# 7.2: Diagonalization

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##### Outcomes
1. Determine when it is possible to diagonalize a matrix.
2. When possible, diagonalize a matrix.

## Similarity and Diagonalization

We begin this section by recalling the definition of similar matrices. Recall that if $$A,B$$ are two $$n\times n$$ matrices, then they are similar if and only if there exists an invertible matrix $$P$$ such that $A=P^{-1}BP\nonumber$

In this case we write $$A \sim B$$. The concept of similarity is an example of an equivalence relation.

##### Lemma $$\PageIndex{1}$$: Similarity is an Equivalence Relation

Similarity is an equivalence relation, i.e. for $$n \times n$$ matrices $$A,B,$$ and $$C$$,

1. $$A \sim A$$ (reflexive)
2. If $$A \sim B$$, then $$B \sim A$$ (symmetric)
3. If $$A \sim B$$ and $$B \sim C$$, then $$A \sim C$$ (transitive)
Proof

It is clear that $$A\sim A$$, taking $$P=I$$.

Now, if $$A\sim B,$$ then for some $$P$$ invertible, $A=P^{-1}BP\nonumber$ and so $PAP^{-1}=B\nonumber$ But then $\left( P^{-1}\right) ^{-1}AP^{-1}=B\nonumber$ which shows that $$B\sim A$$.

Now suppose $$A\sim B$$ and $$B\sim C$$. Then there exist invertible matrices $$P,Q$$ such that $A=P^{-1}BP,\ B=Q^{-1}CQ\nonumber$ Then, $A=P^{-1} \left( Q^{-1}CQ \right)P=\left( QP\right) ^{-1}C\left( QP\right)\nonumber$ showing that $$A$$ is similar to $$C$$.

Another important concept necessary to this section is the trace of a matrix. Consider the definition.

##### Definition $$\PageIndex{1}$$: Trace of a Matrix

If $$A=[a_{ij}]$$ is an $$n\times n$$ matrix, then the trace of $$A$$ is $\mathrm{trace}(A) = \sum_{i=1}^n a_{ii}.\nonumber$

In words, the trace of a matrix is the sum of the entries on the main diagonal.

##### Lemma $$\PageIndex{2}$$: Properties of Trace

For $$n\times n$$ matrices $$A$$ and $$B$$, and any $$k\in\mathbb{R}$$,

1. $$\mathrm{trace}(A+B)=\mathrm{trace}(A) + \mathrm{trace}(B)$$
2. $$\mathrm{trace}(kA)=k\cdot\mathrm{trace}(A)$$
3. $$\mathrm{trace}(AB)=\mathrm{trace}(BA)$$

The following theorem includes a reference to the characteristic polynomial of a matrix. Recall that for any $$n \times n$$ matrix $$A$$, the characteristic polynomial of $$A$$ is $$c_A(x)=\det(xI-A)$$.

##### Theorem $$\PageIndex{1}$$: Properties of Similar Matrices

If $$A$$ and $$B$$ are $$n\times n$$ matrices and $$A\sim B$$, then

1. $$\det(A) = \det(B)$$
2. $$\mathrm{rank}(A) = \mathrm{rank}(B)$$
3. $$\mathrm{trace}(A)= \mathrm{trace}(B)$$
4. $$c_A(x)=c_B(x)$$
5. $$A$$ and $$B$$ have the same eigenvalues

We now proceed to the main concept of this section. When a matrix is similar to a diagonal matrix, the matrix is said to be diagonalizable. We define a diagonal matrix $$D$$ as a matrix containing a zero in every entry except those on the main diagonal. More precisely, if $$d_{ij}$$ is the $$ij^{th}$$ entry of a diagonal matrix $$D$$, then $$d_{ij}=0$$ unless $$i=j$$. Such matrices look like the following. $D = \left[ \begin{array}{ccc} \ast & & 0 \\ & \ddots & \\ 0 & & \ast \end{array} \right]\nonumber$ where $$\ast$$ is a number which might not be zero.

The following is the formal definition of a diagonalizable matrix.

##### Definition $$\PageIndex{2}$$: Diagonalizable

Let $$A$$ be an $$n\times n$$ matrix. Then $$A$$ is said to be diagonalizable if there exists an invertible matrix $$P$$ such that $P^{-1}AP=D\nonumber$ where $$D$$ is a diagonal matrix.

