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.

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- Feb 21st 2011, 11:09 PMBorkborkmathPut the following matrices in canonical form?
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. - Feb 22nd 2011, 03:38 AMtonio
- Feb 22nd 2011, 08:08 AMBorkborkmath
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. - Feb 22nd 2011, 08:48 AMtonio

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

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 - Feb 22nd 2011, 08:53 AMTheEmptySet
Since you mentioned the Jordan Canonical form.

The first thing you need is the characteristic polynomial

a)

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

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

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

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

$\displaystyle m(\lambda)=(1-\lambda)^a$ where $\displaystyle a=1,2,3$

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

Since $\displaystyle (A-I) \ne 0$ matrix the minimum polynomial cannot be $\displaystyle (\lambda -1)$ Now try $\displaystyle (\lambda -1)^2=\lambda^2-2\lambda+1$ and check that $\displaystyle A^2-2A+I=0$this gives that $\displaystyle a=2$

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

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

Also do you need the matricies $\displaystyle P \text{ and } P^{-1}$ such that

$\displaystyle P^{-1}AP=J$ where $\displaystyle J$ is the JCF?

See if you can do the next one! - Feb 26th 2011, 10:07 AMBorkborkmath
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]?