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6.2 Diagonalization of Matrices

Given an $n \times n$ $n \times n$ matrix A, we may ask how many linearly independent eigenvectors the matrix A has. In Section 6.1, we saw several examples (with $n = 2$ $n = 2$ and $n = 3$ $n = 3$ ) in which the $n \times n$ $n \times n$ matrix A has n linearly independent eigenvectors—the largest possible number. By contrast, in Example 5 of Section 6.1, we saw that the $2 \times 2$ $2 \times 2$ matrix

A = [\begin{array}{r} 2 & 3 \\ 0 & 2 \end{array}]

$A = [\begin{array}{r} 2 & 3 \\ 0 & 2 \end{array}]$

has the single eigenvalue $λ = 2$ $λ = 2$ corresponding to the single eigenvector $v = {[\begin{array}{r} 1 & 0 \end{array}]}^{T} .$ $v = {[\begin{array}{r} 1 & 0 \end{array}]}^{T} .$

Something very nice happens when the $n \times n$ $n \times n$ matrix A does have n linearly independent eigenvectors. Suppose that the eigenvalues $λ_{1}, λ_{2}, \dots, λ_{n}$ $λ_{1}, λ_{2}, \dots, λ_{n}$ (not necessarily distinct) of A correspond to the n linearly independent eigenvectors $v_{1}, v_{2}, \dots, v_{n}$ $v_{1}, v_{2}, \dots, v_{n}$ , respectively. Let

P = [\begin{array}{c} | & | & | \\ v_{1} & v_{2} & \dots & v_{n} \\ | & | & | \end{array}]

$P = [\begin{array}{c} | & | & | \\ v_{1} & v_{2} & \dots & v_{n} \\ | & | & | \end{array}]$ (1)

be the $n \times n$ $n \times n$ matrix having these eigenvectors as its column vectors. Then

A P = A [\begin{array}{c} | & | & | \\ v_{1} & v_{2} & \dots & v_{n} \\ | & | & | \end{array}] = [\begin{array}{c} | & | & | \\ A v_{2} & A v_{2} & \dots & A v_{n} \\ | & | & | \end{array}]

$A P = A [\begin{array}{c} | & | & | \\ v_{1} & v_{2} & \dots & v_{n} \\ | & | & | \end{array}] = [\begin{array}{c} | & | & | \\ A v_{2} & A v_{2} & \dots & A v_{n} \\ | & | & | \end{array}]$

and hence

A P = [\begin{array}{c} | & | & | \\ λ_{1} v_{1} & λ_{2} v_{2} & \dots & λ_{n} v_{n} \\ | & | & | \end{array}],

$A P = [\begin{array}{c} | & | & | \\ λ_{1} v_{1} & λ_{2} v_{2} & \dots & λ_{n} v_{n} \\ | & | & | \end{array}],$ (2)

because ${Av}_{j} = λ_{j} v_{j}$ ${Av}_{j} = λ_{j} v_{j}$ for each $j = 1, 2, \dots, n$ $j = 1, 2, \dots, n$ . Thus the product matrix AP has column vectors $λ_{1} v_{1}, λ_{2} v_{2}, \dots, λ_{n} v_{n}$ $λ_{1} v_{1}, λ_{2} v_{2}, \dots, λ_{n} v_{n}$ .

Now consider the diagonal matrix

D = [\begin{array}{c} λ_{1} & 0 & \dots & 0 \\ 0 & λ_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & λ_{n} \end{array}],

$D = [\begin{array}{c} λ_{1} & 0 & \dots & 0 \\ 0 & λ_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & λ_{n} \end{array}],$ (3)

whose diagonal elements are the eigenvalues corresponding (in the same order) to the eigenvectors forming the columns of P. Then

\begin{array}{rcl} P D & = & [\begin{array}{c} | & | & | \\ v_{1} & v_{2} & \dots & v_{n} \\ | & | & | \end{array}] [\begin{array}{c} λ_{1} & 0 & \dots & 0 \\ 0 & λ_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & λ_{n} \end{array}] \\ = & [\begin{array}{c} | & | & | \\ λ_{1} v_{2} & λ_{2} v_{2} & \dots & λ_{n} v_{n} \\ | & | & | \end{array}], \end{array}

$\begin{array}{rcl} P D & = & [\begin{array}{c} | & | & | \\ v_{1} & v_{2} & \dots & v_{n} \\ | & | & | \end{array}] [\begin{array}{c} λ_{1} & 0 & \dots & 0 \\ 0 & λ_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & λ_{n} \end{array}] \\ = & [\begin{array}{c} | & | & | \\ λ_{1} v_{2} & λ_{2} v_{2} & \dots & λ_{n} v_{n} \\ | & | & | \end{array}], \end{array}$ (4)

because the product of the ith row of P and the jth column of D is simply the product of $λ_{j}$ $λ_{j}$ and the ith component of $v_{j}$ $v_{j}$ .