Notice that the above equation can be rearranged as $$A=PDP^{-1}$$. Suppose we wanted to compute $$A^{100}$$. By diagonalizing $$A$$ first it suffices to then compute $$\left(PDP^{-1}\right)^{100}$$, which reduces to $$PD^{100}P^{-1}$$. This last computation is much simpler than $$A^{100}$$. While this process is described in detail later, it provides motivation for diagonalization.

## Diagonalizing a Matrix

The most important theorem about diagonalizability is the following major result.

##### Theorem $$\PageIndex{2}$$: Eigenvectors and Diagonalizable Matrices

An $$n\times n$$ matrix $$A$$ is diagonalizable if and only if there is an invertible matrix $$P$$ given by $P=\left[\begin{array}{cccc} X_{1} & X_{2} & \cdots & X_{n} \end{array} \right]\nonumber$ where the $$X_{k}$$ are eigenvectors of $$A$$.

Moreover if $$A$$ is diagonalizable, the corresponding eigenvalues of $$A$$ are the diagonal entries of the diagonal matrix $$D$$.

Proof

Suppose $$P$$ is given as above as an invertible matrix whose columns are eigenvectors of $$A$$. Then $$P^{-1}$$ is of the form $P^{-1}=\left[\begin{array}{c} W_{1}^{T} \\ W_{2}^{T} \\ \vdots \\ W_{n}^{T} \end{array} \right]\nonumber$ where $$W_{k}^{T}X_{j}=\delta _{kj},$$ which is the Kronecker’s symbol defined by $\delta _{ij}=\left\{ \begin{array}{c} 1 \text{ if }i=j \\ 0\text{ if }i\neq j \end{array} \right.\nonumber$

Then \begin{aligned} P^{-1}AP & = \left[\begin{array}{c} W_{1}^{T} \\ W_{2}^{T} \\ \vdots \\ W_{n}^{T} \end{array} \right] \left[\begin{array}{cccc} AX_{1} & AX_{2} & \cdots & AX_{n} \end{array} \right] \\ & = \left[\begin{array}{c} W_{1}^{T} \\ W_{2}^{T} \\ \vdots \\ W_{n}^{T} \end{array} \right] \left[\begin{array}{cccc} \lambda _{1}X_{1} & \lambda _{2}X_{2} & \cdots & \lambda _{n}X_{n} \end{array} \right] \\ &= \left[\begin{array}{ccc} \lambda _{1} & & 0 \\ & \ddots & \\ 0 & & \lambda _{n} \end{array} \right] \end{aligned}

Conversely, suppose $$A$$ is diagonalizable so that $$P^{-1}AP=D.$$ Let $P=\left[\begin{array}{cccc} X_{1} & X_{2} & \cdots & X_{n} \end{array} \right]\nonumber$ where the columns are the $$X_{k}$$ and $D=\left[\begin{array}{ccc} \lambda _{1} & & 0 \\ & \ddots & \\ 0 & & \lambda _{n} \end{array} \right]\nonumber$ Then $AP=PD=\left[\begin{array}{cccc} X_{1} & X_{2} & \cdots & X_{n} \end{array} \right] \left[\begin{array}{ccc} \lambda _{1} & & 0 \\ & \ddots & \\ 0 & & \lambda _{n} \end{array} \right]\nonumber$ and so $\left[\begin{array}{cccc} AX_{1} & AX_{2} & \cdots & AX_{n} \end{array} \right] =\left[\begin{array}{cccc} \lambda _{1}X_{1} & \lambda _{2}X_{2} & \cdots & \lambda _{n}X_{n} \end{array} \right]\nonumber$ showing the $$X_{k}$$ are eigenvectors of $$A$$ and the $$\lambda _{k}$$ are eigenvectors.

Notice that because the matrix $$P$$ defined above is invertible it follows that the set of eigenvectors of $$A$$, $$\left\{ X_1, X_2, \cdots, X_n \right\}$$, form a basis of $$\mathbb{R}^n$$.

We demonstrate the concept given in the above theorem in the next example. Note that not only are the columns of the matrix $$P$$ formed by eigenvectors, but $$P$$ must be invertible so must consist of a wide variety of eigenvectors. We achieve this by using basic eigenvectors for the columns of $$P$$.

