The Jordan form for 4x4 matrix

Question

Suppose that a 4x4 matrix $A$ has vectors $v_a,b_b,v_c,v_d$ that satisfy $$Av_a=v_c, Av_c=-v_a, \ \ Av_b=-2v_b,\ \ Av_d=-2v_d+3v_b.$$

I want to find the Jordan canonical form of $A$, and the matrix $P$ such that $A=PJP^{-1}.$

I feel the question is ambiguous because when saying $A$ has vectors $v_a,b_b,v_c,v_d$ such that ... does it mean that $A's$ columns or rows consist of those vectors or $A$ is a matrix which is unknown but satisfies those condition above. So I am really greatful fo any help or hints

Answer 1

It means exactly what it says, not more and not less. In matrix notation that gives $$ A[v_a,v_c,v_b,v_d]= [v_a,v_c,v_b,v_d] \begin{bmatrix} 0&-1&0&0\\ 1&0&0&0\\ 0&0&-2&3\\ 0&0&0&-2 \end{bmatrix} $$ In this form the task of identifying the Jordan normal form is greatly reduced. The transformation matrix has to use the basis matrix $V=[v_a,v_c,v_b,v_d]$ with the unidentified column vectors of the basis used.

Answer 2

the given relations mean that, in the basis $\mathbf V=\{v_a,v_b,v_c,v_d\}$ (the vectors must be linearly independent) The linear transformation $A$ is represented by the matrix $$ A=\begin{bmatrix} 0&0&-1&0\\ 0&-2&0&3\\ 1&0&0&0\\ 0&0&0&-2 \end{bmatrix} $$ Remember that in that basis $\mathbf V$ the vectors are represented as $v_a=[1,0,0,0]$, $v_b=[0,1,0,0]$, $v_c=[0,0,1,0]$, $v_d=[0,0,0,1]$.

The Jordan form $>J$ of this matrix is given by: $$ J=S^{-1}AS $$ with (from WolframAlpha)

$$ J=\begin{bmatrix} -2&1&0&0\\ 0&-2&0&0\\ 0&0&-i&0\\ 0&0&0&i \end{bmatrix} \qquad S=\begin{bmatrix} 0&0&-i&i\\ 1&0&0&0\\ 0&0&1&1\\ 0&1/3&0&0 \end{bmatrix} $$

this means that the eigenvalue $\lambda_1=-2$ has an eigenvector $$\nu_1=[0,1,0,0]^T=v_b:$$ (as we can see from the third condition), and a generalized eigenvector $$\nu_2=\left[0,0,0,\frac{1}{3}\right]^T= \frac{1}{3}v_d$$.

The other eigenvalues and eigenvectors are: $$ \lambda_2= i \qquad \nu_2=[i,0,1,0]^T $$

$$ \lambda_3= -i \qquad \nu_3=[-i,0,-1,0]^T $$

If you know the components of the vectors of the basis $\mathbf V$ in another basis you can express the eigenvectors in this other basis.