Finally, upon comparing the results in (2) and (4), we see that

A P = P D .

$A P = P D .$ (5)

But the matrix P is invertible, because its n column vectors are linearly independent. So we may multiply on the right by $P^{- 1}$ $P^{- 1}$ to obtain

A = P D P^{- 1} .

$A = P D P^{- 1} .$ (6)

Equation (6) expresses the $n \times n$ $n \times n$ matrix A having n linearly independent eigenvectors in terms of the eigenvector matrix P and the diagonal eigenvalue matrix D. It can be rewritten as $D = P^{- 1} A P$ $D = P^{- 1} A P$ , but the form in (6) is the one that should be memorized.

Example 1

In Example 1 of Section 6.1 we saw that the matrix

A = [\begin{array}{r} 5 & - 6 \\ 2 & - 2 \end{array}]

$A = [\begin{array}{r} 5 & - 6 \\ 2 & - 2 \end{array}]$

has eigenvalues $λ_{1} = 2$ $λ_{1} = 2$ and $λ_{2} = 1$ $λ_{2} = 1$ corresponding to the linearly independent eigenvectors $v_{1} = {[\begin{array}{l} 2 & 1 \end{array}]}^{T}$ $v_{1} = {[\begin{array}{l} 2 & 1 \end{array}]}^{T}$ and $v_{2} = {[\begin{array}{l} 3 & 2 \end{array}]}^{T},$ $v_{2} = {[\begin{array}{l} 3 & 2 \end{array}]}^{T},$ respectively. Then

P = [\begin{array}{r} 2 & 3 \\ 1 & 2 \end{array}], D = [\begin{array}{r} 2 & 0 \\ 0 & 1 \end{array}], and P^{- 1} = [\begin{array}{r} 2 & - 3 \\ - 1 & 2 \end{array}] .

$P = [\begin{array}{r} 2 & 3 \\ 1 & 2 \end{array}], D = [\begin{array}{r} 2 & 0 \\ 0 & 1 \end{array}], and P^{- 1} = [\begin{array}{r} 2 & - 3 \\ - 1 & 2 \end{array}] .$

\begin{array}{rcl} P D P^{- 1} & = & [\begin{array}{r} 2 & 3 \\ 1 & 2 \end{array}] [\begin{array}{r} 2 & 0 \\ 0 & 1 \end{array}] [\begin{array}{r} 2 & - 3 \\ - 1 & 2 \end{array}] \\ = & [\begin{array}{r} 4 & 3 \\ 2 & 2 \end{array}] [\begin{array}{r} 2 & - 3 \\ - 1 & 2 \end{array}] = [\begin{array}{r} 5 & - 6 \\ 2 & - 2 \end{array}] = A, \end{array}

$\begin{array}{rcl} P D P^{- 1} & = & [\begin{array}{r} 2 & 3 \\ 1 & 2 \end{array}] [\begin{array}{r} 2 & 0 \\ 0 & 1 \end{array}] [\begin{array}{r} 2 & - 3 \\ - 1 & 2 \end{array}] \\ = & [\begin{array}{r} 4 & 3 \\ 2 & 2 \end{array}] [\begin{array}{r} 2 & - 3 \\ - 1 & 2 \end{array}] = [\begin{array}{r} 5 & - 6 \\ 2 & - 2 \end{array}] = A, \end{array}$

in accord with Eq. (6).

Similarity and Diagonalization

The following definition embodies the precise relationship in (6) between the original matrix A and the diagonal matrix D.

Note that this relationship between A and B is symmetric, for if $B = P^{- 1} A P$ $B = P^{- 1} A P$ , then $A = Q^{- 1} B Q$ $A = Q^{- 1} B Q$ for some invertible matrix Q—just take $Q = P^{- 1}$ $Q = P^{- 1}$ .

An $n \times n$ $n \times n$ matrix A is called diagonalizable if it is similar to a diagonal matrix D; that is, there exist a diagonal matrix D and an invertible matrix P such that $A = P D P^{- 1}$ $A = P D P^{- 1}$ , and so

P^{- 1} A P = D .

$P^{- 1} A P = D .$ (8)

The process of finding the diagonalizing matrix P and the diagonal matrix D in (8) is called diagonalization of the matrix A. In Example 1 we showed that the matrices

A = [\begin{array}{l} 5 & - 6 \\ 2 & - 2 \end{array}] and D = [\begin{array}{l} 2 & 0 \\ 0 & 1 \end{array}]

$A = [\begin{array}{l} 5 & - 6 \\ 2 & - 2 \end{array}] and D = [\begin{array}{l} 2 & 0 \\ 0 & 1 \end{array}]$

are similar, and hence that the $2 \times 2$ $2 \times 2$ matrix A is diagonalizable.

