DIAGONALIZATION OF A SQUARE MATRIX

Just as the coordinate matrix of a vector depends on the basis chosen for the associated vector space, the matrix of a linear operator on a vector space depends on the basis for the vector space. It is often favorable to work with as simple a matrix as possible, for instance, diagonal matrix. The concepts of eigenvalues and eigenvectors assist us to select a basis for a vector space on which a linear operator is defined so that the linear operator has a diagonal matrix as its representation.

As described in Example 2 of the article First Encounter with Eigenvalues and Eigenvectors, the matrix $A = \begin{bmatrix}2 & -7 \\ -2 & -3 \end{bmatrix}$ has two eigenspaces E₁ and E₂, whose bases are $\begin{Bmatrix} \begin{bmatrix} 7 \\ -2 \end{bmatrix} \end{Bmatrix}$ and $\begin{Bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \end{Bmatrix}$ , respectively. The matrix A can be viewed as the matrix of a linear operator T : $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ defined by $T(\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}) = \begin{bmatrix} 2x_1 - 7x_2 \\ -2x_1 - 3x_2 \end{bmatrix}$ where the standard basis $\begin{Bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \: \begin{bmatrix} 0 \\ 1 \end{bmatrix} \end{Bmatrix}$ is chosen for $\mathbb{R}^2$ . In other words, A is the standard matrix for T. If we “change” the basis to a “new” basis whose members are the linearly independent eigenvectors that span the eigenspaces of A, i.e. $\begin{Bmatrix} \begin{bmatrix} 7 \\ -2 \end{bmatrix}, \: \begin{bmatrix} 1 \\ 1 \end{bmatrix} \end{Bmatrix}$ , then the matrix of T with respect to the “new” basis will be the diagonal matrix $A = \begin{bmatrix}4 & 0 \\ 0 & -5 \end{bmatrix}$ . This article shall focus on how to find the diagonal matrix and provide the theorem justifying the procedure.

The Diagonalization Theorem

An n×n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. In fact, A = PDP^-1, with D a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P.

Example 1

Is the matrix $A = \begin{bmatrix} 5 & 6 & 2 \\ 0 & -1 & -8 \\ 1 & 0 & -2 \end{bmatrix}$ diagonalizable?

Answer

As described in Example 3 of the article First Encounter with Eigenvalues and Eigenvectors, there are only 2 linearly independent eigenvectors while the matrix A is of order 3. By the Diagonalization Theorem above, it can be concluded that A is not diagonalizable.

Example 2

Consider the matrix $A = \begin{bmatrix}2 & -7 \\ -2 & -3 \end{bmatrix}$ . Find the matrix P such that P^-1AP is a diagonal matrix. Let the diagonal matrix be D. Verify that the diagonal entries of D are eigenvalues of A, conforming with the Diagonalization Theorem.

Answer

In Example 2 of the article First Encounter with Eigenvalues and Eigenvectors, it is shown that the basis vectors of the eigenspaces of A are $\begin{bmatrix} 7 \\ -2 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ . According to the theorem, P must be as follows.

$P = \begin{bmatrix}7 & 1 \\ -2 & 1 \end{bmatrix}$ .

The matrix P will diagonalize A.

To verify that D = P^-1AP is a diagonal matrix, first determine P^-1. It is a simple matter to show that $P^{-1} = \frac{1}{9} \begin{bmatrix} 1 & -1 \\ 2 & 7 \end{bmatrix}$ . It follows that:

$D = P^{-1} AP = \frac{1}{9} \begin{bmatrix} 1 & -1 \\ 2 & 7 \end{bmatrix} \begin{bmatrix}2 & -7 \\ -2 & -3 \end{bmatrix} \begin{bmatrix}7 & 1 \\ -2 & 1 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 36 & 0 \\ 0 & -45 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & -5 \end{bmatrix}$ .

The diagonal entries of D is 4 and -5. They are the eigenvalues of A. (See Example 2 of the article mentioned above.)

Example 3

Find a matrix P that diagonalizes $A = \begin{bmatrix}2 & 1 & 1 \\ 0 & -4 & -2 \\ 0 & 0 & 3 \end{bmatrix}$ . Then, determine diagonal matrix produced by P.

