Prove Cayley-Hamilton theorem

Question

I have a exercise in my linear algebra textbook:

Let $c_2\lambda^2+c_1\lambda +c_0=0$ be the characteristic equation for the matrix $$A=\begin{pmatrix}1&3\\3&1\end{pmatrix}$$

Prove that $c_2A^2+c_1A +c_0I=0$

This is Cayley-Hamilton theorem.

My solution:

If a root $\lambda$ exists, then it is the eigen value for the eigen vector $\vec{v}$. If we multiply $c_2A^2+c_1A +c_0I$ with $\vec{v}$ then we get: $$(c_2\lambda^2+c_1\lambda +c_0)\vec{v}$$ and since the eigen vector is not the zero vector and $c_2\lambda^2+c_1\lambda +c_0=0$ is true, $c_2A^2+c_1A +c_0I=0$ is also true.

Is this enough to prove the theorem and solve the problem?

Answer 1

I wouldn't say that you've got a proof yet, but you're not far off.

The problem is that you have shown that the eigenvector $v$ with eigenvalue $\lambda$ is in the kernel of $c_2 A^2 + c_1 A + c_0 I$ (i.e. $(c_2 A^2 + c_1 A + c_0 I) \vec{v} = 0$), but a priori the entire matrix $c_2 A^2 + c_1 A + c_0 I$ need not be zero if it maps a single vector nonzero to zero (I suppose unless $A$ were a $1 \times 1$).

On the other hand for this specific $2\times2$ matrix, if you knew that $c_2 A^2 + c_1 A + c_0 I$ mapped two linearly independent vectors to zero then it must be zero (if you like, for dimension reasons). So, what if we actually calculate the characteristic polynomial of $A$, and find that it has two distinct roots (it does)? Then the eigenvectors corresponding to these two roots have to be linearly independent (the roots are $4$ and $-2$, and they must be eigenvalues). Then we can use your argument for each eigenvalue/vector, and then this would complete the proof.

Answer 2

You have a specific matrix given: $$ A=\begin{pmatrix}1&3\\3&1\end{pmatrix}. $$ Its characteristic equation is given by $$ p(\lambda):=\det(\lambda I - A) = \begin{vmatrix}\lambda-1&-3\\-3&\lambda-1\end{vmatrix} = (\lambda-1)^2-(-3)^2 = \lambda^2-2\lambda-8. $$ Now, $$ A^2=\begin{pmatrix}1&3\\3&1\end{pmatrix}\begin{pmatrix}1&3\\3&1\end{pmatrix} = \begin{pmatrix}10&6\\6&10\end{pmatrix} $$ so $$ p(A) = A^2 - 2A - 8I = \begin{pmatrix}10&6\\6&10\end{pmatrix} - 2 \begin{pmatrix}1&3\\3&1\end{pmatrix} - 8 \begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}0&0\\0&0\end{pmatrix}. $$