Solve $BAB^{-1}=J$ for matrix $B$

Question

I have 3 Matrices : $A , B$ and $J.$ I know matrices $A$ and $J. J$ is the real jordan normal form of $A.$ The relationship between the three matrices is given by $BAB^{-1} =J.$

How can I find the matrix $B.$

For example,

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

$J= \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$BAB^{-1}=J$ can be written as $BA-JB=0$. That's as far as I got.

Thanks for your help.

Answer 1

$BA - JB = 0$ can be considered as a $9 \times 9$ system of linear equations in the entries of $B$. Solve that system.

Answer 2

$\det(A-\lambda I) = 0$

$-(\lambda-1)(\lambda+i)(\lambda-i)$

Complex eigenvalues Here is a little trick

$(B^{-1} A B)^2 = J^2\\ B^{-1} A^2 B = J^2$

And $A^2$ has real eigenvalues and the same matrices $B, B^{-1}$

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

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

And we know that $-1,-1,1$ will be the eigenvalues of $A^2.$ Find the associated vectors.

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

$B = \begin{bmatrix}-1&0&1\\0&1&1\\1&0&1\end{bmatrix}$ And I will leave it to you to find $B^{-1}$

Answer 3

The columns of the transform matrix $B$ are the real and imaginary parts of the eigenvector corresponding to the eigenvalue $\pm i$ (that's the first two columns!) then the eigenvector corresponding to the final eigenvalue of 1.

Calculate these to show that $B= \begin{bmatrix} 1 && 0 && 1 \\ 0 && -1 && 1 \\ -1 && 0 && 1 \\ \end{bmatrix} $