Q:Given
$$
A =
\pmatrix{6 & 1 & 1\\ -1 & 4 & 2\\ 0 & 0 & 5}
$$
Find the characteristic polynomial of $A$
And Let $f : U → U$ be a linear map represented by the matrix $A$ with respect to a certain basis, where $U$ is a 3-dimensional vector space over $\mathbb{C}$. Given that the 5-eigenspace of $f$
is 1-dimensional, determine the Jordan canonical form of A. Justify your answer briefly
A: I've worked out the characteristic polynomial as
$$(t-5)(t^2-10t+25)$$ where the root is $t = +5$ for all the $t$ above.
But what is the method to work out the $JCF$, why is the characteristic polynomial necessary to do this, and what is the reason for the question to specify dimension.
Best Answer
$$A-5I=\begin{bmatrix} 1&1&1\\-1&-1&2\\0&0&0 \end{bmatrix}$$ This matrix has rank $1$, hence we know there'll be $\color{red}1$ Jordan block (necessarily of size $3$), i.e. $A$ will have the form: $$J=\begin{bmatrix} 5&1&0\\0&5&1\\0&0&5 \end{bmatrix}$$ To find a Jordan basis, we consider the increasing sequence: $$0\varsubsetneq\ker(A-5I)\varsubsetneq\ker(A-5I)^2\varsubsetneq\ker(A-5I)^3=\mathbf R^3.$$ This sequence is increasing since $\;(A-5I)^2=\begin{bmatrix} 0&0&3\\0&0&-3\\0&0&0 \end{bmatrix}$ has rank $1$, hence $\dim \ker(A-5I)^2=2$. Also $(A-5I)^3=0$.
Start from a vector $e_3*\in\ker(A-5I)^3\setminus\ker(A-5I)^2$. The only condition is its third coordinate is non-zero, say $e_3=(0,0,1)$.
Set $\;u_2=(A-5I)e_3$. We have $e_2\ne 0,\, e_2\in\ker(A-5I)^2$, then $e_1=(A-5I)e_2$. $e_1\neq 0$ and $e_1\in\ker(A-5I)$, in other words, $e_1$ is an eigenvector.
Further more $\mathcal B=(e_1,e_2,e_3)$ is, by construction, a Jordan basis, since
$Ae_1=5e_1$,
$(A-5I)e_2=e_1\iff Ae_2=5e_2+e_1$,