# Finding the Jordan canonical form a characteristic polynomial

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.

• Isn't the characteristic polynomial $-(\lambda -5)^3$? You have to find two generalized eigenvectors, Are you familiar with defective matrices, algebraic/geometric multiplicity, generalized eigenvectors and chaining? – Moo Apr 16 '16 at 12:30
• I get $(6-\lambda)(4-\lambda)(5-\lambda) + 5-\lambda = (5-\lambda)((6-\lambda)(4-\lambda)+1)$ and the second degree polynomial looks like a conjugate – mathreadler Apr 16 '16 at 12:39
• My CP agrees with the OPs, I was just showing that you have triple eigenvalue. – Moo Apr 16 '16 at 13:14
• I was taught that the characterised polynomial is $det(t*I-A)$ where I is the identity matrix, I then expanded on the bottom right element when I removed $(t-5)$ and then continued to calculate the determinant. I am wrong or is my polynomial correct? – HELP Apr 16 '16 at 13:48
• a keyword to glance is the companion matrix for a square one – janmarqz Apr 20 '16 at 17:35

$$A-5I=\begin{bmatrix} 1&1&1\\-1&-1&2\\0&0&0 \end{bmatrix}$$ 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

• $Ae_1=5e_1$,

• $(A-5I)e_2=e_1\iff Ae_2=5e_2+e_1$,

• $(A-5I)e_3=e_2\iff Ae_3=5e_3+e_2$.
• "This matrix has rank 1" should be "This matrix has defect 1 (and rank 2)". – user35603 Apr 16 '16 at 16:57
• I'm still confused with the method, it's the sequence part I'm unsure of – HELP Apr 16 '16 at 17:16
• I illustrated a general theorem:for any eigenvalue $\lambda$ of an endomorphism $u$ of a vector space $E$ of dimension $n$, the sequence of the successive kernels $\ker(u-\lambda\operatorname{id}_E)^k$ is first increasing until $k=$ a certain integer $r$, then stabilises. You calculate a Jordan basis, starting from a maximal linearly independent set in $\ker(u-\lambda\operatorname{id}_E)^r\setminus\ker(u-\lambda\operatorname{id}_E)^{r-1}$, then cascading down to a basis of the eigenspace. I hope I'm clear enough… – Bernard Apr 16 '16 at 17:52