Definition

Let $A\in {\mathbb{C}}^{n\times n}$. $A$ is said to be diagonalizable if there exist $P$ and $D$ in ${\mathbb{C}}^{n\times n}$ such that $D$ is a diagonal matrix and $A=PD{P}^{-1}$.

Testing if a matrix is diagonalizable

$A$ is diagonalizable if and only if for every eigenvalue $\lambda$ of $A$, the algebraic multiplicity of $\lambda$ is equal to the geometric multiplicity of $\lambda$.

An equivalent characterization is that the sum of the geometric multiplicities of the eigenvalues of $A$ is $n$.

Examples

1. Let $A=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right].$ Note that $p_A=(1-\lambda {)}^2$. Hence, the only eigenvalue of $A$ is $1$. Now $A-I=\left[\begin{array}{cc}0& 2\\ 0& 0\end{array}\right].$ The nullity of this matrix is $1$, implying that the geometric multiplicity is $1$, not $2$. So $A$ is not diagonalizable.

2. Let $A=\left[\begin{array}{ccc}4& 0& -2\\ 2& 5& 4\\ 0& 0& 5\end{array}\right].$ Note that $p_A=(4-\lambda )(5-\lambda {)}^2$. So the eigenvalues are $4$ and $5$.

   The geometric multiplicity of $4$ is $1$ since the geometric multiplicity cannot be $0$ and cannot exceed the algebraic multiplity of $4$, which is $1$.

   Now, note that $A-5I=\left[\begin{array}{ccc}-1& 0& -2\\ 2& 0& 4\\ 0& 0& 0\end{array}\right]$. This matrix has rank $1$ since the second row is $-2$ times the first row and the third row is a row of $0$'s. Hence, the nullity of $A-5I$ is $2$, implying that the geometric multiplicity of $5$ is $2$.

   So $A$ is diagonalizable as the sum of the geometric multiplities is $3$.

Finding $P$ and $D$

Now that we know the matrix $A$ in the second example above is diagonalizable, how do we actually find $P$ and $D$?

First, observe that $A=PD{P}^{-1}$ can be rewritten as $AP=PD$. Column $i$ of the left-hand side is given by $A{P}_i$ where $P_i$ denotes the $i$ column of $P$ and column $i$ of the right-hand side is given by $P{D}_i$. Since $D$ is a diagonal matrix, $P{D}_i=D_{i,i}{P}_i$. Thus $P_i$ must be an eigenvector with eigenvalue $D_{i,i}$. Since we need $P$ to be invertible, we need $n$ linearly independent eigenvectors. This is why we need the sum of the geometric multiplicities to equal $n$.

To form $D$, first list all the eigenvalues (including multiple appearances) in any order. (Hence, an eigenvalue with algebraic multiplicity $k$ will appear $k$ times in the list.) Then form $D$ with this list of values as the diagonal.

For our example, the list could be $4,5,5$. So we can set $D=\left[\begin{array}{ccc}4& 0& 0\\ 0& 5& 0\\ 0& 0& 5\end{array}\right]$.

To form $P$, first find a basis for each eigenspace. To form the first column of $P$, take a vector from the basis for the eigenspace of the first diagonal entry of $D$ and remove that vector from future inclusion into $P$. For each subsequent column $i$, take a vector from the basis for the eigenspace of the $i$th diagonal entry of $D$ and remove it from future includsion into $P$.

For our example, we first need to find a basis for the eigenspace of $4$ and a basis for the eigenspace of $5$.

For the eigenvalue $4$, we want to find a basis for the nullspace of $A-4I=\left[\begin{array}{ccc}0& 0& -2\\ 2& 1& 4\\ 0& 0& 1\end{array}\right]$. Row-reducing this matrix to RREF gives the matrix $\left[\begin{array}{ccc}1& \frac{1}{2}& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]$. Hence, all solutions to $\left[\begin{array}{ccc}1& \frac{1}{2}& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$ are of the form $\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}-\frac{1}{2}s\\ s\\ 0\end{array}\right]=s\left[\begin{array}{c}-\frac{1}{2}\\ 1\\ 0\end{array}\right]$ since $x_2$ is a free variable. Thus, the single vector $\left[\begin{array}{c}-\frac{1}{2}\\ 1\\ 0\end{array}\right]$ forms a basis for the eigenspace.

