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QUESTION: Let A$ \in M_{n\times n} (\mathbb R)$ such that $A^2 = -A^3$. Write $e^{tA}$ as a finite sum.

I've been trying to solve this. I don't know much about the matrix exponential as my only exposure to it was this semester in my Differential Equations course. My teacher gave us this problem as a challenge.

What I've done so far:

$ A^n = -A^{n-1} $ if n is odd and $n\geqslant3$

$A^n = (-1)^{n/2}A^{3n/2}$ if n is even and $n\geqslant2$

Then I put $n=2k+1$ or $n=2k$ and it gives:

$A^n = -A^{2k}=-(-1)^kA^{3k}$ or $A^n = (-1)^k A^{3k}$

Then i tried to compute the matrix exponential(A+I+sum of even powers+sum of odd powers) to see if I would reach anywhere:

$$\sum_{n=0}^\infty \frac{(tA)^n}{n!}= I + tA + \sum_{k=0}^\infty [\frac {(-1)^kt^{2k}A^{3k}}{(2k)!} -\frac {(-1)^k t^{2k+1}A^{3k}}{(2k+1)!}]$$

Now I'm stuck....Any ideas?

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  • $\begingroup$ $A^3 = -1A^2$, then $A^4 = A(A^3) = A (-A^2)=-A^3= A^2$ and similarly for all powers over $3$ you just need to write the sum with $I,A,A^2$ $\endgroup$
    – Stefan
    Commented Jun 23, 2018 at 16:32
  • $\begingroup$ Thanks, I'm so dumb! $\endgroup$ Commented Jun 23, 2018 at 16:38

2 Answers 2

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$$ A^2 = -A^3 \implies A^3 = -A^2 $$

Multiply by $A$ to get $$ A^4 = -A^3 =A^2$$

Multiply by $A$ to get $$ A^5 = A^3 = -A^2$$

and $$A^6 = -A^3 = A^2$$

Thus we have $$A^{2k} = A^2$$ and $$A^{2k+1} = -A^2$$

Now you can do the rest.

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  • $\begingroup$ Thank you! I was over thinking this! I'm so dumb! Thank you so much! $\endgroup$ Commented Jun 23, 2018 at 16:41
  • $\begingroup$ @PedroFernandes Thank you for your attention. Sometimes, when we are tired, we make problems harder than they really are. $\endgroup$ Commented Jun 23, 2018 at 19:42
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Here is an alternative way if you wish to play with the spectrum of $A$ instead.

It is given that the polynomial $x^2+x^3 = x^2(x+1)$ annihilates the matrix $A$ so the minimal polynomial $m_A$ divides $x^2(x+1)$.

The possibilities are:

  • $m_A(x) = x$ which gives $A = 0$ so $e^{tA} = e^0 =I$.
  • $m_A(x) = x+1$ which gives $A = -I$ so $e^{tA} = e^{-tI} = e^{-t}I$.
  • $m_A(x) = x^2$ which gives $A^2 = 0$ so $e^{tA} = I + tA$ directly from the series.
  • $m_A(x) = x(x+1)$ so we know that for any entire function $f$ holds $f(A) = f(0)P_1 + f(-1)P_2$ where $P_1, P_2$ are fixed polynomials in $A$. Plugging in $f(x) = x$ and $f(x) = x+1$ we get $$\begin{cases} A = -P_2 \\ A+I = P_1 \\ \end{cases} $$

    so $f(A) = f(0)(A+I) - f(-1)A$. For $f(x) = e^{tx}$ we get $$e^{tA} = e^{0} (A+I) - e^{-t}A = A(1-e^{-t}) + I$$

  • $m_A(x) = x^2(x+1)$ so we know that for any entire function $f$ holds $f(A) = f(0)P_1 + f'(0)P_2 + f(-1)P_3$ where $P_1, P_2, P_3$ are fixed polynomials in $A$. Plugging in $f(x) = 1$, $f(x) = x$ and $f(x) = x^2$ we get $$\begin{cases} I = P_1+P_3 \\ A = P_2 - P_3 \\ A^2 = P_3 \end{cases} $$

    so $P_1 = I- A^2$, $P_2 = A^2+A$ and $P_3 = A^2$. Therefore $f(A) = f(0)(I- A^2) + f'(0)(A^2+A) + f(-1)A^2$ so for $f(x) = e^{tx}$ we get $$e^{tA} = e^0(I- A^2) + 0\cdot e^0(A^2+A) + e^{-t}A^2 = A^2(e^{-t} - 1) + I$$

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