Let $f:R^3 \to R^3$ be a function which matrix with respect to the standard basis is:
$$
A = \begin{pmatrix}
-4 & -8 & 8 \\
1 & 2 & -2 \\
-1 & -2 & 2 \\
\end{pmatrix}$$
Then:
$$
B = \begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}$$Find an invertible matrix $S$ such that $ A = SBS^{-1}$. Is that matrix unique?
The teacher said that if $A$ and $B$ are similar then they have the same characteristic polynomial and therefore the same eigenvalues.
So if the diagonal matrix of $A$ is $ D_A = H_1^{-1}AH_1$ and the diagonal matrix of $B$ is $ D_B = H_2^{-1}BH_2$ I have:
$ D_B = D_A $ so $H_2^{-1}BH_2 = H_1^{-1}AH_1$
from which:
$ H_1H_2^{-1}BH_2H_1^{-1} = A $
So the matrix $ S$ is:
$ S = H_1H_2^{-1}$
My problem is that the matrices $A,B$ can not be diagonalized so how do I find the matrices $ H_1$ and $H_2$ ?
From what I found online this problem should be related to Jordan but we did't study it so I should solve it without Jordan.
Best Answer
Note that $Be_3=e_1$, $Be_1=Be_2=0$. Pick a "random" vector such as $v_3=\begin{pmatrix}1\\0\\0\end{pmatrix}$ and let $v_1=Av_3=\begin{pmatrix}-4\\1\\-1\end{pmatrix}$. Then verify $Av_1=0$ and pick an independant $v_2$ in $\ker A$, for example $v_2=\begin{pmatrix}0\\1\\1\end{pmatrix}$. Then $B$ represents, with respect to the basis $v_1,v_2,v_3$, the same linear map as does $A$ with respect to the standard basis. Let $S=\begin{pmatrix}v_1&v_2&v_3\end{pmatrix}=\begin{pmatrix}-4&0&1\\1&1&0\\-1&1&0\end{pmatrix}$. Then $$ SB=\begin{pmatrix}-4&0&1\\1&1&0\\-1&1&0\end{pmatrix}\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}=\begin{pmatrix}0&0&-4\\0&0&1\\0&0&-1\end{pmatrix}$$ and $$ AS = \begin{pmatrix}-4&-8&8\\1&2&-2\\-1&-2&2\end{pmatrix}\begin{pmatrix}-4&0&1\\1&1&0\\-1&1&0\end{pmatrix}=\begin{pmatrix}0&0&-4\\0&0&1\\0&0&-1\end{pmatrix}$$ Hence $A=SBS^{-1}$.
Regarding uniqueness or not: note that almost any vector could be chosen for $v_3$ and also many different choices for $v_2$ would be valid.