How to solve for P in a similarity when matrices aren't diagonalizable

When matrices $A,B$ are similar and they are both diagonalizable, as $$\left\{ \begin{array}{l} P_1^{-1}AP_1=Λ\\ P_2^{-1}BP_2=Λ \end{array} \right. \Longrightarrow (P_1P_2^{-1})^{-1}A(P_1P_2^{-1})=B$$

then we know when $P=P_1P_2^{-1}$, we can get $P^{-1}AP=B$. We just solve for the eigenvectors of $A$ and $B$, and then we can easily get the $P_1$ and $P_2$.

My question is about how to proceed when $A$ and $B$ are similar yet not both diagonalizable. How can we solve for a matrix $P$ so that $B = P^{-1}A P$?

I have two example in following:

First

$A=\left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right)$ and $B=\left( \begin{array}{cc} -\frac{12}{7} & -\frac{32}{7} \\ \frac{9}{14} & \frac{12}{7} \\ \end{array} \right)$. How to get the $P$? Actually I know the $P= ( \begin{smallmatrix} 2&10\\3&8 \end{smallmatrix} )$

Second

$A=\left( \begin{array}{ccc} 27 & 48 & 81 \\ -6 & 0 & 0 \\ 1 & 0 & 3 \\ \end{array} \right)$ and $B=\left( \begin{array}{ccc} 52 & 62 & 24 \\ 4 & 8 & 0 \\ -80 & -\frac{197}{2} & -30 \\ \end{array} \right)$ .How to solve the $P$?Actually the $P=\left( \begin{smallmatrix} 8 & 10 & 0 \\ 5 & 3 & 4 \\ 0 & 2 & 0 \\ \end{smallmatrix} \right)$

• In your example, $A, B$ are not similar. Jul 1, 2016 at 21:14
• Let me try to restate your Question in my words. Suppose $A,B$ are similar but not diagonalizable. How can we compute a matrix $P$ which accomplishes the similarity transformation $B = P^{-1}A P$? Jul 1, 2016 at 21:15
• @xpaul Oh,I can sure they are similar.you can calculate $P^{-1}AP$,it equal to $B$ exactly.
– mayi
Jul 1, 2016 at 21:18
• @xpaul: Notice the denominator in the lower left corner is $14$, which cancels a factor of two in the upper right corner. So the trace and determinant are zero for both matrices. I think they must be similar. Jul 1, 2016 at 21:19
• @hardmath Yes,you get my meaning.Thanks.
– mayi
Jul 1, 2016 at 21:20

Lets take this the other way

First of all we find one eigenvector the same way we normally would

$B\mathbf v_1 = \lambda \mathbf v_1$

In this case $\lambda = 0$ and $\mathbf v_1 = \begin{bmatrix}8\\-3\end{bmatrix}$

we have one vector in P, we need to find the other such that.

$BP = P\Lambda + P\begin{bmatrix}0&1\\0&0\end{bmatrix}\\ (B-\lambda I) P = P\begin{bmatrix}0&1\\0&0\end{bmatrix}\\ (B-\lambda I) \mathbf v_2 = \mathbf v_1$

Since $\lambda$ equals $0,$ you have made life easier on us... but not that much easier..

$B \mathbf v_2 = \mathbf v_1\\ B^2 \mathbf v_2 = B\mathbf v_1 = 0\\$

But $B^2 = 0$ which gives us a lot of latitude to find suitable $\mathbf v_2$ There is not a unique solution. Any solution will suffice.

$-\frac {12}7 v_{2,1} -\frac{32}{7} v_{2,2} = 8\\ 3 v_{2,1} + 8 v_{2,2} = -14\\ 3\cdot6 + 8 \cdot -4 = -14$

$P = \begin{bmatrix}8& 6\\-3&-4\end{bmatrix}$

$BP = PA$ or $P^{-1} B P = A$

Update....

If you can't diagonalize the matrix, can you put it into "Jordan Normal" form?

$P^{-1}A P = \begin{bmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 1 \\ 0 & 0 & \lambda_2 \\ \end{bmatrix}$

If you can put two matrices into the same Jordan Normal form, they are similar.

$A = \begin{bmatrix} 27 & 48 & 81 \\ -6 & 0 & 0 \\ 1 & 0 & 3 \\ \end{bmatrix}$

The characteristic equation is: $(\lambda - 6)(\lambda-12)^2 = 0$

We find the first two eigenvectors using the normal approach.

$A \begin{bmatrix} 3\\-3\\1\end{bmatrix} = \begin{bmatrix} 18\\-18\\6\end{bmatrix}$

$A \begin{bmatrix} 18\\-9\\2\end{bmatrix} = \begin{bmatrix} 216\\-108\\24\end{bmatrix}$

the $3^{rd}$ eigenvector....

$(A -12 I)^2 = \begin{bmatrix} 18&144&486\\-18&-144&-486\\6&48&162\end{bmatrix}$

Any vector in the kernel of the above is parallel to the second eigenvector. Now you just need to scale it!

$\begin{bmatrix} 3\\3\\-1\end{bmatrix}$ is in the kernel.

$A \begin{bmatrix} 3\\3\\-1\end{bmatrix} = \begin{bmatrix}108\\-54\\12\end{bmatrix}= 6 \begin{bmatrix} 18\\-9\\2\end{bmatrix}$

$P = \begin{bmatrix} 3&18&\frac 12\\-3&9&\frac 12\\1&2&\frac 16\end{bmatrix}$

• Thank for your solution,I get a lot from it.But maybe the first $A$ have some special.Can you make this solution more universal?Then I'll accept it and this post can help other people which have same confusion with me also.I'm look forward that.
– mayi
Jul 2, 2016 at 13:31