Advanced Linear Algebra: Foundations to Frontiers

Subsection 9.2.5 Diagonalizing a matrix

The algebraic eigenvalue problem or, more simply, the computation of eigenvalues and eigenvectors is often presented as the problem of diagonalizing a matrix. We make that link in this unit.

Why is this important? Consider the equality \(w = A v.\) Notice that we can write \(w \) as a linear combination of the columns of \(X \text{:}\)

In other words, \(X^{-1} w \) is the vector of coefficients when \(w \) is writen in terms of the basis that consists of the columns of \(X \text{.}\) Similarly, we can write \(v \) as a linear combination of the columns of \(X \text{:}\)

Now, since \(X \) is nonsingular, \(w = A v\) is equivalent to \(X^{-1} w = X^{-1} A X X^{-1} v ,\) and hence \(X^{-1} w = D ( X^{-1} v ).\)

How is diagonalizing a matrix related to eigenvalues and eigenvectors? Let's assume that \(X^{-1} A X = D \text{.}\) We can rewrite this as

and partition

Multiplying this out yields

We conclude that

which means that the entries on the diagonal of \(D \) are the eigenvalues of \(A \) and the corresponding eigenvectors are found as columns of \(X \text{.}\)

Now assume that the eigenvalues of \(A \in \Cmxm \) are given by \(\{ \lambda_0, \lambda_1, \ldots , \lambda_{m-1} \} \text{,}\) where eigenvalues are repeated according to their algebraic multiplicity. Assume that there are \(m \) linearly independent vectors \(\{ x_0, x_1, \ldots, x_{m-1}\}\) such that \(A x_j = \lambda_j x_j \text{.}\) Then

Hence, if \(X = \left( \begin{array}{c | c | c | c } x_0 \amp x_1 \amp \cdots \amp x_{m-1} \end{array} \right) \) and \(D = \diag{ \lambda_0, \lambda_1, \ldots, \lambda_{m-1} } \) then \(X^{-1} A X = D \text{.}\) In other words, if \(A \) has \(m \) linearly independent eigenvectors, then \(A\) is diagonalizable.

These insights are summarized in the following theorem:

Here are some classes of matrices that are diagonalizable:

• Diagonal matrices.

  If \(A \) is diagonal, then choosing \(X = I \) and \(A = D \) yields \(X^{-1} A X= D \text{.}\)

• Hermitian matrices.

  If \(A \) is Hermitian, then the spectral decomposition theorem tells us that there exists unitary matrix \(Q \) and diagonal matrix \(D \) such that \(A = Q D Q^H \text{.}\) Choosing \(X = Q \) yields \(X^{-1} A X= D \text{.}\)

• Triangular matrices with distinct diagonal elements.

  If \(U \) is upper triangular and has distinct diagonal elements, then by Homework 9.2.3.4 we know we can find an eigenvector associated with each diagonal element and by design those eigenvectors are linearly independent. Obviously, this can be extended to lower triangular matrices as well.