Now we ask under what conditions a given square matrix is diagonalizable. In deriving Eq. (6), we showed that if the $n \times n$ $n \times n$ matrix A has n linearly independent eigenvectors, then A is diagonalizable. The converse of this statement is also true.

Proof

It remains only to show that, if the $n \times n$ $n \times n$ matrix A is diagonalizable, then it has n linearly independent eigenvectors. Suppose that A is similar to the diagonal matrix D with diagonal elements $d_{1}, d_{2}, \dots, d_{n}$ $d_{1}, d_{2}, \dots, d_{n}$ , and let

P = [v_{1} v_{2} \dots v_{n}]

$P = [v_{1} v_{2} \dots v_{n}]$

be an invertible matrix such that $D = P^{- 1} A P$ $D = P^{- 1} A P$ . Then

A P = A [v_{1} v_{2} \dots v_{n}] = [A v_{1} A v_{2} \dots A v_{n}]

$A P = A [v_{1} v_{2} \dots v_{n}] = [A v_{1} A v_{2} \dots A v_{n}]$

and

P D = [d_{1} v_{1} d_{2} v_{2} \dots d_{n} v_{n}],

$P D = [d_{1} v_{1} d_{2} v_{2} \dots d_{n} v_{n}],$

by essentially the same computation as in Eq. (4). But $A P = P D$ $A P = P D$ because $D = P^{- 1} A P$ $D = P^{- 1} A P$ , so it follows that

A v_{j} = d_{j} v_{j}

$A v_{j} = d_{j} v_{j}$

for each $j = 1, 2, \dots, n$ $j = 1, 2, \dots, n$ . Thus the vectors $v_{1}, v_{2}, \dots, v_{n}$ $v_{1}, v_{2}, \dots, v_{n}$ are eigenvectors of A associated with the eigenvalues $d_{1}, d_{2}, \dots, d_{n}$ $d_{1}, d_{2}, \dots, d_{n}$ , respectively. And it follows from Theorem 2 in Section 3.6 and Theorem 2 in Section 4.3 that these n eigenvectors of the matrix A are linearly independent, because they are the column vectors of the invertible matrix P.

Remark

It is important to remember not only the fact that an $n \times n$ $n \times n$ matrix A having n linearly independent eigenvectors is diagonalizable, but also the specific diagonalization $A = P D P^{- 1}$ $A = P D P^{- 1}$ in Eq. (6), where the matrix P has the n eigenvectors as its columns, and the corresponding eigenvalues are the diagonal elements of the diagonal matrix D.

Example 2

In Example 5 of Section 6.1 we saw that the matrix

A = [\begin{array}{l} 2 & 3 \\ 0 & 2 \end{array}]

$A = [\begin{array}{l} 2 & 3 \\ 0 & 2 \end{array}]$

has only one eigenvalue, $λ = 2,$ $λ = 2,$ and that (to within a constant multiple) only the single eigenvector $v = {[1 0]}^{T}$ $v = {[1 0]}^{T}$ is associated with this eigenvalue. Thus the $2 \times 2$ $2 \times 2$ matrix A does not have $n = 2$ $n = 2$ linearly independent eigenvectors. Hence Theorem 1 implies that A is not diagonalizable.

Example 3

In Example 6 of Section 6.1 we saw that the matrix

A = [\begin{array}{r} 3 & 0 & 0 \\ - 4 & 6 & 2 \\ 16 & - 15 & - 5 \end{array}]

$A = [\begin{array}{r} 3 & 0 & 0 \\ - 4 & 6 & 2 \\ 16 & - 15 & - 5 \end{array}]$

has the following eigenvalues and associated eigenvectors:

\begin{array}{l} λ_{1} = 3 : v_{1} = {[1 0 2]}^{T} \\ λ_{1} = 1 : v_{2} = {[0 2 - 5]}^{T} \\ λ_{1} = 0 : v_{3} = {[0 1 - 3]}^{T} . \end{array}

$\begin{array}{l} λ_{1} = 3 : v_{1} = {[1 0 2]}^{T} \\ λ_{1} = 1 : v_{2} = {[0 2 - 5]}^{T} \\ λ_{1} = 0 : v_{3} = {[0 1 - 3]}^{T} . \end{array}$

It is obvious (why?) that the three eigenvectors $v_{1}, v_{2}, v_{3}$ $v_{1}, v_{2}, v_{3}$ are linearly independent, so Theorem 1 implies that the $3 \times 3$ $3 \times 3$ matrix A is diagonalizable. In particular, the inverse of the eigenvector matrix

P = [v_{1} v_{2} v_{3}] = [\begin{array}{r} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 2 & - 5 & - 3 \end{array}]

$P = [v_{1} v_{2} v_{3}] = [\begin{array}{r} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 2 & - 5 & - 3 \end{array}]$

P^{- 1} = [\begin{array}{r} 1 & 0 & 0 \\ - 2 & 3 & 1 \\ 4 & - 5 & - 2 \end{array}],

$P^{- 1} = [\begin{array}{r} 1 & 0 & 0 \\ - 2 & 3 & 1 \\ 4 & - 5 & - 2 \end{array}],$

and the diagonal eigenvalue matrix is

D = [\begin{array}{r} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{array}] = [\begin{array}{l} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}] .

