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5.2 Diagonalization: Class Notes

Definition

A square matrix $A$ is said to be diagonalizable if it is similar to some diagonal matrix; that is, if there exists an invertible matrix $P$ such that $P^{-1}AP$ is diagonal. In this case, the matrix $P$ is said to diagonalize A.

Theorem 5.2.1

If $A$ is an $n\times n$ matrix, then $A$ is diagonalizable if and only if $A$ has $n$ linearly independent eigenvectors.

#5

Find a matrix $P$ that diagonalizes $A$, and check your work by computing $P^{-1}AP$.

$A=\left[\begin{array}{cc}1& 0\\ 6& -1\end{array}\right]$

Solution

$$det(\lambda I-A)=0\Rightarrow \left|\begin{array}{cc}\lambda -1& 0\\ -6& \lambda +1\end{array}\right|=0\Rightarrow {\lambda }^2-1=0\Rightarrow \lambda =\pm 1$$

Find an eigenvector for each eigenvalue.

$${\lambda }_1=-1:\left[\begin{array}{cc}-2& 0\\ -6& 0\end{array}\right]\to \left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\Rightarrow x_1=0,\text{so}{\overrightarrow{p}}_1=\left[\begin{array}{c}0\\ 1\end{array}\right]t$$

$${\lambda }_2=1:\left[\begin{array}{cc}0& 0\\ -6& 2\end{array}\right]\Rightarrow 3x_1=x_2,\text{so}{\overrightarrow{p}}_2=\left[\begin{array}{c}1\\ 3\end{array}\right]t$$

Form the matrix whose columns are the eigenvectors, $P=\left[{\overrightarrow{p}}_1\right|{\overrightarrow{p}}_2]$:

$$P=\left[\begin{array}{cc}0& 1\\ 1& 3\end{array}\right]$$

Find $P^{-1}$ (here $det(P)=-1$):

$${\displaystyle P^{-1}=\frac{1}{det(P)}\left[\begin{array}{cc}3& -1\\ -1& 0\end{array}\right]=\left[\begin{array}{cc}-3& 1\\ 1& 0\end{array}\right]}$$

Find $P^{-1}AP$:

$$\begin{array}{cc}\hfill {\displaystyle P^{-1}AP}& {\displaystyle =\left[\begin{array}{cc}-3& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 6& -1\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 3\end{array}\right]}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left[\begin{array}{cc}3& -1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 3\end{array}\right]}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]}\hfill \end{array}$$

This is a diagonal matrix.

Notes

• The diagonal entries of $P^{-1}AP$ are the eigenvalues.
• The columns of $P$ are the respective eigenvectors.
• The order of these is connected: the first column of $P$ goes with the first diagonal entry, and the second column with the second diagonal entry.

What makes this work?

$$P^{-1}AP=D\Rightarrow AP=PD$$

Here,

$$\left[\begin{array}{cc}1& 0\\ 6& -1\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 3\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 3\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]$$

Column by column:

$$A{\overrightarrow{p}}_1=\left[\begin{array}{cc}1& 0\\ 6& -1\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ -1\end{array}\right]=-1\left[\begin{array}{c}0\\ 1\end{array}\right]$$

$$A{\overrightarrow{p}}_2=\left[\begin{array}{cc}1& 0\\ 6& -1\end{array}\right]\left[\begin{array}{c}1\\ 3\end{array}\right]=\left[\begin{array}{c}1\\ 3\end{array}\right]=1\left[\begin{array}{c}1\\ 3\end{array}\right]$$

So $\left[A{\overrightarrow{p}}_1\right|A{\overrightarrow{p}}_2]=\left[{\lambda }_1{\overrightarrow{p}}_1\right|{\lambda }_2{\overrightarrow{p}}_2]$: corresponding columns satisfy $A\overrightarrow{x}=\lambda \overrightarrow{x}$.

