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given is a matrix A with

$\begin{pmatrix} a & 1 & 0 & \cdots & 0 \\ 0 & a & 1 & \cdots & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & \cdots & a\end{pmatrix}$

The characteristic polynomial should by given by:

$P_A(t) = (a-t)^n$.

All Eigenvalues are $a$, so the algebraic multiplicities for each eigenvalue is $n$.

The geometrical multiplicity for each eigenvalue is then $1$, since for eigenvalue $a$ I have eigenvector $(1, 0, ..., 0)$.

Now I have to determine the minimal polynomial $m_A(t)$.

It should be $m_A(t) \in T := \{ (a-t), (a-t)^2, (a-t)^3, \cdots, (a-t)^n \}$.

At least $m_A((a-t)^n) = 0$ since $P_A(A) = (a-A)^n = 0$ right?

1) The matrix $(a\cdot E_n - A)^1$ has Rank $n-1$.

2) The matrix $(a \cdot E_n - A)^2$ has Rank $n-2$. and so on.

What I need is a matrix with rank $0 = n-n$.

This means

3) The matrix $(a \cdot E_n - A)^n = 0$. So the minimal polynomial is the characteristic polynomial.

How can I say - in a shorter and more mathematical way than I did - that with each new factor, the matrix-product $(a \cdot E_n - A)^i$ only lowers its rank down 1 number?

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  • $\begingroup$ I'm confused what $m_A((a-t)^n)$ means. It seems you're evaluating the characteristic polynomial, which doesn't make sense. To find the minimal polynomial you can show that $(a-A)^k\neq 0$ for $k<n$. $\endgroup$ – Ian Coley Apr 9 '14 at 8:28
  • $\begingroup$ Aside from Ian's point, which I agree with, the argument is pretty mathematical in my opinion. You could say that $(a\cdot E_n - A)$ takes $e_1$ to $0$, and $e_i$ to $-e_{i-1}$ for $i>1$, so $(a\cdot E_n - A)^{n-1}\cdot e_n=(-1)^{n-1}e_1$, so $(a\cdot E_n -A)^{n-1}$ is not the zero matrix, which leaves only $(a - t)^n$ as a possibility for the minimal polynomial. But, I think your argument is fine. $\endgroup$ – Callus Apr 9 '14 at 8:29
  • $\begingroup$ Thank you very much. The algebraic and geometric multiplicities were computed correctly? $\endgroup$ – Vazrael Apr 9 '14 at 8:50
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We have

$$B:=A-aE_n=\begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \cdots & \cdots & \cdots & \cdots & 1 \\ 0 & 0 & 0 &\cdots & 0\end{pmatrix}$$ and let $\left\{v_k\right\}=\left\{(0,\ldots,0,1,0,\ldots,0)^T\right\}$ the standard basis then we see that $$Bv_k=v_{k-1}$$ hence by induction $$B^{n-1}v_n=v_1\ne0$$ hence $$B^{n-1}\ne0$$ and then the minimal polynomial of $A$ is its characteristic polynomial.

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