Calculate ${e^{At}}$ of $A = \left( {\begin{array}{*{20}{c}} i&j&k\\ i&j&k\\ i&j&k \end{array}} \right)$ knowing that $i+j+k=0 $

Question

How to calculate ${e^{At}}$ for a matrix $A = \left( {\begin{array}{*{20}{c}} i&j&k\\ i&j&k\\ i&j&k \end{array}} \right)$ knowing that $i+j+k=0$

answer:

if you calculate $A^2$ you get

$${A^2} = \left( {\begin{array}{*{20}{c}} {i\left( {j + k + i} \right)}&{j\left( {i + j + k} \right)}&{k\left( {i + j + k} \right)}\\ {i\left( {j + k + i} \right)}&{j\left( {i + j + k} \right)}&{k\left( {i + j + k} \right)}\\ {i\left( {j + k + i} \right)}&{j\left( {i + j + k} \right)}&{k\left( {i + j + k} \right)} \end{array}} \right) = 0$$ This is a nilpotent matrix with order 2 and hence easy to get the exponential.

Not saying anything. You tell us. What would that imply for the exponential? — Phicar, Feb 12 '22 at 11:10
I didn't actually, but I would guess is because you did not show any work on your question. — Phicar, Feb 12 '22 at 11:12
More generally one could diagonalize $A=T D T^{-1}$ and then compute $\exp(A)$ as $T\exp(D) T^{-1}$. — Maximilian Janisch, Feb 12 '22 at 11:24

score 1 · Accepted Answer · answered Feb 12 '22 at 11:21

In general, aplying the definition of matrix exponential is hard, but if you know that your matrix is nilpotent $A^2=0$ then your definition becomes trivial: $$e^{At}=\sum_{n=0}^\infty \frac{A^nt^n}{n!}=I_3+At=\left( {\begin{array}{*{20}{c}} 1+it&jt&kt\\ ti&1+tj&tk\\ ti&tj&1+tk \end{array}} \right)$$

Calculate ${e^{At}}$ of $A = \left( {\begin{array}{*{20}{c}} i&j&k\\ i&j&k\\ i&j&k \end{array}} \right)$ knowing that $i+j+k=0 $

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