# 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. Dec 7, 2016 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. Dec 7, 2016 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. Dec 8, 2016 at 9:42
• What about brute forcing? :-) Dec 8, 2016 at 12:24
• good luck with that=) Dec 8, 2016 at 12:31

It seems that you are 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 diagonal 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 asks 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? Dec 8, 2016 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. Dec 8, 2016 at 18:09