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Given $λ^n + a_{n-1}λ^{n-1} + ... + a_0$ is the characteristic polynomial of a matrix $\mathbf{A}$ I have to proof or disprove $\mathbf{A}^n + a_{n-1}*\mathbf{A^{n-1}} + ... + a_0\mathbf{I} = 0$

What is wrong with this proof?

$λ_i^n + a_{n-1}λ_i^{n-1} + ... + a_0 = 0$ with $λ_i$ an eigenvalue of $\mathbf{A}$ with corresponding eigenvector $\mathbf{v_i}$

From this equation follows the vector equation $λ_i^n\mathbf{v_i} + a_{n-1}λ_i^{n-1}\mathbf{v_i} + ... + a_0\mathbf{v_i} = 0$

From the definition of the eigenvalue follows $\mathbf{A}^n\mathbf{v_i} + a_{n-1}*\mathbf{A^{n-1}}\mathbf{v_i} + ... + a_0\mathbf{v_i} = 0$

Distribution $(\mathbf{A}^n + a_{n-1}*\mathbf{A^{n-1}} + ... + a_0\mathbf{I})\mathbf{v_i} = 0$

$\mathbf{v_i}$ can't be zero -> $\mathbf{A}^n + a_{n-1}*\mathbf{A^{n-1}} + ... + a_0\mathbf{I} = 0$

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    $\begingroup$ That is the Cayley–Hamilton theorem. $\endgroup$
    – lhf
    Aug 5, 2019 at 11:07
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    $\begingroup$ Given that it's the statement of the Cayley-Hamilton Theorem (which is nontrivial to prove), if the problem was given as a HW problem, it's not really fair to the student. $\endgroup$
    – quasi
    Aug 5, 2019 at 11:14

2 Answers 2

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In your last step you seem to divide by a vector, which is not allowed.

The matrix $M = \mathbf{A}^n + a_{n-1}*\mathbf{A^{n-1}} + ... + a_0\mathbf{I}$ is zero if and only if $Mv = 0$ for all vectors $v$. You proved this only for eigenvectors of $\mathbf{A}$. If $\mathbf{A}$ had a basis of eigenvectors, your proof would be sufficient, but $\mathbf{A}$ might not have any eigenvectors at all.

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  • $\begingroup$ How do you get to "not any eigenvectors at all"? You might not be able to select a full basis of eigenvectors, but since $I,A,...,A^n$ are linearly dependent, there is a minimal polynomial $A^m+...+c_1A+c_0I=0$ , and for any root of the minimal polynomial $(A-\lambda I)$ has a non-trivial kernel. $\endgroup$ Aug 5, 2019 at 11:13
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    $\begingroup$ Consider the rotation by e.g. $\pi/2$ in the plane, i.e. $A = \begin{pmatrix} \cos(\pi/2) & -\sin(\pi/2) \\ \sin(\pi/2) & \cos(\pi/2) \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ as a map from $\mathbb R^2$ to itself. Then the characteristic polynomial is $\chi_A(x) = x^2 + 1$ and does not have a real root. It is also intuitively clear that a rotation by 90 degrees cannot have an eigenvector, because no vector is just "stretched" by some factor. $\endgroup$ Aug 5, 2019 at 11:21
  • $\begingroup$ Ok, then this proof method would have to apply to the complexification of the vector space, so that complex eigenvalues and eigenvectors become admissible. $\endgroup$ Aug 5, 2019 at 11:23
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    $\begingroup$ In the complex case your argument is correct and the characteristic polynomial always has a root and thereby an eigenpair. However, the field might be neither $\mathbb C$ nor $\mathbb R$, so a complexification is not possible in general. $\endgroup$ Aug 5, 2019 at 11:27
  • $\begingroup$ @iljusch No, it's not correct in the complex case. Yes, in that case $A$ must have an eigenvector, but all that the argument proves is that $0$ is the only eigenvalue of $p(A)$, which does not (immediately) imply $p(A)=0$. (For example, $0$ is the only complex eigenvalue of $\begin{bmatrix}0&1\\0&0\end{bmatrix}$.) $\endgroup$ Aug 5, 2019 at 16:28
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Unless the field $K$ of scalars is algebraically closed, there may not be an eigenvalue $\lambda_i\in K$.

But even if we assume $K$ is algebraically closed (e.g., $K=\mathbb{C}$), there's still an error in your last line.

Assuming $\mathbf{v_i}$ is an eigenvector, the equation $$ (\mathbf{A}^n + a_{n-1}\mathbf{A^{n-1}} + ... + a_0\mathbf{I})\mathbf{v_i} = 0 $$ doesn't imply that $$\mathbf{A}^n + a_{n-1}\mathbf{A^{n-1}} + ... + a_0\mathbf{I}=0$$ It only implies that it's singular (i.e., its null space is nonzero).

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