Proof of Theorem 5.2.1

($\Rightarrow$) If $A$ is diagonalizable, then there is an invertible matrix $P$ and a diagonal matrix $D$ such that $P^{-1}AP=D$ (by the definition). Then $AP=PD$. Let

$$P=\left[\begin{array}{cccc}{\overrightarrow{p}}_1& {\overrightarrow{p}}_2& \cdots & {\overrightarrow{p}}_n\end{array}\right]\text{and}D=\left[\begin{array}{ccc}{\lambda }_1& & 0\\ & \ddots & \\ 0& & {\lambda }_n\end{array}\right].$$

Then

$$AP=A\left[\begin{array}{cccc}{\overrightarrow{p}}_1& {\overrightarrow{p}}_2& \cdots & {\overrightarrow{p}}_n\end{array}\right]=\left[\begin{array}{cccc}A{\overrightarrow{p}}_1& A{\overrightarrow{p}}_2& \cdots & A{\overrightarrow{p}}_n\end{array}\right]$$

and

$$PD=\left[\begin{array}{cccc}{\overrightarrow{p}}_1& {\overrightarrow{p}}_2& \cdots & {\overrightarrow{p}}_n\end{array}\right]\left[\begin{array}{ccc}{\lambda }_1& & 0\\ & \ddots & \\ 0& & {\lambda }_n\end{array}\right]=\left[\begin{array}{cccc}{\lambda }_1{\overrightarrow{p}}_1& {\lambda }_2{\overrightarrow{p}}_2& \cdots & {\lambda }_n{\overrightarrow{p}}_n\end{array}\right]$$

Matching columns, $AP=PD$ gives $A{\overrightarrow{p}}_i={\lambda }_i{\overrightarrow{p}}_i,1\le i\le n$.

Since $P$ is invertible, $\{{\overrightarrow{p}}_i\}$ is a linearly independent set (Equivalent Statements), so each ${\overrightarrow{p}}_i\ne \overrightarrow{0}$. So the $n$ column vectors ${\overrightarrow{p}}_i$ are linearly independent eigenvectors of $A$.

($\Leftarrow$) Suppose $A$ has linearly independent eigenvectors ${\overrightarrow{p}}_1,{\overrightarrow{p}}_2,\dots ,{\overrightarrow{p}}_n$ with corresponding eigenvalues ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$. Let $P=\left[\begin{array}{cccc}{\overrightarrow{p}}_1& {\overrightarrow{p}}_2& \cdots & {\overrightarrow{p}}_n\end{array}\right]$ and $D=\left[\begin{array}{ccc}{\lambda }_1& & 0\\ & \ddots & \\ 0& & {\lambda }_n\end{array}\right]$. Then

$$AP=\left[\begin{array}{cccc}A{\overrightarrow{p}}_1& A{\overrightarrow{p}}_2& \cdots & A{\overrightarrow{p}}_n\end{array}\right]=\left[\begin{array}{cccc}{\lambda }_1{\overrightarrow{p}}_1& {\lambda }_2{\overrightarrow{p}}_2& \cdots & {\lambda }_n{\overrightarrow{p}}_n\end{array}\right]=PD$$

$\{{\overrightarrow{p}}_i\}$ is an independent set, so $P$ is invertible. Then $AP=PD\Rightarrow P^{-1}AP=D$, so $A$ is diagonalizable.

Tip: The statement is about the total number of eigenvectors, not eigenvalues.

#17

Compute the matrix $A^{10}$ by first diagonalizing $A$.

