 Diagonalization of a Matrix (with Examples)

Master of Science in Mathematics
Lecture Notes. Accessed on: 2019-10-21 17:45:10

An $n\times n$ matrix $A$ is called diagonalizable if $A$ is similar to some diagonal matrix $D.$ If the matrix of a linear transformation $T$ with respect to some basis is diagonal then we call $T$ diagonalizable.

Diagonalization Theorem

Theorem. An $n\times n$ matrix $A$ is diagonalizable if and only if it has $n$ linearly independent eigenvectors. In that case, the diagonal matrix $D$ is similar to $A$ and is given by \begin{equation} \label{diagmat} D= \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots& \vdots \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix} \end{equation} where $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $A.$ If $C$ is a matrix whose columns are linearly independent eigenvectors of $A$, then $D=C^{-1}A C.$

Corollary. Let $T$ be a linear transformation given by $T(x)=A x$ where $A$ is a square matrix. If $\mathcal{D}=(v_1, \ldots,v_n)$ is an eigenbasis for $T$, with $A v_i=\lambda_i v_i$, then the $\mathcal{D}$-matrix $D$ of $T$ given in \eqref{diagmat} is $$D=[ v_1, \ldots, v_n]^{-1}A [ v_1, \ldots, v_n].$$

Corollary. A matrix $A$ is diagonalizable if and only if there exists an eigenbasis for $A.$ In particular, if an $n\times n$ matrix $A$ has $n$ distinct eigenvalues, then $A$ is diagonalizable.

Example. Let $T: \mathcal{P}_2\to \mathcal{P}_2$ be the linear transformation defined by $$T(a_0+a_1 x+a_2x^2)=(a_0+a_1+a_2)+(a_1+a_2)x+a_2x^2.$$ Show that $T$ is not diagonalizable. The matrix of $T$ with respect to the usual basis $(, x, x^2)$ for $\mathcal{P}_2$ is easily seen to be $$A= \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}.$$

The characteristic polynomial is $f_A(x)=-(x-1)^3$ since $A$ is upper triangular. So $T$ has only one (repeated) eigenvalue $\lambda=1.$ A nonzero polynomial $g$ with $g(x)=a_0+a_1 x+a_2 x^2$ is an eigenvector if and only if $$\label{notdiageq} \begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix}a_0 \\ a_1 \\ a_2\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}.$$

Thus $a_1=0$ and $a_2=0$, so there is only one linearly independent eigenvector for $\lambda=1.$ Thus $T$ is not diagonalizable.

Examples

Example. Let $T:\mathcal{P}_2\to \mathcal{P}_2$ be the linear transformation defined by \begin{equation} T(f(x))=x^2f”(x)+(3x-2)f'(x)+5 f(x). \end{equation}

Find a basis for $\mathcal{P}_2$ such that the matrix representation of $T$ with respect to $\mathcal{B}$ is diagonal. Since $T(x^2)=13x^2-4x$, $T(x)=8x-2$, and $T(1)=5$ the matrix representation of $T$ with respect to the basis $\mathcal{B}=(x^2,x,1)$ is $$A=\begin{bmatrix}13 & 0 & 0 \\ -4 & 8 & 0 \\ 0 & -2 & 5 \end{bmatrix}.$$ Hence \begin{equation} f_T(x)=f_A(x)=\begin{vmatrix} x-13 & 0 & 0 \\ 4 & x-8 & 0 \\ 0 & 2 & x-5 \end{vmatrix}=(x-13)(x-8)(x-5). \end{equation}

The eigenvalues of $T$ are $\lambda_1=13$, $\lambda_2=8$ and $\lambda_3=5.$ Solving each of the homogenous systems $(A-13I_3) x= 0$, $(A-8I_3) x= 0$, and $(A-5I_3) x= 0$ yields the eigenvectors $v_1=5x^2-4x+1$, $v_2=3x-2$, and $v_3=1$, respectively. Notice $v_1, v_2, v_3$ are 3 linearly independent vectors, so $T$ is diagonalizable. We let $\mathcal{D}=( v_1, v_2, v_3)$ and since $T( v_1)=13 v_1$,$T( v_2)=8 v_2$, and $T( v_3)=5 v_3$, the matrix representation of $T$ with respect to $\mathcal{D}$ is the diagonal matrix $$\begin{bmatrix}13 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 5 \end{bmatrix}$$

The method of diagonalization is illustrated with the following example.