Answer

$A - \lambda I = \begin{bmatrix} 2 - \lambda & 1 & 1 \\ 0 & -4 - \lambda & -2 \\ 0 & 0 & 3 - \lambda \end{bmatrix}$

Since (A – λI) is a triangular matrix, its determinant is the product of all its diagonal entries. Thus, the characteristic equation of A is (2-λ)(-4-λ)(3-λ) = 0. It follows that the eigenvalues are 2, -4, and 3. [It is consistent with a theorem that says: “The eigenvalues of a triangular matrix are the entries on the main diagonal.”]. We proceed to find the matrix P that diagonalizes A. The Diagonalization Theorem implies that we have to find the eigenvectors of A.

If λ = 2 then $A - \lambda I = \begin{bmatrix} 0 & 1 & 1 \\ 0 & -6 & -2 \\ 0 & 0 & 1 \end{bmatrix}$ , which has the same solution set as $\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$ . (Stated another way, the two matrices are equivalent.) If $\vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ , then the matrix will correspond to the system of linear equations represented as $\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ . The only free variable in this system is x₁. Therefore, we set x₁ = t where $t \in \mathbb{R}$ . This yields the eigenspace $E_1 = \begin{Bmatrix} t \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \Bigr| t \in \mathbb{R} \end{Bmatrix}$ .

If λ = -4 then $A - \lambda I = \begin{bmatrix} 6 & 1 & 1 \\ 0 & 0 & -2 \\ 0 & 0 & 7 \end{bmatrix}$ , which is equivalent to $\begin{bmatrix} 1 & 1/6 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$ . If $\vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ , then the matrix will correspond to the system of linear equations represented as $\begin{bmatrix} 1 & 1/6 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ . The only free variable in this system is x₂. To solve the system of equations, set x₂ = -6t where $t \in \mathbb{R}$ . Consequently, x₁ = t and the resulting eigenspace is $E_2 = \begin{Bmatrix} t \begin{bmatrix} 1 \\ -6 \\ 0 \end{bmatrix} \Bigr| t \in \mathbb{R} \end{Bmatrix}$ .

If λ = 3 then $A - \lambda I = \begin{bmatrix} -1 & 1 & 1 \\ 0 & -7 & -2 \\ 0 & 0 & 0 \end{bmatrix}$ , which is equivalent to $\begin{bmatrix} 1 & 0 & -5/7 \\ 0 & 1 & 2/7 \\ 0 & 0 & 0 \end{bmatrix}$ . If $\vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ , then the matrix will correspond to the system of linear equations represented as $\begin{bmatrix} 1 & 0 & -5/7 \\ 0 & 1 & 2/7 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ . The only free variable in this system is x₃. To solve the system of equations, set x₃ = 7t where $t \in \mathbb{R}$ . Accordingly, x₁ = 5t, x₂ = -2t and the eigenspace produced is $E_3 = \begin{Bmatrix} t \begin{bmatrix} 5 \\ -2 \\ 7 \end{bmatrix} \Bigr| t \in \mathbb{R} \end{Bmatrix}$ .

Note that the three basis vectors spanning E₁, E₂, and E₃, i.e. $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \: \begin{bmatrix} 1 \\ -6 \\ 0 \end{bmatrix}, \: \begin{bmatrix} 5 \\ -2 \\ 7 \end{bmatrix}$ , are linearly independent. (See the additional note at the end of this article.) The Diagonalization Theorem implies that the diagonalizing matrix is $P = \begin{bmatrix} 1 & 1 & 5 \\ 0 & -6 & -2 \\ 0 & 0 & 7 \end{bmatrix}$ and P yields the diagonal matrix D whose diagonal entries are the eigenvalues corresponding to the eigenvectors in P. Thus, the diagonal matrix is $D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ .

Additional Note

Theorem

If $\vec{v}_1, \vec{v}_2, \cdots , \vec{v}_k$ are eigenvectors of A corresponding to distinct eigenvalues λ₁, λ₂, …, λ_k, then $\begin{Bmatrix} \vec{v}_1, \vec{v}_2, \cdots , \vec{v}_k \end{Bmatrix}$ is a linearly independent set.

On 25/04/2022

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