MHB 25.3 Find the Jordan Normal Form of A

[karush]

View attachment 8908
one book example
$\textsf{Suppose that A is a matrix whose characteristic polynomial is}$
$(\lambda-2)^2(\lambda + 1)^2,
\quad \dim\left(E_2\right)=1
\quad \dim\left(E_{-1}\right)=2$
$\textsf{Find the Jordan Normal Form of A Find the Jordan Normal Form of A}$
$$\quad \dim\left(E_2\right)=1 \quad \dim\left(E_{-1}\right)=2$$
$\textsf{First how do we get A got this from W|A}$
$$\begin{bmatrix}2&0&0&0\\0&2&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}
=(\lambda-2)^2(\lambda + 1)^2$$
$\textsf{but don't think this the right direction... not sure how we use}$
$$ \dim\left(E_2\right)=1 \quad \dim\left(E_{-1}\right)=2$$

I had this posted on another forum but there were no replys

 

[StoneTemplePython]

one book example
$\textsf{Suppose that A is a matrix whose characteristic polynomial is}$
$(\lambda-2)^2(\lambda + 1)^2,
\quad \dim\left(E_2\right)=1
\quad \dim\left(E_{-1}\right)=2$

Can you first take this information and state the minimal polynomial?

 

[karush]

Can you first take this information and state the minimal polynomial?

$(\lambda-2)^2(\lambda + 1)^2,
\quad \dim\left(E_2\right)=1
\quad \dim\left(E_{-1}\right)=2$
Im not sure but
$\left[\begin{array}{r}2 \end{array}\right]$
and
$\left[\begin{array}{r}-1&0\\0&-1 \end{array}\right]$
and then
$\left[\begin{array}{r}2&0&0\\0&-1&0\\0&0&-1 \end{array}\right]$
doesn't look like it

 

[StoneTemplePython]

$(\lambda-2)^2(\lambda + 1)^2,
\quad \dim\left(E_2\right)=1
\quad \dim\left(E_{-1}\right)=2$
Im not sure but
$\left[\begin{array}{r}2 \end{array}\right]$
and
$\left[\begin{array}{r}-1&0\\0&-1 \end{array}\right]$
and then
$\left[\begin{array}{r}2&0&0\\0&-1&0\\0&0&-1 \end{array}\right]$
doesn't look like it

A minimal polynomial is still a polynomial, so it would look something like

$p(x) = x^n +a_1 x^{n-1} + ... + a_{n-1}x + a_n$

(where monic form -- leading coefficient of one-- is standard for linear algebra). What you've written isn't a polynomial.

So, let's take a step back. Jordan forms are typically treated near the end of a linear algebra book after 'everything' else has been covered. There are a few ways to get your arms around them. My preference is to start with Cayley Hamilton, and then ask "can we do better? ".

So what do you know / what text are you using? I am particularly interested in whether you've covered things like Cayley Hamilton yet, and how you got comfortable with such a proof.

 

[karush]

Actually no I haven't

This is our last homework on linear Algebra
We start next week in differential equations

 

[StoneTemplePython]

I'm afraid there is much more for me to say here. You haven't answered the what do you know question -- i.e. there isn't really a way for me to guess what tools you have to work with.

For what its worth, I have seen that this kind of thing tends to happen when people don't study linear algebra on its own and instead get tidbits thrown at them in a diff eq or multivariable calc class. The stuff you are doing with Jordan Forms is conceptually very simple if you know how to work with upper triangular matrices. But again, without knowing what book you are using and what topics have been introduced it is not feasible for me to guess what you have to work with here.

 

[karush]

i got it and turned it in

 

[HOI]

Good! I hope you get a good grade! Since this particular matrix is "upper triangular", its eigenvalues are the numbers on the diagonal: 2, -1, and -1. The set of all eigenvectors for a given eigenvalue form a subspace (its "eigenspace"). The subspace for eigenvalue 2 is 1 dimensional but the eigenspace for eigenvalue might have dimension 2 or 1, depending on the number of independent eigenvectors. If there are two independent eigenvectors corresponding to eigenvalue -1 then the matrix is "diagonalizable" with the eigenvalues on the diagonal:
\begin{bmatrix}2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{bmatrix}.

If all eigenvectors corresponding to an eigenvalue are multiples of one vector, so the "eigenspace" has dimension 1, the best we could do is the "Jordan Normal Form":
\begin{bmatrix}2 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{bmatrix}.

It's easy to find a single eigenvector corresponding to eigenvalue 2: let \begin{bmatrix} x \\ y \\ z\end{bmatrix}. Then we must have \begin{bmatrix} 2 & -1 & 3 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}2x- y+ 3z \\ -y \\ -z\end{bmatrix}= 2\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}2x \\ 2y \\ 2z \end{bmatrix}. So we must have 2x- y+ 3z= 2x, -y= 2y, and -z= 2z. The last two equations give y= z= 0 so the first equation becomes 2x= 2x which is true for all x. Any eigenvector corresponding to eigenvalue 2 is of the form \begin{bmatrix} x \\ 0 \\ 0 \end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}.

So we look for eigenvectors corresponding to eigenvalue -1. If \begin{bmatrix}x \\ y \\ z \end{bmatrix} is such an eigenvector then we must have
\begin{bmatrix} 2 & -1 & 3 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}2x- y+ 3z \\ -y \\ -z\end{bmatrix}= -\begin{bmatrix} x \\ y \\ z\end{bmatrix}= \begin{bmatrix}-x \\ -y \\-z\end{bmatrix}.

So we must have 2x- y+ 3z= -x, -y= -y, and -z= -z. The last two equations are true for any y and z. 2x- y+ 3z= -x is the same as 3x- y+ 3z= 0 so y= 3x+ 3z. We can write any eigenvector as \begin{bmatrix} x \\ 3x+ 3z \\ z\end{bmatrix}= x\begin{bmatrix}1 \\ 3 \\ 0 \end{bmatrix}+ z\begin{bmatrix}0 \\ 3 \\ 1\end{bmatrix}. That is, the "eigenspace" corresponding to eigenvalue -1 has dimension 2 and is spanned by the two vectors \begin{bmatrix}1 \\ 3 \\ 0 \end{bmatrix} and \begin{bmatrix} 0 \\ 3 \\ 1 \end{bmatrix}. Since there are 3 independent eigenvectors, this matrix is diagonalizable.

Now, form the matrix, P, having those eigenvector as columns, P= \begin{bmatrix}1 & 1 & 0 \\0 & 3 & 3 \\ 0 & 0 & 1\end{bmatrix}. It's inverse matrix is P^{-1}= \begin{bmatrix}1 & -\frac{1}{3} & 1 \\ 0 & \frac{1}{3} & -1 \\ 0 & 0 & -1 \end{bmatrix}.

Then we have P^{-1}AP= \begin{bmatrix}1 & -\frac{1}{3} & 1 \\ 0 & \frac{1}{3} & -1 \\ 0 & 0 & -1 \end{bmatrix}\begin{bmatrix}2 & -1 & 3 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 1 & 0 \\0 & 3 & 3 \\ 0 & 0 & 1\end{bmatrix}
=\begin{bmatrix}2 & -\frac{2}{3} & 2 \\ 0 & -\frac{1}{3} & 1 \\ 0 & 0 & -1\end{bmatrix}\begin{bmatrix}1 & 1 & 0 \\0 & 3 & 3 \\ 0 & 0 & 1 \end{bmatrix}
= \begin{bmatrix}2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{bmatrix}.