Finding an invertible $P$ such that $PAP^{-1}$ is in rational form

Question

Let $A=\begin{bmatrix}1 & 3 & 3\\3 & 1 & 3\\-3 & -3 & -5\end{bmatrix}.$

I've computed the characteristic polynomial $\text{char(x)}=(x-1)(x+2)^2$ and minimum polynomial $\text{min(x)}=(x-1)(x+2)$. Also, I know that the rational form of $A$ will be:

$\begin{bmatrix}0 & 2 & 0\\1 & -1 & 0\\ 0 & 0 & 2\end{bmatrix}.$

Where I'm stuck is trying to find a $P$ such that $PAP^{-1}$ will equal the rational canonical form. I understand finding the basis for the kernel of the invariant spaces, but I get stuck after that.

Answer 1

Let $A\in M_n(K)$, $p(x)$ be its characteristic polynomial and $m(x)$ be its minimal polynomial. Recall (for those who do not know) that the rational canonical form of $A\in M_n(K)$ satisfies (in particular) that follows:

$F=diag(F_1,\cdots,F_k)\in M_n(K)$ where $F_i$ is the companion matrix of $p_i(x)\in K[x]$, $p_i$ is a divisor of $p_{i+1}$, $p_1\cdots p_k=p$ and $p_k=m$.

Here $p_1(x)=x+2,p_2(x)=(x-1)(x+2)$.

We take $e_1=[1,0,-1]^T\in\ker(A+2I)$.

For $e_2$ we choose a random vector, for example, $e_2=[1,0,0]^T$. It does not seem very random; yet, if it is not convenient, then pick another vector.

Let $e_3=Ae_2=[1,3,-3]^T$ (since $e_1,e_2,e_3$ is a basis, our choice was convenient) ; then $Ae_3=2e_2-e_3$ and we are done.

Answer 2

You have the caracteristic equation and know that $1,-2,-2$ are your eigenvalues.

If \lambda is an eigenvalue a v is the associated eigenvector

$A\mathbf v = \lambda v\\ (A-\lambda I)\mathbf v = \mathbf 0$

We need to find the kernel of $(A-I)$ and $(A+2I)$

$A+I = \begin{bmatrix} 0&3&3\\3&0&3\\-3&-3&-6\end{bmatrix}$

What is the kernel of this matrix?

$\mathbf v_1 = \begin{bmatrix} 1\\1\\-1\end{bmatrix}\\ A\mathbf v_1 = \mathbf v_1$

$A-2I = \begin{bmatrix} 3&3&3\\3&3&3\\-3&-3&-3\end{bmatrix}$

$\mathbf v_2 = \begin{bmatrix} 1\\-1\\0\end{bmatrix}$ $\mathbf v_3 = \begin{bmatrix} 1\\0\\-1\end{bmatrix}$

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

and

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

Answer 3

Doug M's method is the best. But here is a more "elementary" method: You want to find matrix $P= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$ so that $PAP^{-1}= B$. That is the same as $PA= BP$ so $\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$$\begin{bmatrix}1 & 3 & 3 \\ 3 & 3 & 0 \\ -3 & -3 & 5\end{bmatrix}= $$\begin{bmatrix}0 & 2 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$\begin{bmatrix}a & b & c \\ d & e & f\\ g & h & i \end{bmatrix}$.

Doing the Multiplication on both sides, $\begin{bmatrix}a+ 3b- 3c & 3a+ 3b- 3c & 3a+ 5c \\ d+ 3e- 3f & 3d+ 3e- 3c & 3d+ 5f \\ g+ 3h- 3i & 3g+ 3e- 3h & 3g+ 5i \end{bmatrix}$$= \begin{bmatrix}2d & 2e & 2f \\ a- d & b- e & c- f \\ 2g & 2h & 2i\end{bmatrix}$.

In order that those two matrices be equal, they must have the corresponding components equal.

We have a+ 3b- 3c= 2d, 3a+ 3b- 3= 2e, 3a+ 5c= 2f, d+ 3e- 3f= a- d, 3d+ 3e- 3c= b- e, 3d+ 5f= c- f, g+ 3h- 3i= 2g, 3g+ 3e- 3h= 2h, and 3g+ 5i= 2i.

Solve those 9 linear equations for the 9 values, a, b, c, d, e, f, g, h, i.

(I did say that Doug M's method was best!)