MHB -m99.54 List possible Jordan Canonical forms for A.

[karush]

nmh{2000}
Let A be a matrix with characteristic polynomial
$ p_A(t) = (t − 1)^3(t − 5)^2(t − 6)$
(a) List the possible Jordan Canonical forms for A.
(b) Suppose all eigenspaces are one dimensional. What is the Jordan form for A in this case?

 

[karush]

$$\begin{bmatrix} 6 \end{bmatrix}$$
$$\begin{bmatrix} 5&0 \\ 0&5 \end{bmatrix}$$
$$\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$$
$$\begin{bmatrix}
6&0&0&0&0&0\\
0&5&0&0&0&0\\
0&0&5&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&1&0\\
0&0&0&0&0&1
\end{bmatrix}$$

 

[I like Serena]

Suppose $3$ is an eigenvalue with (algebraic) multiplicity $4$.
Then the possible corresponding Jordan blocks for eigenvalue $3$ are:
$$\begin{bmatrix}3 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{bmatrix}
$$

 

[karush]

Suppose $3$ is an eigenvalue with (algebraic) multiplicity $4$.
Then the possible corresponding Jordan blocks for eigenvalue $3$ are:
$$\begin{bmatrix}3 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{bmatrix}
$$

Then the possible corresponding Jordan blocks for eigenvalue $5$ are:
\begin{bmatrix} 5&0 \\ 0&5 \end{bmatrix} and \begin{bmatrix} 5&1 \\ 0&5 \end{bmatrix}so p_A(t) = (t − 1)^3(t − 5)^2(t − 6) has 2 possible Jordan Canonical forms for A.?what would more?

 

[I like Serena]

Then the possible corresponding Jordan blocks for eigenvalue $5$ are:
\begin{bmatrix} 5&0 \\ 0&5 \end{bmatrix} and \begin{bmatrix} 5&1 \\ 0&5 \end{bmatrix}so p_A(t) = (t − 1)^3(t − 5)^2(t − 6) has 2 possible Jordan Canonical forms for A.?what would more?

There are a bit more, since eigenvalue $1$ also has multiple possibilities for the corresponding Jordan block.

 

[karush]

There are a bit more, since eigenvalue $1$ also has multiple possibilities for the corresponding Jordan block.

$\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}
and
\begin{bmatrix} 1&1&0\\0&1&0\\0&0&1 \end{bmatrix}
and
\begin{bmatrix} 1&0&0\\0&0&1\\0&0&1 \end{bmatrix} $so now we can put the possible combos together?

 

[I like Serena]

$\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}
and
\begin{bmatrix} 1&1&0\\0&1&0\\0&0&1 \end{bmatrix}
and
\begin{bmatrix} 1&0&0\\0&0&1\\0&0&1 \end{bmatrix} $so now we can put the possible combos together?

I'm afraid that 3rd matrix is not correct.
Instead we should have:
$\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}
and
\begin{bmatrix} 1&1&0\\0&1&0\\0&0&1 \end{bmatrix}
and
\begin{bmatrix} 1&1&0\\0&1&1\\0&0&1 \end{bmatrix} $

And yes, now we can put the possible combos together.
There should be $2\times 3=6$ of them.

 

[karush]

I hope anyway,, I go blind looking at this...

$\textit{List the possible Jordan Canonical forms for A.}$\\ \\
$\textit{For eigenvalue 6 are}$
$$\begin{bmatrix} 6 \end{bmatrix}$$
$\textit{For eigenvalue 5 are}$
$$\begin{bmatrix} 5&0 \\ 0&5 \end{bmatrix},
\quad
\begin{bmatrix} 5&1 \\ 0&5 \end{bmatrix}$$
$\textit{ For eigenvalue 1 are}$
$$\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}
, \quad
\begin{bmatrix} 1&1&0\\0&1&0\\0&0&1 \end{bmatrix}
, \quad
\begin{bmatrix} 1&1&0\\0&1&1\\0&0&1 \end{bmatrix}$$
$\textit{ So the 6 possible Jordan Canonical Forms for A are:}$
$$\begin{bmatrix}
6&0&0&0&0&0\\0&5&0&0&0&0\\0&0&5&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\end{bmatrix},\quad
\begin{bmatrix}
6&0&0&0&0&0\\0&5&0&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\end{bmatrix},\quad
\begin{bmatrix}
6&0&0&0&0&0\\0&5&0&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&1\\0&0&0&0&0&1
\end{bmatrix}$$
$$\begin{bmatrix}
6&0&0&0&0&0\\0&5&1&0&0&0\\0&0&5&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\end{bmatrix},\quad
\begin{bmatrix}
6&0&0&0&0&0\\0&5&1&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\end{bmatrix},\quad
\begin{bmatrix}
6&0&0&0&0&0\\0&5&1&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&1\\0&0&0&0&0&1
\end{bmatrix}$$

