Similar Matrices

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In Section ?? we discussed solutions to differential equations $\dot {X}=CX$ for three classes of matrices $C$ . See Table ??. We stated that in a certain sense every $2\times 2$ matrix can be thought of as a member of one of these families. In this section we show that every $2\times 2$ matrix is similar to one of the matrices in that table (see Theorem ??), where similarity is defined as follows.

The $n\times n$ matrices $B$ and $C$ are similar if there exists an invertible $n\times n$ matrix $P$ such that $C = P\inv BP.$

Our present interest in similar matrices stems from the fact that if we know the solutions to the system of differential equations $\dot {Y}=CY$ in closed form, then we know the solutions to the system of differential equations $\dot {X}=BX$ in closed form. More precisely,

Suppose that $B$ and $C=P\inv BP$ are similar matrices. If $Y(t)$ is a solution to the system of differential equations $\dot {Y}=CY$ , then $X(t)=PY(t)$ is a solution to the system of differential equations $\dot {X}=BX$ .

Proof: Since the entries in the matrix $P$ are constants, it follows that $\frac {dX}{dt} = P\frac {dY}{dt}.$ Since $Y(t)$ is a solution to the $\dot {Y}=CY$ equation, it follows that $\frac {dX}{dt} = PCY.$ Since $Y=P\inv X$ and $PCP\inv = B$ , $\frac {dX}{dt} = PCP\inv X = BX.$ Thus $X(t)$ is a solution to $\dot {X}=BX$ , as claimed. $\blacksquare$

Invariants of Similarity

Let $A$ and $B$ be similar $2\times 2$ matrices. Then

$\begin{eqnarray*} p_A(\lambda) & = & p_B(\lambda),\\ \det(A) & = & \det(B),\\ \trace(A) & = & \trace(B), \end{eqnarray*}$

and the eigenvalues of $A$ and $B$ are equal.

Proof: The determinant is a function on $2\times 2$ matrices that has several important properties. Recall, in particular, from Chapter ??, Theorem ?? that for any pair of $2\times 2$ matrices $A$ and $B$ : $\begin{equation} \label{e:detprod} \det(AB) = \det(A)\det(B), \end{equation}$ and for any invertible $2\times 2$ matrix $P$ $\begin{equation} \label{e:detinv} \det(P\inv) = \frac{1}{\det(P)}. \end{equation}$
Let $P$ be an invertible $2\times 2$ matrix so that $B=P\inv AP$ . Using (??) and (??) we see that
$\begin{eqnarray*} p_B(\lambda) & = & \det(B-\lambda I_2) \\ & = & \det(P\inv AP-\lambda I_2) \\ & = & \det(P\inv(A-\lambda I_2)P) \\ & = & \det(A-\lambda I_2) \\ & = & p_A(\lambda). \end{eqnarray*}$
Hence the eigenvalues of $A$ and $B$ are the same. It follows from (??) and (??) of Section ?? that the determinants and traces of $A$ and $B$ are equal. $\blacksquare$

For example, if $A = \mattwo {-1}{0}{0}{1} \AND P = \mattwo {1}{2}{1}{1},$ then $P\inv = \mattwo {-1}{2}{1}{-1}$ and $P\inv AP = \mattwo {3}{4}{-2}{-3}.$ A calculation shows that $\det (P\inv AP)=-1=\det (A) \AND {\rm tr}(P\inv AP)=0={\rm tr}(A),$ as stated in Lemma ??.

Similarity and Matrix Exponentials

We introduce similarity at this juncture for the following reason: if $C$ is a matrix that is similar to $B$ , then $e^C$ can be computed from $e^B$ . More precisely:

Let $C$ and $B$ be $n\times n$ similar matrices, and let $P$ be an invertible $n\times n$ matrix such that $C=P\inv BP.$ Then $\begin{equation} \label{e:similarexp} e^C = P\inv e^BP. \end{equation}$

Proof: Note that for all powers of $k$ we have $(P\inv BP)^k = P\inv B^kP.$ Next verify (??) by computing $\begin{align*} e^C &=\sum^{\infty}_{k=0} \frac{1}{k!}C^k = \sum^{\infty}_{k=0} \frac{1}{k!}(P\inv BP)^k \\ &= \sum^{\infty}_{k=0} \frac{1}{k!}P\inv B^kP = P\inv\left(\sum^{\infty}_{k=0} \frac{1}{k!}B^k\right)P = P\inv e^B P. \end{align*}$ $\blacksquare$

Classification of $2\times 2$ Matrices

We now classify all $2\times 2$ matrices up to similarity.

