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Find the minimal and characteristic polynomials of $$A = \begin{bmatrix}λ & 1&0&...&0\\0 & λ&1&...&0\\...&...&λ&...&1\\0&...&...&...&λ\end{bmatrix}$$

The characteristic polynomial is $(\lambda-x)^n=0$. But how to get the minimal polynomial? The minimal polynomial should be the monic polynomial $\phi$ of least degree satisfying $\phi(A)=0$.

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    $\begingroup$ The degree of $(\lambda-x)$ in the minimal polynomial is the size of the largest Jordan block for $\lambda$. $\endgroup$ – amd May 24 '17 at 7:59
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The characteristic polynomial and the minimal polynomial are equal.

Set $0 \leq k \leq n-1. $

$(1)$ Note that the column $A^ke_1$, where $e_1$ is the first vector of the standard base, has a non-zero entry in it's $k-1$ row and zeroes above it. Can be proved by induction.

$(2)$ $A^ke_1$ is not a linear combination of $A^0e_1, Ae_1,\dots,A^{k-1}e_1$.

$(3)$ $m_A|c_A \implies \deg m_A \leq \deg c_A.$

Let $m_A(X) = \sum_{i=0}^da_iX^i$, where $a_d=1$.

So $m_A(A) = \sum_{i=0}^da_iA^i = 0$, hence $\sum_{i=0}^da_iA^ie_1 = 0$

Rearranging: $\sum_{i=0}^{d-1}-a_iA^ie_1 = A^de_1$

By $(2)$, it can't be that $d\leq n-1$, hence $\deg m_A=d \geq n.$

$c_A$ and $m_A$ are monic polynomials, so $\deg m_A=n= \deg c_A$

$$\implies m_A=c_A$$

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  • $\begingroup$ How does the matrix relate to the column $A^ke_1$ $\endgroup$ – stedmoaoa May 24 '17 at 12:25
  • $\begingroup$ @stedmoaoa There is no direct relation, I'm just using it later in the proof. $\endgroup$ – Itay4 May 24 '17 at 12:27
  • $\begingroup$ Can you be more specific about $A^ke_1$ $\endgroup$ – stedmoaoa May 24 '17 at 12:34
  • $\begingroup$ @stedmoaoa Generally, the product of any matrix $M$ and $e_1$ is the first column of $M$. In our case, check what is $A^k$, and the product will be it's first column, and that's what we need for the proof. $\endgroup$ – Itay4 May 24 '17 at 12:37
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Hint The minimal polynomial divides the characteristic polynomial. In this case the latter is $c_A(x) = (x - \lambda)^n$, so the former must have the form $m_A(x) = (x - \lambda)^k$ for some $k \leq n$. Before attempting a proof for general $n$, it might be useful to work out the case $n = 3$ to see better what's going on.

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  • $\begingroup$ The minimal is the same as the characteristic but how to generalize it? $\endgroup$ – stedmoaoa May 24 '17 at 12:31
  • $\begingroup$ As you can see from doing the computation evaluating $(A - \lambda I)^{n - 1}$ does not give the zero matrix, so like you say, the minimal polynomial must be $(x - \lambda)^n$. Can you show that this is true for general $n$? This is manageable because in this case the matrices $(A - \lambda I)^k$ have a simple form $\endgroup$ – Travis May 24 '17 at 12:36
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Hint: Check matrix $A-\lambda {I}$ for nilpotency.

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