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Let $A$ be the following $3 \times 3$ matrix:

$$ A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix} $$

I'm supposed to calculate $A^n$, where $n \in \Bbb R$, $\exp(tA)$ and $\sin(\pi A)$. Obviously $A$ is not diagonalizable. Since we haven't had anything about Jordan decomposition in the lecture, I'm not sure how to solve this.

The eigenvalues $\lambda_1 = -1 , \lambda_{2,3} = 1$ can be read off. I tried to expand the two eigenvectors into a orthonormal basis, i.e.:

$$ \mathbf{x}_{\lambda_1} = \begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix} \qquad \mathbf{x}_{\lambda_2} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ \end{pmatrix} \qquad \mathbf{x}_3 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ \end{pmatrix} $$

But I'm rather unsure how to continue. I suspect that

$$ A^n = \begin{pmatrix} 1 & n & 0 \\ 0 & 1 & 0 \\ 0 & 0 & (-1)^n \\ \end{pmatrix} \qquad \text{for} \qquad n \in \Bbb N_0, $$

But how to expand this to $n \in \Bbb R$? In general, how can I solve such a problem of matrix functions, if I've not heard anything about Jordan decomposition?

EDIT: Thanks for your help. I could show that the above mentioned matrix for $A^n$ is correct even for $n \in \Bbb Z$. The two other functions are straightforward then. If someone has an idea or hint about $A^n$ for $n \notin \Bbb Z$, i would appreciate it.

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    $\begingroup$ It already is in Jordan normal form. Decompose as $A=D+N$, where $D$ is diagonal, and $N$ is nilpotent and they commute. In such case, $\exp(tA)=\exp(tD)\exp(tN)$. $\endgroup$ – Bernard Nov 20 '15 at 20:12
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Like you said, $$ A^n = \begin{bmatrix} 1 & n & 0 \\ 0 & 1 & 0 \\ 0 & 0 & (-1)^n \\ \end{bmatrix} \qquad \text{for} \qquad n \in \Bbb N.$$

Then $$ \exp(tA)=\sum_{n=0}^\infty\frac{t^nA^n}{n!} =\begin{bmatrix} \sum_{n=0}^\infty \frac{t^n}{n!} &\sum_{n=0}^\infty n\frac{t^n}{n!}&0\\0&\sum_{n=0}^\infty \frac{t^n}{n!}&0\\ 0&0&\sum_{n=0}^\infty \frac{(-1)^nt^n}{n!} \end{bmatrix} =\begin{bmatrix} e^{t}&te^t&0\\ 0&e^t&0\\ 0&0&e^{-t} \end{bmatrix}. $$ You can play the same game for the sine.

About the powers, you could define $$ A^t=\begin{bmatrix}1&t&0\\0&1&0\\0&0&e^{\pi i t}\end{bmatrix}. $$ This agrees with the integer powers of $A$ and satisfies the exponential property $A^{t+s}=A^{t}A^{s}$. It is important to notice that for non-integer $t$ this choice is rather arbitrary and not the result of a calculation.

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  • $\begingroup$ Yes, but what about $A^n$ for $n \notin \Bbb N$? I suppose that for $n \in \Bbb Z$ I can find it rather easy, using the inverse $A^{-1}$ for negative $n$ but I'm not sure what to do about $n \notin \Bbb Z$. $\endgroup$ – root Nov 20 '15 at 20:21
  • $\begingroup$ I didn't address that because I find it tricky. Usually one defines arbitrary powers by $r^s=\exp(s\log r)$. But $A$ is not an exponential: you cannot have $e^t=1$ and $te^t=1$. That means that it has no logarithm. I cannot immediately think of another way of addressing $A^s$ for arbitrary real $s$. $\endgroup$ – Martin Argerami Nov 20 '15 at 20:24
  • $\begingroup$ To be honest, I suspect that this is a typo in the exercise sheet, but nevertheless it would be interesting. $\endgroup$ – root Nov 20 '15 at 20:26
  • $\begingroup$ I included a possible way of doing the powers of $A$. $\endgroup$ – Martin Argerami Nov 21 '15 at 0:04
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Here's another strategy, take it or leave it. You can write $ tA = tD+tB$ where $tD$ is a diagonal matrix and $tB$ is an upper triangular "nilpotent" matrix. Nilpotent matrices have the special property that $B^k = 0$ for some finite $k$ (in your case $k=2$). Then, because any matrix commutes with a diagonal matrix, the following formula holds:

$$ \exp(t(D+B)) = \exp(tD)\exp(tB) $$

You can compute $\exp(tD)$ easily, and $\exp(tB)$ will just be a polynomial of $tB$ since the series for $\exp()$ will terminate after finitely many terms. In fact, because $B^2=0$ for your matrix, $\exp(tB) = I+tB$. Then you just multiply $\exp(tD)$ with $I+tB$ to get the result.

Note: the exponential formula above only holds because the matrices commute. $\exp(A+B)\neq \exp(A)\exp(B)$ in general!

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  • $\begingroup$ The word you are looking for is "nilpotent". "Idempotent" means $B^2=B$. $\endgroup$ – Ian Nov 20 '15 at 20:31
  • $\begingroup$ Right, thanks. Edited. $\endgroup$ – icurays1 Nov 20 '15 at 20:32

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