Diagonal Matrices

Finding powers of a diagonal matrix, is very simple.

$${\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]}^k=\left[\begin{array}{cc}{a}^k& 0\\ 0& b^k\end{array}\right]$$

Diagonalizable

If $A$ is similar to diagonal matrix $D$ ($A=PD{P}^{-1}$) then A is diagonalizable.

$$\begin{array}{rl}A& =PD{P}^{-1}\\ A& =\left[\begin{array}{ccc}{v}_1& v_2\dots & v_n\end{array}\right]\left[\begin{array}{cccc}{\lambda }_1& & & \\ & {\lambda }_2& & \\ & & \ddots & \\ & & & {\lambda }_n\end{array}\right]{\left[\begin{array}{ccc}{v}_1& v_2\dots & v_n\end{array}\right]}^{-1}\end{array}$$

IMPORTANT

$v_1\dots v_n$ are eigenvectors ${\lambda }_1\dots {\lambda }_n$ are eigenvalue $A$ is diagonalizable if and only if $A$ has $n$ linearly independent eigenvectors. Invertibility has no effect on diagonalizability

Proof

---

Example

Diagonalize $A=\left[\begin{array}{cc}2& 6\\ 0& -1\end{array}\right]$ Eigenvalue $=2,-1$ Eigenvectors $=\left[\begin{array}{c}1\\ 0\end{array}\right],\left[\begin{array}{c}2\\ -1\end{array}\right]$

$$\begin{array}{rl}A& =PD{P}^{-1}\\ & =\left[\begin{array}{cc}1& 2\\ 0& -1\end{array}\right]\left[\begin{array}{cc}2& 0\\ 0& -1\end{array}\right]{\left[\begin{array}{cc}1& 2\\ 0& -1\end{array}\right]}^{-1}\\ & =\left[\begin{array}{cc}1& 2\\ 0& -1\end{array}\right]\left[\begin{array}{cc}2& 0\\ 0& -1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& -1\end{array}\right]\end{array}$$

Note

This is a special case and is not always true,
${\left[\begin{array}{cc}1& 2\\ 0& -1\end{array}\right]}^{-1}=\left[\begin{array}{cc}1& 2\\ 0& -1\end{array}\right]$

If $A$ is $n\times n$ and has $n$ distinct eigenvalues, then $A$ is diagonalizable. $A$ does not have to have $n$ distinct eigenvalues for $A$ to be diagonalizable.

Properties

• $A$ is diagonalizable if and only if $A$ has $n$ linearly independent eigenvectors.
• Invertibility has no effect on diagonalizability.
• If A has $n$ distinct eigenvalues, then $A$ is diagonalizable. (The converse is not necessarily true)

Find Diagonalizability

$A=PD{P}^{-1}$ Finding $D$ is straight forward. It is simply the eigenvalue of $A$. $P$ MUST be Invertible.

• If $\sum g_i=n$, $A$ is diagonalizable.
• If $g_i=a_i$ for all $i$, $A$ is diagonalizable.
• If the eigenvectors of $A$ form a basis in ${\mathbb{R}}^n$, $A$ is diagonalizable.

Repeated Eigenvalue

Diagonalize $A=\left[\begin{array}{ccc}7& 4& 16\\ 2& 5& 8\\ -2& -2& -5\end{array}\right]$ with eigenvalues ${\lambda }_1=1,{\lambda }_2={\lambda }_3=3$.

Find eigenvectors of ${\lambda }_1=1$:

$\text{Null}(A-{\lambda }_{1}I)=\left[\begin{array}{c}2\\ 1\\ -1\end{array}\right]$

Find eigenvectors of ${\lambda }_2={\lambda }_3=3$:

$\text{Null}(A-{\lambda }_{2}I)=\left\{\left[\begin{array}{c}-1\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 1\end{array}\right]\right\}$

Find $P$

$$\begin{array}{rl}P& =\left[\begin{array}{ccc}\overrightarrow{v_1}& \overrightarrow{v_2}& \overrightarrow{v_3}\end{array}\right]\\ & =\left[\begin{array}{ccc}2& -1& -4\\ 1& 1& 0\\ -2& 0& 1\end{array}\right]\end{array}$$

Find $D$

$$\begin{array}{rl}D& =\left[\begin{array}{cccc}{\lambda }_1& & & \\ & {\lambda }_2& & \\ & & \ddots & \\ & & & {\lambda }_4\end{array}\right]\\ & =\left[\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]\end{array}$$

---