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What is the characteristic polynomial and minimal polynomial of $A$ and $B$ ?

$A=\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a\end{pmatrix},B=\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{pmatrix}$

My attempt : for $A$) , $ch_A(x) = (x-a)^4$ , $m_A = (x-a)^2(x-a)$

for $B)$ $ch_B(x) = (x-a)^4$ , $m_B = (x-a)^2(x-a)^2$

where $ch$ and $m$ denote the characteristic and minimal polynomial.

Is my answer is correct or not ???

Any hints/solution will be apprciated

thanks u

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  • $\begingroup$ So, you’re saying that the characteristic and minimal polynomials of $B$ are identical. What is $(B-aI)^2$? $\endgroup$ – amd Oct 8 '18 at 3:31
  • $\begingroup$ @amd..$(B- aI)^2 =\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{pmatrix}$ but why u take $(B-aI)^2$ ?? $\endgroup$ – jasmine Oct 8 '18 at 3:41
  • $\begingroup$ Redo $(B-aI)^2$. Also $(A-aI)^2$. $\endgroup$ – Jyrki Lahtonen Oct 8 '18 at 3:53
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    $\begingroup$ You’re saying that $(x-a)^4$ is the minimal polynomial, which means that evaluating this at $x=B$ gives zero, i.e., $(B-aI)^4=0$ and that there’s no smaller power of $B-aI$ that also vanishes. The latter is false. $\endgroup$ – amd Oct 8 '18 at 21:04
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Set

$N_A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}; \tag 1$

then

$N_A^2 = 0; \tag 2$

further note that

$A - aI = N_A; \tag 3$

thus

$(A - aI)^2 = 0, \tag 4$

which means that

$m_A = (x-a)^2, \tag 5$

since $A$ can satisfy no polynomial of degree 1; indeed,

$cA + dI = \begin{bmatrix} ca + d & c & 0 & 0 \\ 0 & ca + d & 0 & 0 \\ 0 & 0 & ca + d & 0 \\ 0 & 0 & 0 & ca + d \end{bmatrix} = 0 \Longleftrightarrow c = d = 0; \tag 6$

thus $m_A(x) = (x - a)^2$ is the polynomial of least degree satisfied by $A$; hence, minimal.

It it easy to see by direct and simple calculation that

$e_1 = (1, 0, 0, 0)^T, \; e_3 = (0, 0, 1, 0)^T, \; e_4 = (0, 0, 0, 1) \tag 7$

are eigenvectors of $A$, each with eigenvalue $a$, and that

$(A - aI) e_2 = e_1, \tag 8$

that is, $e_2 = (0, 1, 0, 0)^T$ is a generalized eigenvector of $A$ corresponding to eigenvalue $a$; therefore $a$ is an eigenvalue of algebraic multiplicity $4$ (tho' of geometric multiplicity $3$); thus the characteristic polynomial of $A$ is

$c_A(x) = (x - a)^4, \tag 9$

which of course may also be had as

$c_A(x) = \det(A - xI). \tag{10}$

As for $B$, we set

$N_B = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}; \tag{11}$

then

$N_B^2 = 0; \tag{12}$

therefore

$(B - aI)^2 = N_B^2 = 0, \tag{13}$

so

$m_B(x) = (x -a)^2; \tag{14}$

we can see that

$\deg m_B(x) \ne 1 \tag{15}$

by more or less the same logic that was used above to show $\deg m_A(x) \ne 1$.

Finally, we see that $e_1$ and $e_3$ are eigenvectors of $A$ corresponding to $a$, and that here $e_2$ and $e_4$ are generalized eigenvectors for $a$; it follows now that $a$ is of algebraic multiplicity $4$, and we conclude that

$c_B(x) = (x - a)^4. \tag{16}$

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    $\begingroup$ thanks u lewis sir ur explaination is very simple and beautiful and very easy to understand $\endgroup$ – jasmine Oct 8 '18 at 6:15
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    $\begingroup$ Thank you my friend and thanks for the "acceptance"! 💣😉😎 $\endgroup$ – Robert Lewis Oct 8 '18 at 6:16
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$A=\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a\end{pmatrix},$ first i break them into two jordan nlock forms that $P=\begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix}$ and $Q=\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}$

now , the characteristic and minimal polynomial of $P = (\lambda -a)^2=f(t)$

and the characteristic and minimal polynomial of $Q = (\lambda -a)^2=g(t)$

Now the minimial polynomial of A $m_A = Lcm(f(t),g(t))=lcm\{(\lambda -a)^2,(\lambda -a)^2\}=(\lambda -a)^2$

similarly for B matrix we get $m_B = Lcm(f'(t),g'(t))=lcm\{(\lambda -a)^2,(\lambda -a)^2\}=(\lambda -a)^2$

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