##### Example $$\PageIndex{1}$$: Diagonalize a Matrix

Let $A=\left[\begin{array}{rrr} 2 & 0 & 0 \\ 1 & 4 & -1 \\ -2 & -4 & 4 \end{array} \right]\nonumber$ Find an invertible matrix $$P$$ and a diagonal matrix $$D$$ such that $$P^{-1}AP=D$$.

###### Solution

By Theorem $$\PageIndex{2}$$ we use the eigenvectors of $$A$$ as the columns of $$P$$, and the corresponding eigenvalues of $$A$$ as the diagonal entries of $$D$$.

First, we will find the eigenvalues of $$A$$. To do so, we solve $$\det \left( \lambda I -A \right) =0$$ as follows. $\det \left( \lambda \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] - \left[\begin{array}{rrr} 2 & 0 & 0 \\ 1 & 4 & -1 \\ -2 & -4 & 4 \end{array} \right] \right) = 0\nonumber$

This computation is left as an exercise, and you should verify that the eigenvalues are $$\lambda_1 =2, \lambda_2 = 2$$, and $$\lambda_3 = 6$$.

Next, we need to find the eigenvectors. We first find the eigenvectors for $$\lambda_1, \lambda_2 = 2$$. Solving $$\left(2I - A \right)X = 0$$ to find the eigenvectors, we find that the eigenvectors are $t\left[\begin{array}{r} -2 \\ 1 \\ 0 \end{array} \right] +s\left[\begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right]\nonumber$ where $$t,s$$ are scalars. Hence there are two basic eigenvectors which are given by $X_1 = \left[\begin{array}{r} -2 \\ 1 \\ 0 \end{array} \right], X_2 = \left[\begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right]\nonumber$

You can verify that the basic eigenvector for $$\lambda_3 =6$$ is $$X_3 = \left[\begin{array}{r} 0 \\ 1 \\ -2 \end{array} \right]$$

Then, we construct the matrix $$P$$ as follows. $P= \left[\begin{array}{rrr} X_1 & X_2 & X_3 \end{array} \right] = \left[\begin{array}{rrr} -2 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & -2 \end{array} \right]\nonumber$ That is, the columns of $$P$$ are the basic eigenvectors of $$A$$. Then, you can verify that $P^{-1}=\left[\begin{array}{rrr} - \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{4} & \frac{1}{2} & - \frac{1}{4} \end{array} \right]\nonumber$ Thus, \begin{aligned} P^{-1}AP &=\left[\begin{array}{rrr} - \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{4} & \frac{1}{2} & - \frac{1}{4} \end{array} \right] \left[\begin{array}{rrr} 2 & 0 & 0 \\ 1 & 4 & -1 \\ -2 & -4 & 4 \end{array} \right] \left[\begin{array}{rrr} -2 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & -2 \end{array} \right] \\ &=\left[\begin{array}{rrr} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6 \end{array} \right] \end{aligned}

You can see that the result here is a diagonal matrix where the entries on the main diagonal are the eigenvalues of $$A$$. We expected this based on Theorem $$\PageIndex{2}$$. Notice that eigenvalues on the main diagonal must be in the same order as the corresponding eigenvectors in $$P$$.

Consider the next important theorem.

##### Theorem $$\PageIndex{3}$$: Linearly Independent Eigenvectors

Let $$A$$ be an $$n\times n$$ matrix, and suppose that $$A$$ has distinct eigenvalues $$\lambda_1, \lambda_2, \ldots, \lambda_m$$. For each $$i$$, let $$X_i$$ be a $$\lambda_i$$-eigenvector of $$A$$. Then $$\{ X_1, X_2, \ldots, X_m\}$$ is linearly independent.

The corollary that follows from this theorem gives a useful tool in determining if $$A$$ is diagonalizable.

##### Corollary $$\PageIndex{1}$$: Distinct Eigenvalues

Let $$A$$ be an $$n \times n$$ matrix and suppose it has $$n$$ distinct eigenvalues. Then it follows that $$A$$ is diagonalizable.

It is possible that a matrix $$A$$ cannot be diagonalized. In other words, we cannot find an invertible matrix $$P$$ so that $$P^{-1}AP=D$$.

