# what is the characteristic polynomial and minimal polynomial of A and B?

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

• So, you’re saying that the characteristic and minimal polynomials of $B$ are identical. What is $(B-aI)^2$? – amd Oct 8 '18 at 3:31
• @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 $(B-aI)^2$ ?? – jasmine Oct 8 '18 at 3:41
• Redo $(B-aI)^2$. Also $(A-aI)^2$. – Jyrki Lahtonen Oct 8 '18 at 3:53
• 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. – amd Oct 8 '18 at 21:04

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

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

• thanks u lewis sir ur explaination is very simple and beautiful and very easy to understand – jasmine Oct 8 '18 at 6:15
• Thank you my friend and thanks for the "acceptance"! 💣😉😎 – Robert Lewis Oct 8 '18 at 6:16

$$A=\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a\end{pmatrix},$$ 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 $$m_B = Lcm(f'(t),g'(t))=lcm\{(\lambda -a)^2,(\lambda -a)^2\}=(\lambda -a)^2$$