Jordan form and canonical base of a matrix

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Jordan form and canonical base of a matrix

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#1 h Feb 26, 2023, 1:51 PM

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Find the Jordan form and a canonical basis of the following matrix $A$ over the field $Z_5$ :
$A = \begin{bmatrix} 2 & 1 & 2 & 0 & 0 \\ 0 & 4 & 0 & 3 & 4 \\ 0 & 0 & 2 & 1 & 2 \\ 0 & 0 & 0 & 4 & 1 \\ 0 & 0 & 0 & 0 & 2 \end{bmatrix}$

This post has been edited 1 time. Last edited by And1viper, Feb 26, 2023, 1:51 PM
Reason: Edited Latex

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#2 h Apr 18, 2025, 12:30 PM

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bro $Z$ is not a field, the problem is faulty

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#3 h Apr 19, 2025, 12:49 AM • 1 Y

Y by aidan0626

Suan_16 wrote:

bro $Z$ is not a field, the problem is faulty

Actually it is a field. $\mathbb{Z}_p$ is a finite field for all primes $p$ .

The eigenvalues are $\lambda_1=\lambda_2=\lambda_3=2$ and $\lambda_4=\lambda_5=4$ .

$A-2I_5=\begin{bmatrix} 0 & 1 & 2 & 0 & 0 \\ 0 & 2 & 0 & 3 & 4 \\ 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$

Observe from rows 3 and 4, we must take none of columns 4 and 5. Then row 2 forces us to take none of column 2. Then row 1 means we must take none of column 3. So $E_2=\text{span}(\mathbf{e}_1)$ . The geometric multiplicity is 1.

$(A-2I_5)^3=\begin{bmatrix} 0 & 4 & 0 & 1 & 3 \\ 0 & 3 & 0 & 1 & 3 \\ 0 & 0 & 0 & 4 & 2 \\ 0 & 0 & 0 & 3 & 4 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$

If we take none of columns 4 and 5, then we must take none of column 2. So the kernel here contains $\mathbf{e}_1$ and $\mathbf{e}_3$ . If we take instead 1 of column 4 and 3 of column 5, which means we have to take none of column 2. So another vector in this kernel is $\mathbf{e}_4+3\mathbf{e}_5$ .

$A-4I_5=\begin{bmatrix} 3 & 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 3 & 4 \\ 0 & 0 & 3 & 1 & 2 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 3 \end{bmatrix}$

From row 4 or 5, we must take none of column 5. Then from row 2, we must take none of column 4. Then from row 3, we must take none of column 3. So the geometric multiplicity is again 1.

$(A-4I_5)^2=\begin{bmatrix} 4 & 3 & 2 & 0 & 3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & 3 & 3 \\ 0 & 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 & 4 \end{bmatrix}$

We still must take none of column 5.
If we take none of columns 3 and 4, then we can take 1 of column 1 and 2 of column 2. So the kernel here contains $\mathbf{e}_1+2\mathbf{e}_2$ . We can take instead 1 of column 3 and 2 of column 4. In that case, we can take 1 of column 1 and 3 of column 2. This gives us $\mathbf{e}_1+3\mathbf{e}_2+\mathbf{e}_3+2\mathbf{e}_4$ .

So the generalized eigenvectors are $v_1=\begin{bmatrix}1\\0\\0\\0\\0\end{bmatrix}$ , $v_2=\begin{bmatrix}0\\0\\1\\0\\0\end{bmatrix}$ , $v_3=\begin{bmatrix}0\\0\\0\\1\\3\end{bmatrix}$ , $v_4=\begin{bmatrix}1\\2\\0\\0\\0\end{bmatrix}$ , and $v_5=\begin{bmatrix}1\\3\\1\\2\\0\end{bmatrix}$

So $A=PJP^{-1}$ where:

$P=\begin{bmatrix} 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 2 & 3 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 3 & 0 & 0 \end{bmatrix}$ , $J=\begin{bmatrix} 2 & 2 & 0 & 0 & 0 \\ 0 & 2 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 4 & 3 \\ 0 & 0 & 0 & 0 & 4 \end{bmatrix}$ , $P^{-1}=\begin{bmatrix} 1 & 2 & 0 & 4 & 2 \\ 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2 \\ 0 & 3 & 0 & 3 & 4 \\ 0 & 0 & 0 & 3 & 4 \end{bmatrix}$

Note that the three positions where we would normally have 1s are instead 2s and a 3. These are chosen so the product comes out correct.

This post has been edited 2 times. Last edited by rchokler, Apr 19, 2025, 3:10 AM

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#4 h Apr 19, 2025, 10:57 AM

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Oh sorry I forgot that it is $Z_5={0,1,2,3,4}$ instead of $Z^5$

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