Example. Let $T:\mathcal{P}_3\to \mathcal{P}_3$ be the linear transformation defined by \begin{equation} T(f(x))=xf'(x)+f(x+1). \end{equation}

Find a basis for $\mathcal{P}_3$ such that the matrix representation of $T$ with respect to $\mathcal{B}$ is diagonal. Since $T(x^3)=4x^3+3x^2+3x+1$, $T(x^2)=3x^2+2x+1$, $T(x)=2x+1$, and $T(1)=1$ the matrix representation of $T$ with respect to the basis $\mathcal{B}=(x^3,x^2,x,1)$ is $$A=\begin{bmatrix} 4 & 0 & 0 & 0 \\ 3 & 3 & 0 & 0 \\ 3 & 2 & 2 & 0 \\ 1 & 1 & 1 & 1 \end{bmatrix}.$$

Since $A$ is lower triangular, $f_T(x)=f_A(x)=(x-4)(x-3)(x-2)(x-1)$; and so the eigenvalues are $\lambda_1=4$, $\lambda_2=3$, $\lambda_3=2$, and $\lambda_4=1.$ Solving for a basis for each eigenspace of $A$ yields $$E_{\lambda_1}=\left(\begin{bmatrix}6 \\ 18 \\ 27 \\ 17\end{bmatrix}\right), \quad E_{\lambda_2}=\left(\begin{bmatrix}0 \\ 2 \\ 4 \\ 3\end{bmatrix}\right), \quad E_{\lambda_3}=\left(\begin{bmatrix}0 \\ 0 \\ 1 \\ 1\end{bmatrix}\right),\quad E_{\lambda_4}=\left(\begin{bmatrix}0 \\ 0 \\ 0 \\ 1\end{bmatrix}\right).$$

By taking the polynomial corresponding to the basis vectors, we let $\mathcal{D}=( v_1, v_2, v_3, v_4)$ where $v_1=6x^3+18x^2+27x+17$, $v_2=2x^2+4x+3$, $v_3=x+1$, and $v_4=1.$ The diagonal matrix $$\begin{bmatrix} 4 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ is the matrix representation of $T$ in $\mathcal{D}$-coordinates and has the eigenvalues of $T$ on its main diagonal. The transition matrix $P$ from $\mathcal{B}$-coordinates to $\mathcal{D}$-coordinates is $$P=\begin{bmatrix} 6 & 0 & 0 & 0 \\ 18 & 2 & 0 & 0 \\ 27 & 4 & 1 & 0 \\ 17 & 3 & 1 & 1 \end{bmatrix}$$ and satisfies the required relation $D=P^{-1}AP$ as can be verified.

Example. If $A$ is similar to $B$, show that $A^n$ is similar to $B^n$, for any positive integer $n.$

Example. Suppose that $C^{-1}AC=D.$ Show that for any integer $n$, $A^n=CD^nC^{-1}.$

Example. Let $a$ and $b$ be real numbers. By diagonalizing $M= \begin{bmatrix} a & b-a \\ 0 & b \end{bmatrix},$ prove that $M^n= \begin{bmatrix} a^n & b^n-a^n \\ 0 & b^n \end{bmatrix}$ for all positive integers $n.$ We need a basis of $\mathbb{R}^2$ consisting of eigenvectors of $M.$ One such basis is $v_1= e_1$ and $v_2= e_1+ e_2$ where $a$ and $b$ are eigenvalues for corresponding to these eigenvectors, respectively. Let $P=\begin{bmatrix} v_1 & v_2\end{bmatrix}$, then the diagonalization is \begin{equation*} D=\begin{bmatrix} v_1 & v_2\end{bmatrix}^{-1}M \begin{bmatrix} v_1 & v_2\end{bmatrix}=\begin{bmatrix}a & 0 \\ 0 & b \end{bmatrix}. \end{equation*} Therefore \begin{equation*} M^n=(PDP^{-1})^n=\underbrace{(PDP^{-1})\cdots (PDP^{-1})}_{n\text{-times}}=PD^n P^{-1} =\begin{bmatrix} a^n & b^n-a^n \\ 0 & b^n \end{bmatrix}. \end{equation*} as desired.

Why is diagonalization useful? Is there orthogonal diagonalization?

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