# Cayley-Hamilton Theorem: characteristic polynomial and distinct eigenvalues

I do not know where to start for solving this exercise although I have the "official" solution. In the solution, I see that some variables are exchanged but I cannot connect the steps to a coherent "story". Some direct help is highly appreciated. I do know how to get the poles from a characteristic equation.

Exercise

Consider a matrix $A \in \mathbb{R}^{n \times n}$ with the characteristic polynomial $$\det(sI-A)=a(s)=s^n+a_{n-1}s^{n-1}+ \cdots + a_1 s+a_0$$

a) Show that if $A$ has distinct eigenvalues $(λ_1, λ_2, \ldots , λ_n)$, the following relationship holds: $$\Lambda^n + a_{n-1}\Lambda^{n-1}+ \cdots + a_0 I = 0$$ with $\Lambda = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & \lambda_n \end{pmatrix}$

b) Now show that $$A^n + a_{n-1}A^{n-1}+ \cdots + a_0 I = 0$$

(This proves the Cayley-Hamilton Theorem for distinct eigenvalues.) Hint: Use the fact that a matrix A with distinct eigenvalues can be written as $A = T ΛT^{−1}$; where $Λ$ is diagonal.

Solution:

a) The characteristic equation is true for all eigenvalues of $A$, $λ_1 . . . λ_n$ $$\lambda^n + a_{n-1}\lambda^{n-1}+\cdots + a_0 = 0$$ $$\Lambda = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & \lambda_n \end{pmatrix}$$ so $\Lambda^n + a_{n-1} \Lambda^{n-1}+ \cdots + a_0 I = 0$

This is the matrix characteristic equation.

b) With distinct eigenvalues and diagonal $Λ$ we have $$A=T\Lambda T^{-1} \\ A^2 = T \Lambda T^{-1} T \Lambda T^{-1} = T \Lambda^2 T^{-1} \\ \vdots \\ A^m=T \Lambda^m T^{-1}$$

Multiply the matrix characteristic equation by $T$ (left) and $T^{-1}$ (right) to obtain $$T \Lambda^n T^{-1} + a_{n-1}T \Lambda^{n-1} T^{-1}+\cdots + a_0 TT^{-1}= 0$$ $$A^n + a_{n-1}A^{n-1}+\cdots+a_0 I = 0$$

• You do not get the roots from the characteristic equation (in most cases you can't), then were given to you by the authors of this exercise. – TZakrevskiy Dec 7 '16 at 20:31
• Why can I not get roots from a characteristic equation? Usually, the ABC formula will help me to get the roots. In case of higher orders of polynomials, Matlab could help me. I am really puzzled what I am supposed to learn from this exercise. – autoship Dec 7 '16 at 22:03
• in general case for a matrix of dimension $5$ the eigenvalues are not expressible in radicals. A couple of decades ago Godunov and his students made a $7\times 7$ matrix with small int eger entries (around $10^4..10^5$) with eigenvalues $0,\pm 1,\pm2,\pm 4$. None of the existing software back then could find those eigenvalues - Mathematica, Maple, Matlab, Reduce, you name it. Whatever the numerical method there will always be matrix that beats that method. – TZakrevskiy Dec 8 '16 at 9:42
• What about brute forcing? :-) – autoship Dec 8 '16 at 12:24
• good luck with that=) – TZakrevskiy Dec 8 '16 at 12:31

It seems that you have done. Let $\det(sI-A)=a(s)=s^n+a_{n-1}s^{n-1}+ \cdots + a_1 s+a_0$ be the characteristic polynomial. We know by assumption that it has exactly $n$ distinct roots, namely: $\lambda_1,\dots,\lambda_n$.

Let $\Lambda = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & \lambda_n \end{pmatrix}\text{ }$ be the diagonal matrix whose entries are the roots of $a(s)$.

We want to show that $\Lambda^n + a_{n-1}\Lambda^{n-1}+ \cdots + a_0 I = 0.$ Since

$$\Lambda^k = \begin{pmatrix} \lambda_1^k & 0 & \cdots & 0 \\ 0 & \lambda_2^k & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & \lambda_n^k \end{pmatrix},$$

thus $$\Lambda^n + a_{n-1}\Lambda^{n-1}+ \cdots + a_0 I= \begin{pmatrix} a(\lambda_1) & 0 & \cdots & 0 \\ 0 & a(\lambda_2) & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & a(\lambda_n) \end{pmatrix}= \begin{pmatrix} 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & 0 \end{pmatrix}$$ because $\lambda_i$ are roots of $a(s)$.

The second part ask to prove the theorem for a matrix $A$ similar to $\Lambda$, i.e.

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

Thus $A = T ΛT^{−1}$; where $Λ$ is diagonal, by definition of similarity; and

$$A^k = T\Lambda^k T^{-1} \quad \forall k\in \Bbb N.$$

Finally: $$A^n + a_{n-1}A^{n-1}+\cdots+a_0 I = 0\Longleftrightarrow T \Lambda^n T^{-1} + a_{n-1}T \Lambda^{n-1} T^{-1}+\cdots + a_0 TT^{-1}= 0$$ $$\Longleftrightarrow T( \Lambda^n + a_{n-1} \Lambda^{n-1} +\cdots + a_0I)T^{-1}= 0 \Longleftrightarrow \Lambda^n + a_{n-1} \Lambda^{n-1} +\cdots + a_0I =T^{-1}0T=0.$$

• and what do I know now with this result? In other words, if the eigenvalues are in diagonal form in a matrix, the result of the polynomial function is zero, correct? Why do we have to write the $T$ and $T^{-1}$ in $A^k = T \Lambda^kT^{-1}$, if $A = \Lambda^k$ anyway? – autoship Dec 8 '16 at 12:36
• If $A=\Lambda$ then $T=I$, but in general there many matrices having the same characteristic polynomial, and only one of these is diagonal. Precisely: if two matrices $A,B$ are similar i.e. $A=TBT^{-1}$ for some $T$, then they have the same characteristic polynomial. So the second part asks you to prove that: if $\Lambda$ is a solution of $a(s)$, then also every similar matrices are solutions. – InsideOut Dec 8 '16 at 18:09