$A=\left[\begin{array}{cc}0& 3\\ 2& -1\end{array}\right]$

Solution

The process: (1) find the eigenvalues ${\lambda }_i$, (2) find the eigenvectors ${\overrightarrow{p}}_i$, (3) form $P$ and $P^{-1}$.

$$\left|\begin{array}{cc}\lambda & -3\\ -2& \lambda +1\end{array}\right|=0\Rightarrow {\lambda }^2+\lambda -6=0\Rightarrow \lambda =-3,2$$

$${\lambda }_1=-3:\left[\begin{array}{cc}-3& -3\\ -2& -2\end{array}\right]\to \left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]\Rightarrow {\overrightarrow{p}}_1=\left[\begin{array}{c}1\\ -1\end{array}\right]t$$

$${\lambda }_2=2:\left[\begin{array}{cc}2& -3\\ -2& 3\end{array}\right]\to \left[\begin{array}{cc}1& -\frac{3}{2}\\ 0& 0\end{array}\right]\Rightarrow {\overrightarrow{p}}_2=\left[\begin{array}{c}3\\ 2\end{array}\right]t$$

$${\displaystyle P=\left[\begin{array}{cc}1& 3\\ -1& 2\end{array}\right],det(P)=5\Rightarrow P^{-1}=\left[\begin{array}{cc}\frac{2}{5}& -\frac{3}{5}\\ \frac{1}{5}& \frac{1}{5}\end{array}\right]}$$

$$P^{-1}AP=D=\left[\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right]=\left[\begin{array}{cc}-3& 0\\ 0& 2\end{array}\right]$$

Note that if $P^{-1}AP=D$, then $A=PD{P}^{-1}$. Then the inner factors cancel in pairs:

$$A^k=(PD{P}^{-1})(PD{P}^{-1})\cdots (PD{P}^{-1})=P{D}^k{P}^{-1}$$

Here $D^{10}=\left[\begin{array}{cc}(-3{)}^{10}& 0\\ 0& 2^{10}\end{array}\right]=\left[\begin{array}{cc}{3}^{10}& 0\\ 0& 2^{10}\end{array}\right]$, so

$${\displaystyle A^{10}=\left[\begin{array}{cc}1& 3\\ -1& 2\end{array}\right]\left[\begin{array}{cc}{3}^{10}& 0\\ 0& 2^{10}\end{array}\right]\left[\begin{array}{cc}\frac{2}{5}& -\frac{3}{5}\\ \frac{1}{5}& \frac{1}{5}\end{array}\right]}$$

Theorem 5.2.3

If $k$ is a positive integer, $\lambda$ is an eigenvalue of a matrix $A$, and $𝐱$ is a corresponding eigenvector, then ${\lambda }^k$ is an eigenvalue of $A^k$ and $𝐱$ is a corresponding eigenvector.

(Not a proof.) Consider $A\overrightarrow{x}=\lambda \overrightarrow{x}$. Then

$$A^2\overrightarrow{x}=A(\lambda \overrightarrow{x})=\lambda A\overrightarrow{x}={\lambda }^2\overrightarrow{x}$$

Example

Let $A=\left[\begin{array}{ccc}3& 2& 4\\ 2& 0& 2\\ 4& 2& 3\end{array}\right]$.

a. Find the eigenvalues of $A$.
b. For each eigenvalue $\lambda$, find the rank of the matrix $\lambda I-A$.
c. Is $A$ diagonalizable? Justify your conclusion.

Solution

a. Expand $det(\lambda I-A)$ along the first row.

$$\left|\begin{array}{ccc}\lambda -3& -2& -4\\ -2& \lambda & -2\\ -4& -2& \lambda -3\end{array}\right|=0$$

$$\begin{array}{cc}\hfill {\displaystyle (\lambda -3)\left|\begin{array}{cc}\lambda & -2\\ -2& \lambda -3\end{array}\right|+2\left|\begin{array}{cc}-2& -2\\ -4& \lambda -3\end{array}\right|-4\left|\begin{array}{cc}-2& \lambda \\ -4& -2\end{array}\right|}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (\lambda -3)({\lambda }^2-3\lambda -4)+2[-2(\lambda -3)-8]-4(4+4\lambda )}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (\lambda -3)(\lambda -4)(\lambda +1)-4(\lambda +1)-16(\lambda +1)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (\lambda +1)[(\lambda -3)(\lambda -4)-4-16]}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (\lambda +1)({\lambda }^2-7\lambda -8)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (\lambda +1{)}^2(\lambda -8)}& {\displaystyle =0}\hfill \end{array}$$

