Put the following matrices in canonical form?

**Borkborkmath** · February 21st 2011, 11:09 PM

a]
(1 0 1)
(0 1 0)
(0 0 1)

b]
(1 0 1)
(0 1 0)
(1 0 1)

c]
(1 0 0 1)
(0 1 1 0)
(0 0 1 0)
(1 0 0 0)

Help please.

**tonio** · February 22nd 2011, 03:38 AM

Originally Posted by Borkborkmath

a]
(1 0 1)
(0 1 0)
(0 0 1)

b]
(1 0 1)
(0 1 0)
(1 0 1)

c]
(1 0 0 1)
(0 1 1 0)
(0 0 1 0)
(1 0 0 0)

Help please.

1) What do you call "canonical form" to? Do you mean Jordan Canonical Form, Smith

Canonical form, Rational Canonical form...?

2) Show some self work and show where you're stuck.

Tonio

**Borkborkmath** · February 22nd 2011, 08:08 AM

I'm not sure which canonical form is required, since my book doesn't say. It just tells me to put those matrices into canonical form.

I know that λI is a canonical form for Jordan matrices and that
(λ 1 0 0)
(0 λ 1 0)
(0 0 λ 1)
(0 0 0 λ)
is another canonical form.

The thing is that I do not know what to do or where to start.

**tonio** · February 22nd 2011, 08:48 AM

Originally Posted by Borkborkmath

I'm not sure which canonical form is required, since my book doesn't say. It just tells me to put those matrices into canonical form.

I know that λI is a canonical form for Jordan matrices and that
(λ 1 0 0)
(0 λ 1 0)
(0 0 λ 1)
(0 0 0 λ)
is another canonical form.

The thing is that I do not know what to do or where to start.

Well, from what you say it seems (just seems!) that they mean the Jordan Canonical Form (JCF), but

in order to be completely sure you must read the book before the part of that question.

And even if it actually is the JCF and you don't know about this, it is too involved a question to write

down an explanation in this forum. You better go to your instructor.

Tonio

**TheEmptySet** · February 22nd 2011, 08:53 AM

Since you mentioned the Jordan Canonical form.

The first thing you need is the characteristic polynomial

a)

$\displaystyle c(\lambda)=|A-\lambda I|=\begin{vmatrix} 1-\lambda & 0 & 1 \\ 0 & 1-\lambda & 0 \\ 0 & 0 & 1-\lambda \end{vmatrix}=(1-\lambda)^3$

Putting $\lambda =1$ back into the matrix gives

$\displaystyle (A-I)=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

The dimension of the nullspace of this matrix is 2 so we need to find the Jordan cannonical form.

Note that the minimum polynomial must be of the form
$m(\lambda)=(1-\lambda)^a$ where $a=1,2,3$

because $m(\lambda) | c(\lambda)$ and have the same roots.

Since $(A-I) \ne 0$ matrix the minimum polynomial cannot be $(\lambda -1)$ Now try $(\lambda -1)^2=\lambda^2-2\lambda+1$ and check that $A^2-2A+I=0$ this gives that $a=2$
The degree of the minimum polynomial gives the size of the largest Jordan block. So the JCF has one 2 by 2 block and one 1 by 1 block. So the JCF is

$\displaystyle \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$

Also do you need the matricies $P \text{ and } P^{-1}$ such that
$P^{-1}AP=J$ where $J$ is the JCF?

See if you can do the next one!

**Borkborkmath** · February 26th 2011, 10:07 AM

Thank you very much for your explanations!
I hadn't had time to come back and check until now, but I think I got b:
[0 0 0]
[0 1 0]
[0 0 2]?

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