Let $C$ and $P=(v_1|v_2)$ be $2\times 2$ matrices where the vectors $v_1$ and $v_2$ are specified below.

Suppose that $C$ has two linearly independent real eigenvectors $v_1$ and $v_2$ with real eigenvalues $\lambda _1$ and $\lambda _2$ . Then $P\inv CP = \mattwoc {\lambda _1}{0}{0}{\lambda _2}.$
Suppose that $C$ has no real eigenvectors and complex conjugate eigenvalues $\sigma \pm i\tau$ where $\tau \neq 0$ . Then $P\inv CP = \mattwo {\sigma }{-\tau }{\tau }{\sigma },$ where $v_1 + iv_2$ is an eigenvector of $C$ associated with the eigenvalue $\lambda _1=\sigma -i\tau$ .
Suppose that $C$ has exactly one linearly independent real eigenvector $v_1$ with real eigenvalue $\lambda _1$ . Then $P\inv CP = \mattwoc {\lambda _1}{1}{0}{\lambda _1},$ where $v_2$ is a generalized eigenvector of $C$ that satisfies $\begin{equation} \label{e:Cw=lw+v} (C-\lambda_1 I_2) v_2 = v_1. \end{equation}$

Proof

The strategy in the proof of this theorem is to determine the $1^{st}$ and $2^{nd}$ columns of $P\inv CP$ by computing (in each case) $P\inv CPe_j$ for $j=1$ and $j=2$ . Note from the definition of $P$ that $Pe_1 = v_1 \AND Pe_2 = v_2.$ In addition, if $P$ is invertible, then $P\inv v_1 = e_1 \AND P\inv v_2 = e_2.$ Note that if $v_1$ and $v_2$ are linearly independent, then $P$ is invertible.

(a) Since $v_1$ and $v_2$ are assumed to be linearly independent, $P$ is invertible. So we can compute $P\inv CPe_1 = P\inv C v_1 = \lambda P\inv v_1 = \lambda e_1.$ It follows that the $1^{st}$ column of $P\inv CP$ is $\vectwoc {\lambda _1}{0}.$ Similarly, the $2^{nd}$ column of $P\inv CP$ is $\vectwoc {0}{\lambda _2}$ thus verifying (a).

(b) Lemma ?? implies that $v_1$ and $v_2$ are linearly independent and hence that $P$ is invertible. Using (??), with $\tau$ replaced by $-\tau$ , $v$ replaced by $v_1$ , and $w$ replaced by $w_1$ , we calculate $P\inv CPe_1 = P\inv Cv_1 = \sigma P\inv v_1 + \tau P\inv v_2 = \sigma e_1 + \tau e_2,$ and $P\inv CPe_2 = P\inv Cv_2 = -\tau P\inv v_1 + \sigma P\inv v_2 = -\tau e_1 + \sigma e_2.$ Thus the columns of $P\inv CP$ are $\vectwo {\sigma }{\tau } \AND \vectwo {-\tau }{\sigma },$ as desired.

(c) Let $v_1$ be an eigenvector and assume that $v_2$ is a generalized eigenvector satisfying (??). By Lemma ?? the vectors $v_1$ and $v_2$ exist and are linearly independent.

For this choice of $v_1$ and $v_2$ , compute $P\inv CPe_1 = P\inv Cv_1 = \lambda _1 P\inv v_1 = \lambda _1 e_1,$ and $P\inv CPe_2 = P\inv Cv_2 = P\inv v_1+\lambda _1 P\inv v_2 = e_1+\lambda _1 e_2.$ Thus the two columns of $P\inv CP$ are: $\vectwoc {\lambda _1}{0} \AND \vectwoc {1}{\lambda _1}.$ $\blacksquare$

Closed Form Solutions Using Similarity

We now use Lemma ??, Theorem ??, and the explicit solutions to the normal form equations Table ?? to find solutions for $\dot {X}=CX$ where $C$ is any $2\times 2$ matrix. The idea behind the use of similarity to solve systems of ODEs is to transform a given system into another normal form system whose solution is already known. This method is very much like the technique of change of variables used when finding indefinite integrals in calculus.

