equation, we have . If is invertible, we can multiply by on the left to give , where . Comparing and , is the transformed representation of , where performs the same linear transformation as but with respect to another basis . We say that is similar to because has properties that are similar to . For example, the trace of , which is defined as , is the same as the trace of .
To show that , we have,
where is the identity matrix and where we have used the identity for the second equality.
Question
Proof that .
Answer
The common properties of similar matrices are useful in explaining certain group theory concepts, e.g. why there are exactly 32 crystallographic point groups.
One of the most common applications of similarity transformations is to transform a matrix to a diagonal matrix . Consider the eigenvalue problem and let the eigenvectors of be the columns of :
Since , where , we have
If the eigenvectors of are linearly independent, then is nonsingular (i.e. invertible). This allows us to multiply on the left by to give , with the diagonal entries of being the eigenvalues of .
Question
Why is nonsingular if the eigenvectors of are linearly independent?
Answer
The eigenvectors are linearly independent if the only solution to is when for all . In other words,
or simply .
We need to show that the only solution to is and this is possible if is invertible such that
Using the same eigenvalue problem, we can show that Hermitian operators are diagonalisable, i.e. (see this article).
Question
Show that a Hermitian matrix can be diagonalised by , i.e. , where is a unitary matrix, and that is also Hermitian.
Answer
A unitary matrix has the property: . If a complete set of orthonormal eigenvectors of are the columns of , we have and because orthonormal eigenvectors are linearly independent. The remaining step is to show that .
For example, . Since is nonsingular, multiplying on the right of gives . So, .
To show that is also Hermitian, we have
.
As mentioned above, . The th column of is , while the th column of is . So, , where is an eigenvector of and is the corresponding eigenvalue. Therefore, the order of the columns of the change of basis matrix corresponds to the order of the diagonal entries in the diagonal matrix.