(b) Suppose all eigenspaces are one dimensional. What is the Jordan form for A in this case?
I suppose the answer to this is"
$$\begin{bmatrix}
6\\5\\5\\1\\1\\1
\end{bmatrix}$$

 

[I like Serena]

I hope anyway,, I go blind looking at this...

$\textit{List the possible Jordan Canonical forms for A.}$\\ \\
$\textit{For eigenvalue 6 are}$
$$\begin{bmatrix} 6 \end{bmatrix}$$
$\textit{For eigenvalue 5 are}$
$$\begin{bmatrix} 5&0 \\ 0&5 \end{bmatrix},
\quad
\begin{bmatrix} 5&1 \\ 0&5 \end{bmatrix}$$
$\textit{ For eigenvalue 1 are}$
$$\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}
, \quad
\begin{bmatrix} 1&1&0\\0&1&0\\0&0&1 \end{bmatrix}
, \quad
\begin{bmatrix} 1&1&0\\0&1&1\\0&0&1 \end{bmatrix}$$
$\textit{ So the 6 possible Jordan Canonical Forms for A are:}$
$$\begin{bmatrix}
6&0&0&0&0&0\\0&5&0&0&0&0\\0&0&5&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\end{bmatrix},\quad
\begin{bmatrix}
6&0&0&0&0&0\\0&5&0&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\end{bmatrix},\quad
\begin{bmatrix}
6&0&0&0&0&0\\0&5&0&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&1\\0&0&0&0&0&1
\end{bmatrix}$$
$$\begin{bmatrix}
6&0&0&0&0&0\\0&5&1&0&0&0\\0&0&5&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\end{bmatrix},\quad
\begin{bmatrix}
6&0&0&0&0&0\\0&5&1&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\end{bmatrix},\quad
\begin{bmatrix}
6&0&0&0&0&0\\0&5&1&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&1\\0&0&0&0&0&1
\end{bmatrix}$$

Correct.

(b) Suppose all eigenspaces are one dimensional. What is the Jordan form for A in this case?
I suppose the answer to this is"
$$\begin{bmatrix}
6\\5\\5\\1\\1\\1
\end{bmatrix}$$

No, they're asking to pick one of the possible forms of A.
To figure out which one, we need to know what the eigenvectors are.
Can you tell what the eigenvectors are for, say, the first form of $A$?

 

[karush]

Correct.
No, they're asking to pick one of the possible forms of A.
To figure out which one, we need to know what the eigenvectors are.Can you tell what the eigenvectors are for, say, the first form of $A$?

Isn't it just 6

 

[karush]

Actually I really don't know what to do

 

[I like Serena]

Isn't it just 6

That is the first eigenvalue of the first form of A. An eigenvector is something different.

An eigenvector $\mathbf v$ is a non-zero vector that corresponds to an eigenvalue $\lambda$ such that:
$$A\mathbf v = \lambda \mathbf v$$

In this case:
$$A\begin{bmatrix}1\\0\\0\\0\\0\\0\end{bmatrix}=6\begin{bmatrix}1\\0\\0\\0\\0\\0\end{bmatrix}$$
Therefore $\lambda=6$ is an eigenvalue with eigenvector $\mathbf v=\begin{bmatrix}1\\0\\0\\0\\0\\0\end{bmatrix}$.

This is the only independent vector with this property, which means that the eigenspace of $6$ has dimension $1$ as requested by the problem statement.