Each term has a factor of $(\lambda +1)$, which we factor out.

b. For ${\lambda }_1=8$:

$$\left[\begin{array}{ccc}5& -2& -4\\ -2& 8& -2\\ -4& -2& 5\end{array}\right]\to \left[\begin{array}{ccc}5& -2& -4\\ 0& 2& -1\\ 0& 0& 0\end{array}\right]$$

Using $R_2\to 5R_2+2R_1$ and $R_3\to R_3-2R_2$:

$$\begin{array}{cc}\hfill {\displaystyle 5R_2+2R_1:}& {\displaystyle (-10,40,-10)+(10,-4,-8)=(0,36,-18)}\hfill \\ \hfill {\displaystyle R_3-2R_2:}& {\displaystyle (-4,-2,5)+(4,-16,4)=(0,-18,9)}\hfill \end{array}$$

Both new rows are multiples of $(0,2,-1)$, which gives the reduced matrix above.

The rank is 2, so the nullity is 1. This will give 1 eigenvector.

For ${\lambda }_2=-1$:

$$\left[\begin{array}{ccc}-4& -2& -4\\ -2& -1& -2\\ -4& -2& -4\end{array}\right]\to \left[\begin{array}{ccc}2& 1& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$

The rank is 1, so the nullity is 2. This will give 2 eigenvectors.

c. 2 + 1 = 3 total eigenvectors (the total number of basis vectors for the eigenspaces is 3), so

Find the eigenvectors and P; find P⁻¹AP.

$${\lambda }_1=8:\left[\begin{array}{ccc}5& -2& -4\\ -2& 8& -2\\ -4& -2& 5\end{array}\right]\to \left[\begin{array}{ccc}5& -2& -4\\ 0& 2& -1\\ 0& 0& 0\end{array}\right]\to \left[\begin{array}{ccc}1& 0& -1\\ 0& 2& -1\\ 0& 0& 0\end{array}\right]$$

So $x_1=x_3$ and $2x_2=x_3$, giving ${\overrightarrow{x}}_1=\left[\begin{array}{c}2\\ 1\\ 2\end{array}\right]t$ and ${\overrightarrow{p}}_1=\left[\begin{array}{c}2\\ 1\\ 2\end{array}\right]$.

$${\lambda }_2=-1:\left[\begin{array}{ccc}-4& -2& -4\\ -2& -1& -2\\ -4& -2& -4\end{array}\right]\to \left[\begin{array}{ccc}2& 1& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\Rightarrow x_1=-\frac{1}{2}{x}_2-x_3$$

$${\displaystyle {\overrightarrow{x}}_2=\left[\begin{array}{c}-\frac{1}{2}s-t\\ s\\ t\end{array}\right]=\left[\begin{array}{c}-\frac{1}{2}\\ 1\\ 0\end{array}\right]s+\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]t}$$

Scaling the first vector by $-2$: ${\overrightarrow{p}}_2=\left[\begin{array}{c}1\\ -2\\ 0\end{array}\right]$, ${\overrightarrow{p}}_3=\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]$.

$$P=\left[\begin{array}{ccc}2& 1& -1\\ 1& -2& 0\\ 2& 0& 1\end{array}\right]$$

Why this works: $PD$ multiplies each column ${\overrightarrow{p}}_i$ of $P$ by its eigenvalue, so

$$PD=\left[\begin{array}{ccc}2& 1& -1\\ 1& -2& 0\\ 2& 0& 1\end{array}\right]\left[\begin{array}{ccc}8& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right]=\left[\begin{array}{ccc}8{\overrightarrow{p}}_1& -1{\overrightarrow{p}}_2& -1{\overrightarrow{p}}_3\end{array}\right]$$

and each column is $\lambda {\overrightarrow{p}}_i=A{\overrightarrow{p}}_i$, which is $A\overrightarrow{x}=\lambda \overrightarrow{x}$.

