# Prove Cayley-Hamilton theorem

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?

• Aren't you just supposed to show that this specific matrix satisfies its characteristic equation? Dec 1, 2021 at 11:10
• I think this is just proof by calculation, no? Dec 1, 2021 at 11:23
• I think that is what the question calls for, but I thought I could be clever and sidestep the calculation. Dec 1, 2021 at 11:39

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.

• Thank you. I thought it felt a little to easy. Dec 1, 2021 at 10:50
• Does this proof only work for symmetrical nXn-matrices which have n eigenvectors. If the matrix only hade one eigenvector, then you can't describe all vectors in eigenvectors and for some vectors $(c_2A^2+c_1A+c_0I)\vec{v}\ne 0$ right? Dec 1, 2021 at 12:39
• @ErikEriksson That's right, in other words what I've given here is a proof of the Cayley-Hamilton theorem for the diagonalizable matrices. (If your matrix is symmetric, then it is always diagonalizable, so this always works.) The whole proof is a bit more complicated. The whole thing follows, for example, by understanding the characteristic polynomial in terms of the generalized eigenspaces of $A$. Dec 1, 2021 at 13:30
• What goes wrong is basically some nilpotency (see Jordan canonical form), and it turns out that the multiplicity of each root of the characteristic polynomial is always enough to kill all of the nilpotent parts. Dec 1, 2021 at 13:30
• (What I'm basically saying here is the equivalent fact that the so-called minimal polynomial of $A$ divides the characteristic polynomial of $A$.) Dec 1, 2021 at 13:37

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 $$p(A) = A^2 - 2A - 8I = \begin{pmatrix}10&6\\6&10\end{pmatrix} - 2 \begin{pmatrix}1&3\\3&1\end{pmatrix} - 8 \begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}0&0\\0&0\end{pmatrix}.$$