1 Diagonalization of a matrix with distinct eigenvalues

Diagonalization means transforming a non-diagonal matrix into an equivalent matrix which is diagonal and hence is simpler to deal with.

A matrix $A$ with distinct eigenvalues has, as we mentioned in Property 3 in HELM booklet  22.1, eigenvectors which are linearly independent. If we form a matrix $P$ whose columns are these eigenvectors, it can be shown that

$det\; <!--nolimits\; -->\left(P\right)\ne 0$

so that $P^{-1}$ exists.

The product $P^{-1}AP$ is then a diagonal matrix $D$ whose diagonal elements are the eigenvalues of $A$ . Thus if ${\lambda }_1,{\lambda }_2,\dots {\lambda }_n$ are the distinct eigenvalues of $A$ with associated eigenvectors $X^{\left(1\right)},X^{\left(2\right)},\dots ,X^{\left(n\right)}$ respectively, then

$P=\left[\begin{array}{ccccccc}\hfill X^{\left(1\right)}\hfill & \hfill \&vellip;\hfill & \hfill X^{\left(2\right)}\hfill & \hfill \&vellip;\hfill & \hfill \&ctdot;\hfill & \hfill \&vellip;\hfill & \hfill X^{\left(n\right)}\hfill \\ \hfill \hfill \\ \hfill \hfill \end{array}\right]$

will produce a product

$P^{-1}AP=D=\left[\begin{array}{cccc}\hfill {\lambda }_1\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {\lambda }_2\hfill & \hfill \dots \hfill & \hfill 0\hfill \\ \hfill \&vellip;\hfill \\ \hfill 0\hfill & \hfill \dots \hfill & \hfill \dots \hfill & \hfill {\lambda }_n\hfill \end{array}\right]$

We see that the order of the eigenvalues in $D$ matches the order in which $P$ is formed from the eigenvectors.

N.B.

1. The matrix $P$ is called the modal matrix of $A$
2. Since $D$ is a diagonal matrix with eigenvalues ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$ which are the same as those of $A$ , then the matrices $D$ and $A$ are said to be similar .
3. The transformation of $A$ into $D$ using

   $P^{-1}AP=D$

   is said to be a similarity transformation.

Example 8

Let $A=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 2\hfill \end{array}\right]$ . Obtain the modal matrix $P$ and calculate the product $P^{-1}AP$ . (The eigenvalues and eigenvectors of this particular matrix $A$ were obtained earlier in this Workbook at page 7.)

Solution

The matrix $A$ has two distinct eigenvalues ${\lambda }_1=-1$ , ${\lambda }_2=5$ with corresponding eigenvectors $X_1=\left[\begin{array}{c}\hfill x\\ \hfill -x\end{array}\right]$ and $X_2=\left[\begin{array}{c}\hfill x\hfill \\ \hfill x\hfill \end{array}\right]$ . We can therefore form the modal matrix from the simplest eigenvectors of these forms:

$P=\left[\begin{array}{cc}\hfill 1& \hfill 1\\ \hfill -1& \hfill 1\end{array}\right]$

(Other eigenvectors would be acceptable e.g. we could use $P=\left[\begin{array}{cc}\hfill 2& \hfill 3\\ \hfill -2& \hfill 3\end{array}\right]$ but there is no reason to over complicate the calculation.)

It is easy to obtain the inverse of this $2\times 2$ matrix $P$ and the reader should confirm that:

$P^{-1}=\frac{1}{\text{det}\left(P\right)}\text{adj}\left(P\right)=\frac{1}{2}{\left[\begin{array}{cc}\hfill 1& \hfill 1\\ \hfill -1& \hfill 1\end{array}\right]}^T=\frac{1}{2}\left[\begin{array}{cc}\hfill 1& \hfill -1\\ \hfill 1& \hfill 1\end{array}\right]$

We can now construct the product $P^{-1}AP$ :

which is a diagonal matrix with entries the eigenvalues, as expected. Show (by repeating the method outlined above) that had we defined $P=\left[\begin{array}{cc}\hfill 1& \hfill 1\\ \hfill 1& \hfill -1\end{array}\right]$ (i.e. interchanged the order in which the eigenvectors were taken) we would find $P^{-1}AP=\left[\begin{array}{cc}\hfill 5& \hfill 0\\ \hfill 0& \hfill -1\end{array}\right]$ (i.e. the resulting diagonal elements would also be interchanged.)