Definitions

If ${\lambda }_0$ is an eigenvalue of an $n\times n$ matrix $A$, then the dimension of the eigenspace corresponding to ${\lambda }_0$ is called the geometric multiplicity of ${\lambda }_0$, and the number of times that $\lambda -{\lambda }_0$ appears as a factor in the characteristic polynomial of $A$ is called the algebraic multiplicity of ${\lambda }_0$.

• Geometric multiplicity: the number of linearly independent eigenvectors that come from the eigenvalue.
• Algebraic multiplicity: the exponent on the factor $(\lambda -{\lambda }_0)$ in the characteristic equation.

Theorem 5.2.4 Geometric and Algebraic Multiplicity

If $A$ is a square matrix, then:

• a) For every eigenvalue of $A$, the geometric multiplicity is less than or equal to the algebraic multiplicity.
• b) $A$ is diagonalizable if and only if its characteristic polynomial can be expressed as a product of linear factors, and the geometric multiplicity of every eigenvalue is equal to the algebraic multiplicity.

#14

Find the geometric and algebraic multiplicity of each eigenvalue of the matrix $A$, and determine whether $A$ is diagonalizable. If $A$ is diagonalizable, then find a matrix $P$ that diagonalizes $A$, and find $P^{-1}AP$.

$A=\left[\begin{array}{ccc}5& 0& 0\\ 1& 5& 0\\ 0& 1& 5\end{array}\right]$

Solution

$A$ is lower triangular, so the characteristic equation is $(\lambda -5{)}^3=0$. So $\lambda =5$ is an eigenvalue with algebraic multiplicity 3. If we can find 3 linearly independent eigenvectors, we can diagonalize $A$.

$$5I-A=\left[\begin{array}{ccc}0& 0& 0\\ -1& 0& 0\\ 0& -1& 0\end{array}\right]$$

This has rank 2, so its nullity is 1: it yields only $\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$ (up to scalar multiples), and the geometric multiplicity is 1.

So $\lambda =5$ has algebraic multiplicity 3 and geometric multiplicity 1. There are not enough eigenvectors.

Theorem 5.2.2

• a) If ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_k$ are distinct eigenvalues of a matrix $A$, and if ${𝐯}_1,{𝐯}_2,\dots ,{𝐯}_k$ are corresponding eigenvectors, then $\{{𝐯}_1,{𝐯}_2,\dots ,{𝐯}_k\}$ is a linearly independent set.
• b) An $n\times n$ matrix with $n$ distinct eigenvalues is diagonalizable.

Similarity

Definition

A transformation of the form $A\to P^{-1}AP$ is called a similarity transformation.

Definition

If $A$ and $B$ are square matrices, then we say that $𝐁$ is similar to $𝐀$ (or that $A$ and $B$ are similar matrices) if there is an invertible matrix $P$ such that $B=P^{-1}AP$.

Example

Prove that if $A$ and $B$ are similar matrices, then $det(A)=det(B)$.

Proof

Since $A$ and $B$ are similar, there is an invertible $P$ with $B=P^{-1}AP$. Then

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle det(B)}& {\displaystyle =det(P^{-1}AP)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =det(P^{-1})det(A)det(P)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{1}{det(P)}det(A)det(P)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =det(A)}\hfill \end{array}}$$

Definition

Any property that is preserved by a similarity transformation is called a similarity invariant and is said to be invariant under similarity.

Similarity invariants

The following are the same for $A$ and $P^{-1}AP$:

• Determinant
• Invertibility
• Rank
• Nullity
• Trace
• Characteristic polynomial
• Eigenvalues
• Eigenspace dimension