$D = [\begin{array}{r} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{array}] = [\begin{array}{l} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}] .$

Therefore, Eq. (6) in the form $P^{- 1} A P = D$ $P^{- 1} A P = D$ yields the diagonalization

\begin{array}{rcl} P^{- 1} AP & = & [\begin{array}{r} 1 & 0 & 0 \\ - 2 & 3 & 1 \\ 4 & - 5 & - 2 \end{array}] [\begin{array}{r} 3 & 0 & 0 \\ - 4 & 6 & 2 \\ 16 & - 15 & - 5 \end{array}] [\begin{array}{r} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 2 & - 5 & - 3 \end{array}] \\ = & [\begin{array}{r} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}] \end{array}

$\begin{array}{rcl} P^{- 1} AP & = & [\begin{array}{r} 1 & 0 & 0 \\ - 2 & 3 & 1 \\ 4 & - 5 & - 2 \end{array}] [\begin{array}{r} 3 & 0 & 0 \\ - 4 & 6 & 2 \\ 16 & - 15 & - 5 \end{array}] [\begin{array}{r} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 2 & - 5 & - 3 \end{array}] \\ = & [\begin{array}{r} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}] \end{array}$

of the matrix A.

The following theorem tells us that any set of eigenvectors associated with distinct eigenvalues (as in Example 3) is automatically linearly independent.

Proof

Our proof will be by induction on k. The theorem is certainly true in the case $k = 1$ $k = 1$ , because any single (nonzero) eigenvector constitutes a linearly independent set. Now assume inductively that any set of $k - 1$ $k - 1$ eigenvectors associated with distinct eigenvalues is linearly independent. Supposing that

c_{1} v_{1} + c_{2} v_{2} + \dots + c_{k} v_{k} = 0,

$c_{1} v_{1} + c_{2} v_{2} + \dots + c_{k} v_{k} = 0,$ (9)

we need to show that $c_{1} = c_{2} = \dots = c_{k} = 0$ $c_{1} = c_{2} = \dots = c_{k} = 0$ . To do this, we will multiply in Eq. (9) by the matrix $A - λ_{1} I$ $A - λ_{1} I$ . First note that

(A - λ_{1} I) v_{j} = A v_{j} - λ_{1} v_{j} = {\begin{array}{l} 0 & if j = 1, \\ (λ_{j} - λ_{1}) v_{j} & if j > 1, \end{array}

$(A - λ_{1} I) v_{j} = A v_{j} - λ_{1} v_{j} = {\begin{array}{l} 0 & if j = 1, \\ (λ_{j} - λ_{1}) v_{j} & if j > 1, \end{array}$

because ${Av}_{j} = λ_{j} v_{j}$ ${Av}_{j} = λ_{j} v_{j}$ for each j. Therefore, the result of multiplying Eq. (9) by $A - λ_{1} I$ $A - λ_{1} I$ is

c_{2} (λ_{2} - λ_{1}) v_{2} + \dots + c_{k} (λ_{k} - λ_{1}) v_{k} = 0 .

$c_{2} (λ_{2} - λ_{1}) v_{2} + \dots + c_{k} (λ_{k} - λ_{1}) v_{k} = 0 .$ (10)

But the $k - 1$ $k - 1$ eigenvectors $v_{2}, v_{3}, \dots, v_{k}$ $v_{2}, v_{3}, \dots, v_{k}$ are linearly independent by the inductive assumption, so each of the scalar coefficients $c_{j} (λ_{j} - λ_{1})$ $c_{j} (λ_{j} - λ_{1})$ here must be zero. Now our hypothesis that the eigenvalues of A are distinct implies that $λ_{j} - λ_{1} \neq 0$ $λ_{j} - λ_{1} \neq 0$ for each $j > 1$ $j > 1$ . It therefore follows from Eq. (10) that $c_{2} = c_{3} = \dots = c_{k} = 0$ $c_{2} = c_{3} = \dots = c_{k} = 0$ . But then Eq. (9) reduces to $c_{1} v_{1} = 0$ $c_{1} v_{1} = 0$ , so it now follows (because $v_{1} \neq 0$ $v_{1} \neq 0$ ) that $c_{1} = 0$ $c_{1} = 0$ as well. Thus we have shown that all the coefficients in Eq. (9) must vanish, and hence that the k eigenvectors $v_{1}, v_{2}, \dots, v_{k}$ $v_{1}, v_{2}, \dots, v_{k}$ are linearly independent. Theorem 2 now follows by induction.