Consider the following example.

##### Example $$\PageIndex{2}$$: A Matrix which cannot be Diagonalized

Let $A = \left[\begin{array}{rr} 1 & 1 \\ 0 & 1 \end{array} \right]\nonumber$ If possible, find an invertible matrix $$P$$ and diagonal matrix $$D$$ so that $$P^{-1}AP=D$$.

###### Solution

Through the usual procedure, we find that the eigenvalues of $$A$$ are $$\lambda_1 =1, \lambda_2=1.$$ To find the eigenvectors, we solve the equation $$\left(\lambda I - A \right) X = 0$$. The matrix $$\left(\lambda I -A \right)$$ is given by $\left[\begin{array}{cc} \lambda - 1 & -1 \\ 0 & \lambda - 1 \end{array} \right]\nonumber$

Substituting in $$\lambda = 1$$, we have the matrix $\left[\begin{array}{cc} 1 - 1 & -1 \\ 0 & 1 - 1 \end{array} \right] = \left[\begin{array}{rr} 0 & -1 \\ 0 & 0 \end{array} \right]\nonumber$

Then, solving the equation $$\left(\lambda I - A\right) X = 0$$ involves carrying the following augmented matrix to its reduced row-echelon form. $\left[\begin{array}{rr|r} 0 & -1 & 0 \\ 0 & 0 & 0 \end{array} \right] \rightarrow \cdots \rightarrow \left[\begin{array}{rr|r} 0 & -1 & 0 \\ 0 & 0 & 0 \end{array} \right]\nonumber$

Then the eigenvectors are of the form $t\left[\begin{array}{r} 1 \\ 0 \end{array} \right]\nonumber$ and the basic eigenvector is $X_1 = \left[\begin{array}{r} 1 \\ 0 \end{array} \right]\nonumber$

In this case, the matrix $$A$$ has one eigenvalue of multiplicity two, but only one basic eigenvector. In order to diagonalize $$A$$, we need to construct an invertible $$2\times 2$$ matrix $$P$$. However, because $$A$$ only has one basic eigenvector, we cannot construct this $$P$$. Notice that if we were to use $$X_1$$ as both columns of $$P$$, $$P$$ would not be invertible. For this reason, we cannot repeat eigenvectors in $$P$$.

Hence this matrix cannot be diagonalized.

The idea that a matrix may not be diagonalizable suggests that conditions exist to determine when it is possible to diagonalize a matrix. We saw earlier in Corollary $$\PageIndex{1}$$ that an $$n \times n$$ matrix with $$n$$ distinct eigenvalues is diagonalizable. It turns out that there are other useful diagonalizability tests.

First we need the following definition.

##### Definition $$\PageIndex{3}$$: Eigenspace

Let $$A$$ be an $$n\times n$$ matrix and $$\lambda\in\mathbb{R}$$. The eigenspace of $$A$$ corresponding to $$\lambda$$, written $$E_{\lambda}(A)$$ is the set of all eigenvectors corresponding to $$\lambda$$.

In other words, the eigenspace $$E_{\lambda}(A)$$ is all $$X$$ such that $$AX = \lambda X$$. Notice that this set can be written $$E_{\lambda}(A) = \mathrm{null}(\lambda I - A)$$, showing that $$E_{\lambda}(A)$$ is a subspace of $$\mathbb{R}^n$$.

Recall that the multiplicity of an eigenvalue $$\lambda$$ is the number of times that it occurs as a root of the characteristic polynomial.

Consider now the following lemma.

##### Lemma $$\PageIndex{3}$$: Dimension of the Eigenspace

If $$A$$ is an $$n\times n$$ matrix, then $\dim(E_{\lambda}(A))\leq m\nonumber$ where $$\lambda$$ is an eigenvalue of $$A$$ of multiplicity $$m$$.

This result tells us that if $$\lambda$$ is an eigenvalue of $$A$$, then the number of linearly independent $$\lambda$$-eigenvectors is never more than the multiplicity of $$\lambda$$. We now use this fact to provide a useful diagonalizability condition.

