MATH 240

Fri. November 8th, 2019


Diagonalization Theorem

An n×nn\times n matrix AA is diagonalizable if and only if AA has nn linearly independent eigenvectors.

In fact, A=PDP1A=PDP^{-1} where DD is some diagonal matrix if and only if the columns of of PP are nn linearly independent eigenvectors of AA. In this case the diagonal entries of DD are eigenvalues of AA that correspond to the eigenvectors in PP.


Ex. Is A=(581007002)A=\begin{pmatrix} 5&-8&1\\0&0&7\\0&0&-2 \end{pmatrix} diagonalizable? If yes, express AA as PDP1PDP^{-1} with DD diagonal.

AA is a triangular matrix, so its eigenvalues are λ1=5, λ2=0, λ3=2\lambda_1=5,\ \lambda_2=0,\ \lambda_3=-2. Since AA is a 3×33\times 3 matrix and it has 33 distinct eigenvalues, we know it is diagonalizable.

Now all we have to do is calculate the eigenvectors of AA to construct PP.

For λ=5\lambda=5:
[A5I0]=(081005700070)x3=0, x2=0, x1 freeNul(A5I)=span{(100)}\begin{gathered} [A-5I\mid0]=\begin{pmatrix} 0&-8&1&0\\0&-5&7&0\\0&0&-7&0 \end{pmatrix}\\\\ x_3=0,\ x_2=0,\ x_1 \text{ free}\\ \text{Nul}(A-5I)=\text{span}\{\begin{pmatrix} 1\\0\\0 \end{pmatrix}\} \end{gathered}

For λ=0\lambda=0:
[A0]=(581000700020)x3=0, x2 free, x1=85x2Nul(A)=span{(8510)}\begin{gathered} [A\mid0]=\begin{pmatrix} 5&-8&1&0\\0&0&7&0\\0&0&-2&0 \end{pmatrix}\\\\ x_3=0,\ x_2\text{ free},\ x_1=\frac{8}{5}x_2\\ \text{Nul}(A)=\text{span}\{\begin{pmatrix} \frac{8}{5}\\1\\0 \end{pmatrix}\} \end{gathered}

For λ=2\lambda=-2:
[A+2I0]=(781002700000)x3 free, x2=72x3, x1=297x3Nul(A+2I)=span{(297721)}\begin{gathered} [A+2I\mid0]=\begin{pmatrix} 7&-8&1&0\\0&2&7&0\\0&0&0&0 \end{pmatrix}\\\\ x_3\text{ free},\ x_2=\frac{-7}{2}x_3,\ x_1=\frac{-29}{7}x_3\\ \text{Nul}(A+2I)=\text{span}\{\begin{pmatrix} \frac{-29}{7}\\\frac{-7}{2}\\1 \end{pmatrix}\} \end{gathered}

P=(1852970172001)D=(500000002)A=PDP1.\begin{aligned} \\ P&=\begin{pmatrix} 1&\frac{8}{5}&\frac{-29}{7}\\0&1&\frac{-7}{2}\\0&0&1 \end{pmatrix}\\ D&=\begin{pmatrix} 5&0&0\\0&0&0\\0&0&-2 \end{pmatrix}\\ A&=PDP^{-1}. \end{aligned}


What happens if a matrix doesn’t have nn distinct eigenvalues?

Theorem.
Let AA be an n×nn\times n matrix whose distinct eigenvalues are λ1,...,λp\lambda_1,...,\lambda_p.


Ex. Determine if A=(1011)A=\begin{pmatrix} 1&0\\1&1 \end{pmatrix} is diagonalizable.

A=(1011)det(AλI)=1λ011λ=(1λ)2=0λ=1A1I=(0010)Nul(A1I)=span{(01)}\begin{aligned} A&=\begin{pmatrix} 1&0\\1&1 \end{pmatrix}\\ \text{det}(A-\lambda I)&=\begin{vmatrix} 1-\lambda&0\\1&1-\lambda \end{vmatrix}\\ &=(1-\lambda)^2=0\\ \lambda&=1\\\\ A-1I&=\begin{pmatrix} 0&0\\1&0 \end{pmatrix}\\ \text{Nul}(A-1I)&=\text{span}\{\begin{pmatrix} 0\\1 \end{pmatrix}\} \end{aligned} AA is not diagonalizable, as λ\lambda has multiplicity 22 in the characteristic equation, but the dimension of its eigenspace is only 11.