If the $n \times n$ $n \times n$ matrix A has n distinct eigenvalues, then by Theorem 2 the n associated eigenvectors are linearly independent, so Theorem 1 implies that the matrix A is diagonalizable. Thus we have the sufficient condition for diagonalizability given in Theorem 3.

In general, however, an $n \times n$ $n \times n$ matrix A can be expected to have fewer than n distinct eigenvalues $λ_{1}, λ_{2}, \dots, λ_{k}$ $λ_{1}, λ_{2}, \dots, λ_{k}$ . If $k < n$ $k < n$ , then we may attempt to diagonalize A by carrying out the following procedure.

Step 1. Find a basis $S_{i}$ $S_{i}$ for the eigenspace associated with each eigenvalue $λ_{i}$ $λ_{i}$ .
Step 2. Form the union S of the bases $S_{1}, S_{2}, \dots, S_{k}$ $S_{1}, S_{2}, \dots, S_{k}$ . According to Theorem 4 in this section, the set S of eigenvectors of A is linearly independent.
Step 3. If S contains n eigenvectors $v_{1}, v_{2}, \dots, v_{n}$ $v_{1}, v_{2}, \dots, v_{n}$ , then the matrix

$P = [v_{1} v_{2} \dots v_{n}]$ $P = [v_{1} v_{2} \dots v_{n}]$

diagonalizes A: that is, $P^{- 1} A P = D$ $P^{- 1} A P = D$ , where the diagonal elements of D are the eigenvalues (repeated as necessary) corresponding to the n eigenvectors $v_{1}, v_{2}, \dots, v_{n}$ $v_{1}, v_{2}, \dots, v_{n}$ .

If the set S—obtained by “merging” the bases for all the eigenspaces of A—contains fewer than n eigenvectors, then it can be proved that the matrix A is not diagonalizable.

Example 4

In Example 7 of Section 6.1, we saw that the matrix

A = [\begin{array}{r} 4 & - 2 & 1 \\ 2 & 0 & 1 \\ 2 & - 2 & 3 \end{array}]

$A = [\begin{array}{r} 4 & - 2 & 1 \\ 2 & 0 & 1 \\ 2 & - 2 & 3 \end{array}]$

has only two distinct eigenvalues, $λ_{1} = 2$ $λ_{1} = 2$ and $λ_{2} = 3$ $λ_{2} = 3$ . We found that the eigenvalue $λ_{1} = 2$ $λ_{1} = 2$ corresponds to a 2-dimensional eigenspace with basis vectors $v_{1} = {[1 1 0]}^{T}$ $v_{1} = {[1 1 0]}^{T}$ and $v_{2} = {[- 1 0 2]}^{T},$ $v_{2} = {[- 1 0 2]}^{T},$ and that $λ_{2} = 3$ $λ_{2} = 3$ corresponds to a 1-dimensional eigenspace with basis vector $v_{3} = {[1 1 1]}^{T} .$ $v_{3} = {[1 1 1]}^{T} .$ By Theorem 4 (or by explicit verification), these three eigenvectors are linearly independent, so Theorem 1 implies that the $3 \times 3$ $3 \times 3$ matrix A is diagonalizable. The diagonalizing matrix

P = [v_{1} v_{2} v_{3}] = [\begin{array}{r} 1 & - 1 & 1 \\ 1 & 0 & 1 \\ 0 & 2 & 1 \end{array}]

$P = [v_{1} v_{2} v_{3}] = [\begin{array}{r} 1 & - 1 & 1 \\ 1 & 0 & 1 \\ 0 & 2 & 1 \end{array}]$

has inverse matrix

P^{- 1} = [\begin{array}{r} - 2 & 3 & - 1 \\ - 1 & 1 & 0 \\ 2 & - 2 & 1 \end{array}],

$P^{- 1} = [\begin{array}{r} - 2 & 3 & - 1 \\ - 1 & 1 & 0 \\ 2 & - 2 & 1 \end{array}],$

so we obtain the diagonalization

\begin{array}{rcl} P^{- 1} A P & = & [\begin{array}{r} - 2 & 3 & - 1 \\ - 1 & 1 & 0 \\ 2 & - 2 & 1 \end{array}] [\begin{array}{r} 4 & - 2 & 1 \\ 2 & 0 & 1 \\ 2 & - 2 & 3 \end{array}] [\begin{array}{r} 1 & - 1 & 1 \\ 1 & 0 & 1 \\ 0 & 2 & 1 \end{array}] \\ = & [\begin{array}{l} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}] = D \end{array}

$\begin{array}{rcl} P^{- 1} A P & = & [\begin{array}{r} - 2 & 3 & - 1 \\ - 1 & 1 & 0 \\ 2 & - 2 & 1 \end{array}] [\begin{array}{r} 4 & - 2 & 1 \\ 2 & 0 & 1 \\ 2 & - 2 & 3 \end{array}] [\begin{array}{r} 1 & - 1 & 1 \\ 1 & 0 & 1 \\ 0 & 2 & 1 \end{array}] \\ = & [\begin{array}{l} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}] = D \end{array}$

of the matrix A.

