If $A$ is a $2\times 2$ matrix with a repeated eigenvalue $r$, then $\mathrm{e}^{At}=\mathrm{e}^{rt}\left[I+(A-rI)t\right]$

Question

If $A$ is a $2\times 2$ matrix with a repeated eigenvalue $r$, show that $\mathrm{e}^{At}=\mathrm{e}^{rt}\left[I+(A-rI)t\right]$.

I have already been able to show that if $A$ is an arbitrary $2\times 2$ matrix $$ \left( \begin{array}\\ a & b \\ c & d \end{array} \right), $$ the repeated eigenvalue $r$ must be equal to $\,r=\dfrac{a+d}{2}$.

From here, my first thought was to take the determinant of the matrix $$\mathrm{e}^{rt}\left[I+(A-rI)t\right],$$ and show that it is equal to $$\det(\mathrm{e}^{At}),$$ but then I started thinking this would be useless since two unequal matrices can have the same determinant.

I've tried directly substituting $r$ into the right-hand side of the equation as well, but with no luck.

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Does anyone have any suggestions on how to proceed from here?

Answer 1

This is much simpler if you observe that what you're trying to prove is invariant under conjugation/similarity. That is, it suffices to show this for some matrix similar to $A$, not necessarily $A$ itself.

Note that if $A$ has a repeated eigenvalue of $r$, then it must be similar either to $\begin{pmatrix}r & 0 \\ 0 & r\end{pmatrix}$ or to $\begin{pmatrix}r & 1 \\ 0 & r\end{pmatrix}$.

Answer 2

They key observation is to write the relation as $$e^{(A-rI)t}=I+(A-rI)t. $$ This is clearly true once you notice that $(A-rI)^2=0$. You can prove this using the Jordan form as Slade mentioned: a $2\times2$ matrix with both eigenvalues equal to zero has its square equal to zero.

Answer 3

As $r$ is an eigenvalue of multiplicity 2 of the $2\times 2$ matrix $A$, then $A$ does not have any other eigenvalues, and its characteristic polynomial should be $$ p(x)=(x-r)^2, $$ and by Cayley-Hamilton Theorem we have that $(A-rI)^2=0$.

Next observe that $$ \mathrm{e}^{tA}=\mathrm{e}^{rt}\mathrm{e}^{t{(A-rI)}}=\mathrm{e}^{rt}\sum_{n=0}^\infty \frac{(A-rI)^n}{n!}=\mathrm{e}^{rt}\sum_{n=0}^1 \frac{(A-rI)^n}{n!}=\mathrm{e}^{rt}\big(I+ (A-rI)\big) $$