##### Theorem $$\PageIndex{4}$$: Diagonalizability Condition

Let $$A$$ be an $$n \times n$$ matrix $$A$$. Then $$A$$ is diagonalizable if and only if for each eigenvalue $$\lambda$$ of $$A$$, $$\dim(E_{\lambda}(A))$$ is equal to the multiplicity of $$\lambda$$.

## Complex Eigenvalues

In some applications, a matrix may have eigenvalues which are complex numbers. For example, this often occurs in differential equations. These questions are approached in the same way as above.

Consider the following example.

##### Example $$\PageIndex{3}$$: A Real Matrix with Complex Eigenvalues

Let $A=\left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 2 & -1 \\ 0 & 1 & 2 \end{array} \right ]\nonumber$ Find the eigenvalues and eigenvectors of $$A$$.

###### Solution

We will first find the eigenvalues as usual by solving the following equation.

$\det \left( \lambda \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right ] - \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 2 & -1 \\ 0 & 1 & 2 \end{array} \right ] \right) =0\nonumber$ This reduces to $$\left( \lambda -1\right) \left( \lambda^{2}-4 \lambda +5\right) =0.$$ The solutions are $$\lambda_1 =1,\lambda_2 = 2+i$$ and $$\lambda_3 =2-i.$$

There is nothing new about finding the eigenvectors for $$\lambda_1 =1$$ so this is left as an exercise.

Consider now the eigenvalue $$\lambda_2 =2+i.$$ As usual, we solve the equation $$\left(\lambda I -A \right) X = 0$$ as given by $\left( \left( 2+i\right) \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right ] - \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 2 & -1 \\ 0 & 1 & 2 \end{array} \right ] \right) X =\left [ \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right ]\nonumber$ In other words, we need to solve the system represented by the augmented matrix $\left [ \begin{array}{crr|r} 1+i & 0 & 0 & 0 \\ 0 & i & 1 & 0 \\ 0 & -1 & i & 0 \end{array} \right ]\nonumber$

We now use our row operations to solve the system. Divide the first row by $$\left( 1+i\right)$$ and then take $$-i$$ times the second row and add to the third row. This yields $\left [ \begin{array}{rrr|r} 1 & 0 & 0 & 0 \\ 0 & i & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right ]\nonumber$ Now multiply the second row by $$-i$$ to obtain the reduced row-echelon form, given by $\left [ \begin{array}{rrr|r} 1 & 0 & 0 & 0 \\ 0 & 1 & -i & 0 \\ 0 & 0 & 0 & 0 \end{array} \right ]\nonumber$ Therefore, the eigenvectors are of the form $t\left [ \begin{array}{r} 0 \\ i \\ 1 \end{array} \right ]\nonumber$ and the basic eigenvector is given by $X_2 = \left [ \begin{array}{r} 0 \\ i \\ 1 \end{array} \right ]\nonumber$

As an exercise, verify that the eigenvectors for $$\lambda_3 =2-i$$ are of the form $t\left [ \begin{array}{r} 0 \\ -i \\ 1 \end{array} \right ]\nonumber$ Hence, the basic eigenvector is given by $X_3 = \left [ \begin{array}{r} 0 \\ -i \\ 1 \end{array} \right ]\nonumber$

As usual, be sure to check your answers! To verify, we check that $$AX_3 = \left(2 - i \right) X_3$$ as follows. $\left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 2 & -1 \\ 0 & 1 & 2 \end{array} \right ] \left [ \begin{array}{r} 0 \\ -i \\ 1 \end{array} \right ] = \left [ \begin{array}{c} 0 \\ -1-2i \\ 2-i \end{array} \right ] =\left( 2-i\right) \left [ \begin{array}{r} 0 \\ -i \\ 1 \end{array} \right ]\nonumber$

Therefore, we know that this eigenvector and eigenvalue are correct.

Notice that in Example $$\PageIndex{3}$$, two of the eigenvalues were given by $$\lambda_2 = 2 + i$$ and $$\lambda_3 = 2-i$$. You may recall that these two complex numbers are conjugates. It turns out that whenever a matrix containing real entries has a complex eigenvalue $$\lambda$$, it also has an eigenvalue equal to $$\overline{\lambda}$$, the conjugate of $$\lambda$$.

This page titled 7.2: Diagonalization is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) via source content that was edited to the style and standards of the LibreTexts platform.