Proof

To simplify the notation, we will illustrate the proof with the typical case $k = 3$ $k = 3$ , with A having three distinct eigenvalues $λ_{1}, λ_{2}$ $λ_{1}, λ_{2}$ , and $λ_{3}$ $λ_{3}$ . Let

\begin{array}{l} S_{1} = {u_{1}, u_{2}, \dots, u_{p}}, \\ S_{2} = {v_{1}, v_{2}, \dots, v_{q}}, and \\ S_{3} = {w_{1}, w_{2}, \dots, w_{r}} \end{array}

$\begin{array}{l} S_{1} = {u_{1}, u_{2}, \dots, u_{p}}, \\ S_{2} = {v_{1}, v_{2}, \dots, v_{q}}, and \\ S_{3} = {w_{1}, w_{2}, \dots, w_{r}} \end{array}$

be bases for the eigenspaces associated with the eigenvalues $λ_{1}, λ_{2}$ $λ_{1}, λ_{2}$ , and $λ_{3}$ $λ_{3}$ . Assuming that a linear combination of the vectors in $S = S_{1} \cup S_{2} \cup S_{3}$ $S = S_{1} \cup S_{2} \cup S_{3}$ vanishes—

\begin{array}{l} a_{1} u_{1} + a_{2} u_{2} + \dots + a_{p} u_{p} \\ + b_{1} v_{1} + b_{2} v_{2} + \dots + b_{q} v_{q} \\ + c_{1} w_{1} + c_{2} w_{2} + \dots + c_{r} w_{r} = 0 \end{array}

$\begin{array}{l} a_{1} u_{1} + a_{2} u_{2} + \dots + a_{p} u_{p} \\ + b_{1} v_{1} + b_{2} v_{2} + \dots + b_{q} v_{q} \\ + c_{1} w_{1} + c_{2} w_{2} + \dots + c_{r} w_{r} = 0 \end{array}$ (11)

—we need to show that the coefficients are all zero. If we write

\begin{array}{rcl} u & = & a_{1} u_{1} + a_{2} u_{2} + \dots + a_{p} u_{p}, \\ v & = & b_{1} v_{1} + b_{2} v_{2} + \dots + b_{q} v_{q}, and \\ w & = & c_{1} w_{1} + c_{2} w_{2} + \dots + c_{r} w_{r}, \end{array}

$\begin{array}{rcl} u & = & a_{1} u_{1} + a_{2} u_{2} + \dots + a_{p} u_{p}, \\ v & = & b_{1} v_{1} + b_{2} v_{2} + \dots + b_{q} v_{q}, and \\ w & = & c_{1} w_{1} + c_{2} w_{2} + \dots + c_{r} w_{r}, \end{array}$

then Eq. (11) takes the simple form

u + v + w = 0 .

$u + v + w = 0 .$ (12)

But the vectors u, v, and w either are zero vectors or are eigenvectors associated with the distinct eigenvalues $λ_{1}, λ_{2}$ $λ_{1}, λ_{2}$ , and $λ_{3}$ $λ_{3}$ . In the latter event, Theorem 2 would imply that u, v, and w are linearly independent. Therefore, Eq. (12) implies that $u = v = w = 0$ $u = v = w = 0$ . Finally, the fact that the ${u_{i}}$ ${u_{i}}$ are linearly independent implies that $a_{1} = a_{2} = \dots = a_{p} = 0$ $a_{1} = a_{2} = \dots = a_{p} = 0$ ; the fact that the ${v_{i}}$ ${v_{i}}$ are linearly independent implies that $b_{1} = b_{2} = \dots = b_{q} = 0$ $b_{1} = b_{2} = \dots = b_{q} = 0$ ; similarly, $c_{1} = c_{2} = \dots = c_{r} = 0$ $c_{1} = c_{2} = \dots = c_{r} = 0$ . Thus we have shown that the coefficients in (11) all vanish, and hence that the vectors in $S = S_{1} \cup S_{2} \cup S_{3}$ $S = S_{1} \cup S_{2} \cup S_{3}$ are linearly independent.

6.2 Problems

In Problems 1 through 28, determine whether or not the given matrix A is diagonalizable. If it is, find a diagonalizing matrix P and a diagonal matrix D such that $P^{- 1} A P = D$ $P^{- 1} A P = D$ .