For the eigenvalue $5$, we want to find a basis for the nullspace of $A-5I=\left[\begin{array}{ccc}-1& 0& -2\\ 2& 0& 4\\ 0& 0& 0\end{array}\right]$. Row-reducing this matrix gives $\left[\begin{array}{ccc}1& 0& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$. Hence, all solutions to $\left[\begin{array}{ccc}1& 0& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$ are of the form $\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}-2t\\ s\\ t\end{array}\right]=s\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}-2\\ 0\\ 1\end{array}\right]$ since $x_2$ and $x_3$ are free variables. Hence, a basis for the nullspace is $\{\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-2\\ 0\\ 1\end{array}\right]\}$.

To form the first column of $P$, we take a vector from the basis for the eigenspace of $4$. There is only one such vector. So $P$ is now $P=\left[\begin{array}{ccc}-\frac{1}{2}& \ast & \ast \\ 1& \ast & \ast \\ 0& \ast & \ast \end{array}\right]$.

For the second column, we take a vector from the basis for the eigenspace of $5$. There are two vectors to choose from. We choose $\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$. So $P$ is now $P=\left[\begin{array}{ccc}-\frac{1}{2}& 0& \ast \\ 1& 1& \ast \\ 0& 0& \ast \end{array}\right]$. And we have no choice but to take the remaining vector in the basis for eigenspace of $5$ to fill the last column of $P$. Hence, $P=\left[\begin{array}{ccc}-\frac{1}{2}& 0& -2\\ 1& 1& 0\\ 0& 0& 1\end{array}\right]$.

Note that $P^{-1}=\left[\begin{array}{ccc}-2& 0& -4\\ 2& 1& 4\\ 0& 0& 1\end{array}\right]$. One can now verify that $A=PD{P}^{-1}$.

Quick Quiz

Exercises

1. Let $A=\left[\begin{array}{cc}1& -2\\ 1& 0\end{array}\right]$, $D=\left[\begin{array}{cc}\frac{1+\sqrt{7}i}{2}& 0\\ 0& \frac{1-\sqrt{7}i}{2}\end{array}\right]$, and $P=\left[\begin{array}{cc}\frac{1+\sqrt{7}i}{2}& \frac{1-\sqrt{7}i}{2}\\ 1& 1\end{array}\right].$

   1. Find $P^{-1}$.  

      $P^{-1}=\left[\begin{array}{cc}-\frac{\sqrt{7}i}{7}& \frac{7+\sqrt{7}i}{14}\\ \frac{\sqrt{7}i}{7}& \frac{7-\sqrt{7}i}{14}\end{array}\right]$

   2. Check that $A=PD{P}^{-1}$.

2. Let $A=\left[\begin{array}{cc}8& -9\\ 4& -4\end{array}\right]$. You are given that $2$ is the only eigenvalue of $A$. Give a basis for the eigenspace of this eigenvalue.

   Note that $A-2I=\left[\begin{array}{cc}6& -9\\ 4& -6\end{array}\right]$ We row-reduce this matrix to RREF: $$\begin{array}{rl}& \left[\begin{array}{cc}6& -9\\ 4& -6\end{array}\right]\\ \stackrel{R_1\leftarrow \frac{1}{6}{R}_1}{\to }& \left[\begin{array}{cc}1& -\frac{3}{2}\\ 4& -6\end{array}\right]\\ \stackrel{R_2\leftarrow R_2-4R_1}{\to }& \left[\begin{array}{cc}1& -\frac{3}{2}\\ 0& 0\end{array}\right]\end{array}$$ Note that the second column is the only nonpivot column. Thus, a basis for the eigenspace has exactly one vector and we can take any nonzero vector in the nullspace of $A-2I$. One such vector is $\left[\begin{array}{c}\frac{3}{2}\\ 1\end{array}\right]$. Hence, a basis for the eigenspace is $\left\{\left[\begin{array}{c}\frac{3}{2}\\ 1\end{array}\right]\right\}$.

   Remark. Since we are given that 2 is the only eigenvalue and the dimension of the eigenspace of this eigevalue is less than the total number of columns of $A$, we see that $A$ is not diagonalizable.