# If $p(\lambda)$ is the characteristic polynomial of a matrix $\bf A$, how to prove or disprove that $p(\mathbf A)=0$?

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 -> $$\mathbf{A}^n + a_{n-1}*\mathbf{A^{n-1}} + ... + a_0\mathbf{I} = 0$$

• That is the Cayley–Hamilton theorem.
– lhf
Aug 5, 2019 at 11:07
• 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. Aug 5, 2019 at 11:14

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.

• 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. Aug 5, 2019 at 11:13
• 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. Aug 5, 2019 at 11:21
• Ok, then this proof method would have to apply to the complexification of the vector space, so that complex eigenvalues and eigenvectors become admissible. Aug 5, 2019 at 11:23
• 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. Aug 5, 2019 at 11:27
• @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 $\begin{bmatrix}0&1\\0&0\end{bmatrix}$.) Aug 5, 2019 at 16:28

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).