$[\begin{array}{r} 5 & - 4 \\ 2 & - 1 \end{array}]$ $[\begin{array}{r} 5 & - 4 \\ 2 & - 1 \end{array}]$
$[\begin{array}{r} 6 & - 6 \\ 4 & - 4 \end{array}]$ $[\begin{array}{r} 6 & - 6 \\ 4 & - 4 \end{array}]$
$[\begin{array}{r} 5 & - 3 \\ 2 & 0 \end{array}]$ $[\begin{array}{r} 5 & - 3 \\ 2 & 0 \end{array}]$
$[\begin{array}{r} 5 & - 4 \\ 3 & - 2 \end{array}]$ $[\begin{array}{r} 5 & - 4 \\ 3 & - 2 \end{array}]$
$[\begin{array}{r} 9 & - 8 \\ 6 & - 5 \end{array}]$ $[\begin{array}{r} 9 & - 8 \\ 6 & - 5 \end{array}]$
$[\begin{array}{r} 10 & - 6 \\ 12 & - 7 \end{array}]$ $[\begin{array}{r} 10 & - 6 \\ 12 & - 7 \end{array}]$
$[\begin{array}{r} 6 & - 10 \\ 2 & - 3 \end{array}]$ $[\begin{array}{r} 6 & - 10 \\ 2 & - 3 \end{array}]$
$[\begin{array}{r} 11 & - 15 \\ 6 & - 8 \end{array}]$ $[\begin{array}{r} 11 & - 15 \\ 6 & - 8 \end{array}]$
$[\begin{array}{r} - 1 & 4 \\ - 1 & 3 \end{array}]$ $[\begin{array}{r} - 1 & 4 \\ - 1 & 3 \end{array}]$
$[\begin{array}{r} 3 & - 1 \\ 1 & 1 \end{array}]$ $[\begin{array}{r} 3 & - 1 \\ 1 & 1 \end{array}]$
$[\begin{array}{r} 5 & 1 \\ - 9 & - 1 \end{array}]$ $[\begin{array}{r} 5 & 1 \\ - 9 & - 1 \end{array}]$
$[\begin{array}{r} 11 & 9 \\ - 16 & - 13 \end{array}]$ $[\begin{array}{r} 11 & 9 \\ - 16 & - 13 \end{array}]$
$[\begin{array}{r} 1 & 3 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}]$ $[\begin{array}{r} 1 & 3 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}]$
$[\begin{array}{r} 2 & - 2 & 1 \\ 2 & - 2 & 1 \\ 2 & - 2 & 1 \end{array}]$ $[\begin{array}{r} 2 & - 2 & 1 \\ 2 & - 2 & 1 \\ 2 & - 2 & 1 \end{array}]$
$[\begin{array}{r} 3 & - 3 & 1 \\ 2 & - 2 & 1 \\ 0 & 0 & 1 \end{array}]$ $[\begin{array}{r} 3 & - 3 & 1 \\ 2 & - 2 & 1 \\ 0 & 0 & 1 \end{array}]$
$[\begin{array}{r} 3 & - 2 & 0 \\ 0 & 1 & 0 \\ - 4 & 4 & 1 \end{array}]$ $[\begin{array}{r} 3 & - 2 & 0 \\ 0 & 1 & 0 \\ - 4 & 4 & 1 \end{array}]$
$[\begin{array}{r} 7 & - 8 & 3 \\ 6 & - 7 & 3 \\ 2 & - 2 & 2 \end{array}]$ $[\begin{array}{r} 7 & - 8 & 3 \\ 6 & - 7 & 3 \\ 2 & - 2 & 2 \end{array}]$
$[\begin{array}{r} 6 & - 5 & 2 \\ 4 & - 3 & 2 \\ 2 & - 2 & 3 \end{array}]$ $[\begin{array}{r} 6 & - 5 & 2 \\ 4 & - 3 & 2 \\ 2 & - 2 & 3 \end{array}]$
$[\begin{array}{r} 1 & 1 & - 1 \\ - 2 & 4 & - 1 \\ - 4 & 4 & 1 \end{array}]$ $[\begin{array}{r} 1 & 1 & - 1 \\ - 2 & 4 & - 1 \\ - 4 & 4 & 1 \end{array}]$
$[\begin{array}{r} 2 & 0 & 0 \\ - 6 & 11 & 2 \\ 6 & - 15 & 0 \end{array}]$ $[\begin{array}{r} 2 & 0 & 0 \\ - 6 & 11 & 2 \\ 6 & - 15 & 0 \end{array}]$
$[\begin{array}{r} 0 & 1 & 0 \\ - 1 & 2 & 0 \\ - 1 & 1 & 1 \end{array}]$ $[\begin{array}{r} 0 & 1 & 0 \\ - 1 & 2 & 0 \\ - 1 & 1 & 1 \end{array}]$
$[\begin{array}{r} 2 & - 2 & 1 \\ - 1 & 2 & 0 \\ - 5 & 7 & - 1 \end{array}]$ $[\begin{array}{r} 2 & - 2 & 1 \\ - 1 & 2 & 0 \\ - 5 & 7 & - 1 \end{array}]$
$[\begin{array}{r} - 2 & 4 & - 1 \\ - 3 & 5 & - 1 \\ - 1 & 1 & 1 \end{array}]$ $[\begin{array}{r} - 2 & 4 & - 1 \\ - 3 & 5 & - 1 \\ - 1 & 1 & 1 \end{array}]$
$[\begin{array}{r} 3 & - 2 & 1 \\ 1 & 0 & 1 \\ - 1 & 1 & 2 \end{array}]$ $[\begin{array}{r} 3 & - 2 & 1 \\ 1 & 0 & 1 \\ - 1 & 1 & 2 \end{array}]$
$[\begin{array}{r} 1 & 0 & - 2 & 0 \\ 0 & 1 & - 2 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \end{array}]$ $[\begin{array}{r} 1 & 0 & - 2 & 0 \\ 0 & 1 & - 2 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \end{array}]$
$[\begin{array}{r} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 2 \end{array}]$ $[\begin{array}{r} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 2 \end{array}]$
$[\begin{array}{r} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 2 \end{array}]$ $[\begin{array}{r} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 2 \end{array}]$
$[\begin{array}{r} 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{array}]$ $[\begin{array}{r} 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{array}]$
Prove: If the matrices A and B are similar and the matrices B and C are similar, then the matrices A and C are similar.
Suppose that the matrices A and B are similar and that n is a positive integer. Prove that the matrices $A^{n}$ $A^{n}$ and $B^{n}$ $B^{n}$ are similar.
Suppose that the invertible matrices A and B are similar. Prove that their inverses $A^{- 1}$ $A^{- 1}$ and $B^{- 1}$ $B^{- 1}$ are also similar.
Show that if the $n \times n$ $n \times n$ matrices A and B are similar, then they have the same characteristic equation and therefore have the same eigenvalues.
Suppose that the $n \times n$ $n \times n$ matrices A and B are similar and that each has n real eigenvalues. Show that $det A = det B$ $det A = det B$ and that $Tr A = Tr B$ $Tr A = Tr B$ . See Problems 38 and 39 in Section 6.1.
Consider the $2 \times 2$ $2 \times 2$ matrix

