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I want to calculate exponential of the matrix which on diagonal has some $a \in \mathbb{R}$ and ones above. The $n\times n$ matrix looks like following

$$ A = \left( \begin{matrix} a & 1 & 0 & 0 & \cdots & 0 \\ 0 & a & 1 & 0 & \cdots & 0 \\ 0 & 0 & a & 1 & 0 & 0 \\ 0 & 0 & 0 & a & \ddots & 0 \\ 0 & 0 & 0 & 0 & \ddots & 1 \\ 0 & 0 & 0 & 0 & 0 & a \end{matrix} \right) $$

I tried to do it by counting determinant of matrix $A-\lambda I$ by the following algorithm :

  1. Divide last row by $a-\lambda$, so the $n$-th row is just $0$ and $1$ in the $n$-th column.

  2. Subtract $(n-1)$-th row by $n$-th row. Then the $1$ in the $n$-th column and $n-1$ row disappears.

  3. Divide $n-1$ row by $a-\lambda$, so the $(n-1)-$ th row is just $0$ and $1$ in the $(n-1)$-th column.

And so on so on. By algorithm above we get matrix with only ones at diagonal, so the determinant of that matrix is just $(a - \lambda)^n$. So we have $n$-th fold eigenvalue equals to $\lambda$. I now I have a problem with derivation of eigenvectors of matrix $A$. Can you give me some advice? Is there any simplest way to calculate that?

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Actually this matrix only has a one dimensional eigenspace, spanned by the first unit vector. This is an example of a 'Jordan'-block, you will find lots of theory if you google it.

Maybe a simpler way to calculate the matrix exponential is to write \begin{align} A=a\mathbb{I}+B \end{align} where $\mathbb{I}$ is the unit matrix and $B$ has only one's above the diagonal. Since $\mathbb{I}$ and $B$ commute, you get \begin{align} \exp(A)=\exp(a\mathbb{I})\cdot \exp(B) \end{align} Calculating the exponential of $a\mathbb{I}$ is straightforward and calculating the exponential of $B$ is also not too difficult, since $B$ is nilpotent, i.e $B^k=0$ for some appropriate $k$.

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  • $\begingroup$ Okey, so calculating $aI$ is really simple. But B is a nilpotent matrix, but using your notation the $k$ that $B^k=0$ equals to $n$. $B^2$ is a matrix $B$ "moved by one unit" to right side, $B^3$ is moved by two etc. But it's nightmare to write this in formal way $n-th$ times. There is no other way that strictly calculation ? $\endgroup$ – John Jun 7 at 13:37
  • $\begingroup$ I would simply prove by induction how the $k$-th power of $B$ looks like. I don't think that this calculation is a 'nightmare'. $\endgroup$ – Jake28 Jun 7 at 14:11

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