MATH 240

Fri. November 8th, 2019

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Diagonalization Theorem

An $n\times n$ matrix $A$ is diagonalizable if and only if $A$ has $n$ linearly independent eigenvectors.

In fact, $A=PDP^{-1}$ where $D$ is some diagonal matrix if and only if the columns of of $P$ are $n$ linearly independent eigenvectors of $A$. In this case the diagonal entries of $D$ are eigenvalues of $A$ that correspond to the eigenvectors in $P$.

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Ex. Is $A=\begin{pmatrix} 5&-8&1\\0&0&7\\0&0&-2 \end{pmatrix}$ diagonalizable? If yes, express $A$ as $PDP^{-1}$ with $D$ diagonal.

$A$ is a triangular matrix, so its eigenvalues are $\lambda_1=5,\ \lambda_2=0,\ \lambda_3=-2$. Since $A$ is a $3\times 3$ matrix and it has $3$ distinct eigenvalues, we know it is diagonalizable.

Now all we have to do is calculate the eigenvectors of $A$ to construct $P$.

For $\lambda=5$:
$\begin{gathered} [A-5I\mid0]=\begin{pmatrix} 0&-8&1&0\\0&-5&7&0\\0&0&-7&0 \end{pmatrix}\\\\ x_3=0,\ x_2=0,\ x_1 \text{ free}\\ \text{Nul}(A-5I)=\text{span}\{\begin{pmatrix} 1\\0\\0 \end{pmatrix}\} \end{gathered}$

For $\lambda=0$:
$\begin{gathered} [A\mid0]=\begin{pmatrix} 5&-8&1&0\\0&0&7&0\\0&0&-2&0 \end{pmatrix}\\\\ x_3=0,\ x_2\text{ free},\ x_1=\frac{8}{5}x_2\\ \text{Nul}(A)=\text{span}\{\begin{pmatrix} \frac{8}{5}\\1\\0 \end{pmatrix}\} \end{gathered}$

For $\lambda=-2$:
$\begin{gathered} [A+2I\mid0]=\begin{pmatrix} 7&-8&1&0\\0&2&7&0\\0&0&0&0 \end{pmatrix}\\\\ x_3\text{ free},\ x_2=\frac{-7}{2}x_3,\ x_1=\frac{-29}{7}x_3\\ \text{Nul}(A+2I)=\text{span}\{\begin{pmatrix} \frac{-29}{7}\\\frac{-7}{2}\\1 \end{pmatrix}\} \end{gathered}$

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

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

Theorem.
Let $A$ be an $n\times n$ matrix whose distinct eigenvalues are $\lambda_1,...,\lambda_p$.

• For $1\leq k\leq p$, the dimension of the eigenspace for $\lambda_k$ is less than or equal to the multiplicity of $\lambda_k$.

• $A$ is diagonalizable iff if the sum of the dimensions of the eigenspaces is $n$. This occurs iff either of the following occur:

  1. The characteristic polynomial factors completely into distinct linear factors.
     (This is the case we dealt with before; this is true because each value’s eigenspace will have a multiplicity & dimension of $1$)
  2. The dimension of the eigenspace for $\lambda_k$ equals the multiplicity of $\lambda_k$ in the characteristic polynomial.

• If $A$ is diagonalizable and $B_k$ is a bsais for the eigenspace corresponding to $\lambda_k$ for each $k$, then the total collection of vectors in the sets $B_1,...,B_p$ forms a basis for $\R^n$.

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

$\begin{aligned} A&=\begin{pmatrix} 1&0\\1&1 \end{pmatrix}\\ \text{det}(A-\lambda I)&=\begin{vmatrix} 1-\lambda&0\\1&1-\lambda \end{vmatrix}\\ &=(1-\lambda)^2=0\\ \lambda&=1\\\\ A-1I&=\begin{pmatrix} 0&0\\1&0 \end{pmatrix}\\ \text{Nul}(A-1I)&=\text{span}\{\begin{pmatrix} 0\\1 \end{pmatrix}\} \end{aligned}$ $A$ is not diagonalizable, as $\lambda$ has multiplicity $2$ in the characteristic equation, but the dimension of its eigenspace is only $1$.