$A = [\begin{array}{r} a & b \\ c & d \end{array}]$ $A = [\begin{array}{r} a & b \\ c & d \end{array}]$

and let $Δ = {(a - d)}^{2} + 4 b c$ $Δ = {(a - d)}^{2} + 4 b c$ . Then show that
1. A is diagonalizable if $Δ > 0$ $Δ > 0$ ;
2. A is not diagonalizable if $Δ < 0$ $Δ < 0$ ;
3. If $Δ = 0$ $Δ = 0$ , then A may be diagonalizable or it may not be.
Let A be a $3 \times 3$ $3 \times 3$ matrix with three distinct eigenvalues. Tell how to construct six different invertible matrices $P_{1}, P_{2}, \dots, P_{6}$ $P_{1}, P_{2}, \dots, P_{6}$ and six different diagonal matrices $D_{1}, D_{2}, \dots, D_{6}$ $D_{1}, D_{2}, \dots, D_{6}$ such that $P_{i} D_{i} {(P_{i})}^{- 1} = A$ $P_{i} D_{i} {(P_{i})}^{- 1} = A$ for each $i = 1, 2, \dots, 6$ $i = 1, 2, \dots, 6$ .
Prove: If the diagonalizable matrices A and B have the same eigenvalues (with the same multiplicities), then A and B are similar.
Given: The diagonalizable matrix A. Show that the eigenvalues of $A^{2}$ $A^{2}$ are the squares of the eigenvalues of A but that A and $A^{2}$ $A^{2}$ have the same eigenvectors.
Suppose that the $n \times n$ $n \times n$ matrix A has n linearly independent eigenvectors associated with a single eigenvalue $λ$ $λ$ . Show that A is a diagonal matrix.
Let $λ_{i}$ $λ_{i}$ be an eigenvalue of the $n \times n$ $n \times n$ matrix A, and assume that the characteristic equation of A has only real solutions. The algebraic multiplicity of $λ_{i}$ $λ_{i}$ is the largest positive integer p(i) such that ${(λ - λ_{i})}^{p (i)}$ ${(λ - λ_{i})}^{p (i)}$ is a factor of the characteristic polynomial $| A - λ I |$ $| A - λ I |$ . The geometric multiplicity of $λ_{i}$ $λ_{i}$ is the dimension q(i) of the eigenspace associated with $λ_{i}$ $λ_{i}$ . It can be shown that $p (i) \geq q (i)$ $p (i) \geq q (i)$ for every eigenvalue $λ_{i}$ $λ_{i}$ . Taking this as already established, prove that the given matrix A is diagonalizable if and only if the geometric multiplicity of each eigenvalue is equal to its algebraic multiplicity.

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Table of Contents for 6.2 Diagonalization of Matrices

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Table of Contents for
6.2 Diagonalization of Matrices