Task!

The matrix $A=\left[\begin{array}{cc}\hfill -1& \hfill 4\hfill \\ \hfill 0& \hfill 3\hfill \end{array}\right]$ has eigenvalues $-1$ and 3 with respective

eigenvectors   $\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right]$ and $\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]$ .

If $P_1=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right],P_2=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right],$ $P_3=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ write down the

products $P_1^{-1}A{P}_1,P_2^{-1}A{P}_2,P_3^{-1}A{P}_3$

(You may not need to do detailed calculations.)

Answer

1.1 Matrix powers

If $P^{-1}AP=D$ then we can obtain $A$ (i.e. make $A$ the subject of this matrix equation) as follows:

Multiplying on the left by $P$ and on the right by $P^{-1}$ we obtain

$P{P}^{-1}AP{P}^{-1}=PD{P}^{-1}$

Now using the fact that   $P{P}^{-1}=P^{-1}P=I$ we obtain

$IAI=PD{P}^{-1}\text{and so}$

$A=PD{P}^{-1}$

We can use this result to obtain the powers of a square matrix, a process which is sometimes useful in control theory. Note that

$A^2=A.A{A}^3=A.A.A.\text{ etc.}$

Clearly, obtaining high powers of $A$ directly would in general involve many multiplications. The process is quite straightforward, however, for a diagonal matrix $D$ , as this next Task shows.

Task!

Obtain $D^2$ and $D^3$ if $D=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 0\\ \hfill 0\hfill & \hfill -2\end{array}\right]$ . Write down $D^{10}$ .

Answer

We now use the relation $A=PD{P}^{-1}$   to obtain a formula for powers of $A$ in terms of the easily calculated powers of the diagonal matrix $D$ :

$A^2=A.A=\left(PD{P}^{-1}\right)\left(PD{P}^{-1}\right)=PD\left(P^{-1}P\right)D{P}^{-1}=PDID{P}^{-1}=P{D}^2{P}^{-1}$

Similarly: $A^3=A^2.A=\left(P{D}^2{P}^{-1}\right)\left(PD{P}^{-1}\right)=P{D}^2\left(P^{-1}P\right)D{P}^{-1}=P{D}^3{P}^{-1}$

The general result is given in the following Key Point:

Example 9

If $A=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 2\hfill \end{array}\right]$ find $A^{23}$ . (Use the results of Example 8.)

Solution

We know from Example 8 that if $P=\left[\begin{array}{cc}\hfill 1& \hfill 1\\ \hfill -1& \hfill 1\end{array}\right]$ then $P^{-1}AP=\left[\begin{array}{cc}\hfill -1& \hfill 0\\ \hfill 0& \hfill 5\end{array}\right]=D$

where $P^{-1}=\frac{1}{2}\left[\begin{array}{cc}\hfill 1& \hfill -1\\ \hfill 1& \hfill 1\end{array}\right]$

$\therefore A=PD{P}^{-1}\text{ and}{A}^{23}=P{D}^{23}{P}^{-1}\text{ using the general result in Key Point 2}$

$\text{ i.e.}A=\left[\begin{array}{cc}\hfill 1& \hfill 1\\ \hfill -1& \hfill 1\end{array}\right]\left[\begin{array}{cc}\hfill -1& \hfill 0\\ \hfill 0& \hfill 5^{23}\end{array}\right]\left[\begin{array}{cc}\hfill 1& \hfill -1\\ \hfill 1& \hfill 1\end{array}\right]$

which is easily evaluated.

Exercise

Find a diagonalizing matrix $P$ if

1. $A=\left[\begin{array}{cc}\hfill 4\hfill & \hfill 2\hfill \\ \hfill -1\hfill & \hfill 1\hfill \\ \hfill \hfill \end{array}\right]$
2. $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill -2\hfill & \hfill 3\hfill \\ \hfill \hfill \end{array}\right]$

Verify, in each case, that $P^{-1}AP$ is diagonal, with the eigenvalues of $A$ as its diagonal elements.

Answer

1. $P=\left[\begin{array}{cc}\hfill -1\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$ , $PA{P}^{-1}=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 3\hfill \end{array}\right]$
2. $P=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -2\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\right]$ ,    $PA{P}^{-1}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]$