MHB 307.8.3 Compute A^n for the Jordan block

[karush]

$\textsf{ Compute $A^n$ for the Jordan block
$\displaystyle A=\begin{bmatrix}
\lambda&0\\0&\lambda
\end{bmatrix}$ and $n=1,2,3,4$ }$
$\textit{Make a guess on what $A^n$ will be for any $n$. }$
$\textit{Note: this matrix is not diagonalizable,compute the matrix powers by hand.}$

ok I think this is down by increasing the dim

 

[HOI]

I think it is done by doing what the problem tells you to do! You are given a 2 by 2 matrix, A, and asks you to find A^2, A^3, and A^4. Do you know how to multiply matrices?

 

[karush]

not sure that souns too easy?

but couldn't find an example

 

[HOI]

An example of what? You are given a Jordan block $A= \begin{bmatrix}\lambda & 0 \\ 0 & \lambda \end{bmatrix}$. You don't say what $\lambda$ is but since this is a "Jordan Block" I imagine it is something like $\begin{bmatrix}a & 1 \\ 0 & a \end{bmatrix}$ (so that A is 4 by 4) or $\begin{bmatrix}a & 1 & 0 \\ 0 & a & 1 \\ 0 & 0 & a \end{bmatrix}$ (so that A is 6 by 6). In the first case, A would be $\begin{bmatrix}a & 1 & 0 & 0 \\0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{bmatrix}$.$A^2= \begin{bmatrix}a & 1 & 0 & 0 \\0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{bmatrix}\begin{bmatrix}a & 1 & 0 & 0 \\0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{bmatrix}= \begin{bmatrix}a^2 & 2a & 0 & 0 \\ 0 & a^2 & 0 & 0 \\ 0 & 0 & a^2 & 2a \\ 0 & 0 & 0 & a^2 \end{bmatrix}$.

Do you see that, with $\lambda= \begin{bmatrix}a & 1 \\ 0 & a \end{bmatrix}$,this is just $A^2= \begin{bmatrix}\lambda^2 & 0 \\ 0 & \lambda^2 \end{bmatrix}$ ? What do you think $A^3$, etc. will be? That is why we work with "block matrices" and is, I suspect, the whole point of this exercise.

 

[karush]

$A^2= \begin{bmatrix}a & 1 & 0 & 0 \\0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{bmatrix}\begin{bmatrix}a & 1 & 0 & 0 \\0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{bmatrix}= \begin{bmatrix}a^2 & 2a & 0 & 0 \\ 0 & a^2 & 0 & 0 \\ 0 & 0 & a^2 & 2a \\ 0 & 0 & 0 & a^2 \end{bmatrix}$.

Do you see that, with $\lambda= \begin{bmatrix}a & 1 \\ 0 & a \end{bmatrix}$,this is just $A^2= \begin{bmatrix}\lambda^2 & 0 \\ 0 & \lambda^2 \end{bmatrix}$ ? What do you think $A^3$, etc. will be? That is why we work with "block matrices" and is, I suspect, the whole point of this exercise.

I suppose we could do this:

$A^3= a^2\begin{bmatrix}a & 1 & 0 & 0 \\0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{bmatrix}
= \begin{bmatrix}a^3 & 2a^2 & 0 & 0 \\
0 & a^3 & 0 & 0 \\
0 & 0 & a^3 & 2a^2 \\
0 & 0 & 0 & a^3\end{bmatrix}
$

$A^4= a^3\begin{bmatrix}a & 1 & 0 & 0 \\0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{bmatrix}
=\begin{bmatrix}a^4& 4 a^3& 0& 0\\0& a^4& 0& 0\\0&0 &a^4 &4 a^3\\0 & 0& 0& a^4\end{bmatrix}$

 

[HOI]

I suppose we could do this:

$A^3= a^2\begin{bmatrix}a & 1 & 0 & 0 \\0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{bmatrix}
= \begin{bmatrix}a^3 & 2a^2 & 0 & 0 \\
0 & a^3 & 0 & 0 \\
0 & 0 & a^3 & 2a^2 \\
0 & 0 & 0 & a^3\end{bmatrix}
$

You mean $A^3= A^2\begin{bmatrix}a & 1 & 0 & 0 \\0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{bmatrix}$

$A^4= a^3\begin{bmatrix}a & 1 & 0 & 0 \\0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{bmatrix}
=\begin{bmatrix}a^4& 4 a^3& 0& 0\\0& a^4& 0& 0\\0&0 &a^4 &4 a^3\\0 & 0& 0& a^4\end{bmatrix}$

 

[karush]

$A^3= A^2\begin{bmatrix}a & 1 & 0 & 0 \\0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{bmatrix}
= \begin{bmatrix}a^3 & 2a^2 & 0 & 0 \\
0 & a^3 & 0 & 0 \\
0 & 0 & a^3 & 2a^2 \\
0 & 0 & 0 & a^3\end{bmatrix}$

$A^4= A^3\begin{bmatrix}a & 1 & 0 & 0 \\0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{bmatrix}
=\begin{bmatrix}a^4& 4 a^3& 0& 0\\0& a^4& 0& 0\\0&0 &a^4 &4 a^3\\0 & 0& 0& a^4\end{bmatrix}$

hopefully

 

[karush]

MHB comments