Jordan canonical form and basis for 4 by 4 matrix with two eigenvalues

Question

So, I am given the matrix in standard basis $$A =\begin{bmatrix}-3&1&3&3\\-10&2&9&9\\-4&0&5&4\\2&1&-3&-2\end{bmatrix}$$ characteristic polynomial is $(λ−1)^3 (λ+1)$ which means eigenvalues are $λ_1= 1$, with multiplicity $3$, and $λ_2= -1$, with multiplicity $1$. How should I go about obtaining the Jordan canonical form and basis of said matrix? I tried calculating the rank of the matrix $A-1\times\operatorname I$, and determining using that, but I can't seem to figure it out.

Answer 1

It's easy to see that $v_1=(0,-3,-2,3)$ is an eigenvector of $A$ with eigenvalue $-1$ and the the eigenspace corresponding to the eigenvalue $1$ is $$ E_1=\{(a+b,a+b,a,b)\mid a,b\in\mathbb{R}\}, $$ which is two-dimensional. Now, let us look for a vector $v=(x,y,z,t)$ such that $A.v=v+w$, for some $w\in E_1$. Note that $$ A.v-v=(-4x+y+3z+3t,-10x+y+9z+9t,-4x+4z+4t,2x+y-3z-3t). $$ In order to determine when such a vector belongs to $E_1$, we can solve the system $$ \left\{\begin{array}{l}-4x+y+3z+3t=-4x+4z+4t+2x+y-3z-3t\\-10x+y+9z+9t=-4x+4z+4t+2x+y-3z-3t.\end{array}\right. $$ One non-zero solution of this system is $(0,1,0,0)$, and $A.(0,1,0,0)=(0,1,0,0)+(1,1,0,1)$. So:

• let $v_2=(1,1,0,1)(\in E_1)$;
• let $v_3=(0,1,0,0)$;
• let $v_4=(1,1,1,0)(\in E_1)$.

With these choices for the $v_i$'s, we have:

• $A.v_1=-v_1$;
• $A.v_2=v_2$;
• $A.v_3=v_3+v_2$;
• $A.v_4=v_4$.

So, if $$ P=\begin{pmatrix}0&1&0&1\\-3&1&1&1\\-2&0&0&1\\3&1&0&0\end{pmatrix} $$ (the columns of $P$ are $v_1$, $v_2$, $v_3$, and $v_4$), then $$ P^{-1}AP=\begin{pmatrix}-1&0&0&0\\0&1&1&0\\0&0&1&0\\0&0&0&1\end{pmatrix}. $$