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Let $$A = \left( {\matrix{ 0 & 1 & 0 & 0 \cr 0 & 0 & 2 & 0 \cr 0 & 0 & 0 & 3 \cr 0 & 0 & 0 & 0 \cr } } \right)$$

The characteristic polynomial is $f_A(x)=x^4$.

Questions:

  1. How do I conclude that $m_A=x^4$? Do I have to evaulate $A^3$ and figure that $A^3\ne 0$?
  2. The jordanian form is: $A = \left( {\matrix{ 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \cr 0 & 0 & 0 & 0 \cr } } \right)$. Why?
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  • $\begingroup$ The characteristic polynomial is $f_A(x)=x^4$. $\endgroup$ – user35603 Dec 20 '14 at 13:01
  • $\begingroup$ @user35603, I've edited the question. $\endgroup$ – AlonAlon Dec 20 '14 at 13:17
  • $\begingroup$ have you computed $ker(A) = \{x: Ax = 0 \}?$ $\endgroup$ – abel Dec 20 '14 at 13:20
  • $\begingroup$ @abel, can you please explain the use of $\ker(A)$? $\endgroup$ – AlonAlon Dec 20 '14 at 13:20
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    $\begingroup$ the number of jordan blocks corresponding to the eigenvalue $0$ is the dimension of $ker(A).$ $\endgroup$ – abel Dec 20 '14 at 13:22
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i will use $e_1, e_2, e_3$ and $e_4$ to stand for the standard basis. with this convention, we have $$Ae_1 = 0, Ae_2 = e_1, Ae_3= 2e_2 \mbox{ and }Ae_4 = 3e_3.$$ so with respect to the basis $\{e_1, e_2, \frac{1}{2}e_3, \frac{1}{6}e_4 \} = \{f_1,f_2,f_3, f_4\}$ in the new $f$ basis, the linear transformation is $$Tf_1 = 0, Tf_2 = f_1, Tf_3 = f_2, Tf_4 = f_3$$ so $T$ has the representation $\pmatrix{0&1&0&0\cr0&0&1&0\cr0&0&0&1\cr0&0&0&0},$ called the jordan form, similar to the matrix in question.

if we permute the basis,e.g,$\{f_4,f_3,f_2,f_1\}$ the same transformation is now represented by $$\pmatrix{0&0&0&0\cr1&0&0&0\cr0&1&0&0\cr0&0&1&0}$$

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  • $\begingroup$ We've learned in class that the $1$-s are below the main diagonal. Can you explain why at this example the $1$-s are above? $\endgroup$ – AlonAlon Dec 20 '14 at 13:50
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(2) As you observe, the characteristic polynomial of $A$ is $x^4$, so it has sole eigenvalue zero (of multiplicity $4$). Now, if we denote by $J_k$ the $k \times k$ Jordan block with eigenvalue $0$, the possible Jordan forms of such a matrix are $$J_4, \quad J_3 \oplus J_1, \quad J_2 \oplus J_2, \quad J_2 \oplus J_1 \oplus J_1, \quad 0.$$

Now, $$\ker A = \langle e_4 \rangle,$$ and in particular $\dim \ker A = 1$, but the only Jordan-form matrix above with $1$-dimensional kernel is $J_4$, and the dimension of the kernel of two similar matrices agrees, so $J_4$ must be the Jordan form of $A$. In fact, as abel observes, the number of Jordan blocks in the Jordan form a matrix with all zero eigenvalues (a nilpotent matrix) is exactly the dimension of its kernel.

(1) Jordan blocks $J_k$ of eigenvalue zero have the property that $J_k^r \neq 0$ for $r < k$ and $J_k^r = 0$ for $r \geq k$. So, $J_4^3 \neq 0$ and hence $A^3 \neq 0$, that is, the minimal polynomial of $A$ must be $x^3$. More generally, the minimal polynomial of a matrix with all zero eigenvalues is $x^r$, where $r$ is the size of the largest Jordan block in its Jordan form.

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  • $\begingroup$ You're welcome, I hope you found it useful. $\endgroup$ – Travis Dec 20 '14 at 19:58

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