We suppose that we are given a system of differential equations $\dot {X}=CX$ and use Theorem ?? to transform $C$ by similarity to one of the normal form matrices listed in that theorem. We then solve the transformed equation, as we did in Section ?? (see Table ??), and use Lemma ?? to transform the solution back to the given system.

For example, suppose that $C$ has a complex eigenvalue $\sigma -i\tau$ with corresponding eigenvector $v+iw$ . Then Theorem ?? states that $B = P\inv CP = \mattwo {\sigma }{-\tau }{\tau }{\sigma },$ where $P=(v|w)$ is an invertible matrix. Using Table ?? the general solution to the system of equations $\dot {Y}=BY$ is: $Y(t) = e^{\sigma t} \mattwo {\cos (\tau t)}{-\sin (\tau t)}{\sin (\tau t)}{\cos (\tau t)} \vectwo {\alpha }{\beta }.$ Lemma ?? states that $X(t) = PY(t)$ is the general solution to the $\dot {X}=CX$ system. Moreover, we can solve the initial value problem by solving $X_0 = PY(0) = P\vectwo {\alpha }{\beta }$ for $\alpha$ and $\beta$ . In particular, $\vectwo {\alpha }{\beta } = P\inv X_0.$ Putting these steps together implies that

$\begin{equation} \label{e:exp0ev} X(t) = e^{\sigma t} P\mattwo{\cos(\tau t)}{-\sin(\tau t)}{\sin(\tau t)}{\cos(\tau t)}P\inv X_0 \end{equation}$ is the solution to the initial value problem.

The Example with Complex Eigenvalues Revisited

Recall the example in (??) $\frac {dX}{dt} = \mattwo {-1}{2}{-5}{-3} X,$ with initial values $X_0=\vectwo {1}{1}.$ This linear system has a complex eigenvalue $\sigma -i\tau =-2-3i$ with corresponding eigenvector $v+iw = \vectwoc {2}{-1-3i}.$ Thus the matrix $P$ that transforms $C$ into normal form is $P = \mattwo {2}{0}{-1}{-3} \AND P\inv = \frac {1}{6}\mattwo {3}{0}{-1}{-2}.$ It follows from (??) that the solution to the initial value problem is

$\begin{eqnarray*} X(t) & = & e^{-2t}P\mattwo{\cos(3t)}{-\sin(3t)}{\sin(3t)}{\cos(3t)}P\inv X_0 \\ & = & \frac{1}{6}e^{-2t}\mattwo{2}{0}{-1}{-3} \mattwo{\cos(3t)}{-\sin(3t)}{\sin(3t)}{\cos(3t)} \mattwo{3}{0}{-1}{-2}\vectwo{1}{1}. \end{eqnarray*}$

A calculation gives

$\begin{eqnarray*} X(t) & = & \frac{1}{2}e^{-2t}\mattwo{2}{0}{-1}{-3} \mattwo{\cos(3t)}{-\sin(3t)}{\sin(3t)}{\cos(3t)}\vectwo{1}{-1} \\ & = & e^{-2t} \vectwoc{\cos(3t)+\sin(3t)}{\cos(3t)-2\sin(3t)}. \end{eqnarray*}$

Thus the solution to (??) that we have found using similarity of matrices is identical to the solution (??) that we found by the direct method.

Solving systems with either distinct real eigenvalues or equal eigenvalues works in a similar fashion.

Exercises

Suppose that the matrices $A$ and $B$ are similar and the matrices $B$ and $C$ are similar. Show that $A$ and $C$ are also similar matrices.

Use (??) in Chapter ?? to verify that the traces of similar matrices are equal.

In Exercises ?? – ?? determine whether or not the given matrices are similar, and why.

$A = \mattwo {1}{2}{3}{4} \AND B = \mattwo {2}{-2}{-3}{8}$ .

$C = \mattwo {2}{2}{2}{2} \AND D = \mattwo {4}{-2}{-2}{4}$ .

Let $B=P\inv AP$ so that $A$ and $B$ are similar matrices. Suppose that $v$ is an eigenvector of $B$ with eigenvalue $\lambda$ . Show that $Pv$ is an eigenvector of $A$ with eigenvalue $\lambda$ .

Which $n\times n$ matrices are similar to $I_n$ ?

Compute $e^A$ where $A = \mattwo {3}{-1}{1}{1}.$ Check your answer using MATLAB.