Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For a matrix $A\in\mathbb{K}^{n\times n}$ where $\mathbb{K}\in\{\mathbb{R},\mathbb{C}\}$ the characteristic polynomial is defined as

$$\chi_A(\lambda) := \text{det}(A-\lambda I_n) = \sum_{k=0}^n c_k \lambda^k$$

with the recursive definition of the coefficients

$$c_n = (-1)^n,\;\;\;\;c_{n-1}=(-1)^{n-1}\text{tr}A,\;\;\;\;c_0 = \text{det}A$$

and the trace of the matrix $A$

$$\text{tr}A := \sum_{k=1}^n a_{kk}$$

But when i try to compute the polynomial my "hand-computed" solution is not the same as the one Mathematica provides.

$$A:=\begin{bmatrix}a&0&0\\0&b&0\\0&0&c\end{bmatrix}$$

leads for me to the polynomial

$$\chi_A^{Hand}(\lambda) = -\lambda^3+\lambda^2 (a+b+c)-\lambda (a+b+c) +abc$$

but calling

CharacteristicPolynomial[{{a, 0, 0}, {0, b, 0}, {0, 0, c}}, \[Lambda]]

in Mathematica leads to the result

$$\chi_A^{Mathematica}(\lambda)=a b c-a b \lambda -a c \lambda -b c \lambda +a \lambda ^2+b \lambda ^2+c \lambda ^2-\lambda ^3$$

Can anyone explain me, what i am doing wrong?

share|improve this question
2  
The answer is that you do not develop correctly the polynomial $(a-\lambda)(b-\lambda)(c-\lambda)$. The correct expression is the one given by Mathematica. –  Did Jul 16 '11 at 14:24
add comment

2 Answers

up vote 1 down vote accepted

Mathematica agrees with your answer. The degree is $n=3$. Your formulas say $c_3=(-1)^3=-1$, and $c_2 = (-1)^2 tr(A) = a+b+c$ and $c_0 = det(A) = a b c$, but the $c_1$ is not determined by your statement, so there is no contradiction. You wrongly concluded that $c_1 = -c_2$.

share|improve this answer
    
Understood. I don't know why i haven't seen this before! But is there any other algorithm or any other solution to make this one "really" recursive? This is just what was given to us in a lecture. –  Christian Ivicevic Jul 16 '11 at 14:29
1  
Yes, there is. I have seen in the book of Carl Meyer. That book also said it is not practical for inexact matrices because of numerical instabilities. Also you may find the Cayley-Hamilton wiki-page useful –  Sasha Jul 16 '11 at 14:35
    
"it is not practical for inexact matrices because of numerical instabilities" - the usual example (due to Jim Wilkinson) is the diagonal matrix with diagonal elements $1,2,\dots,20$; the eigenvalues are easy to compute, but woe betide thee who expands it to its eigenpolynomial... –  J. M. Jul 16 '11 at 15:07
add comment

Your "recursive" definition isn't really recursive. It only describes three of the coefficients. You get the same value for those as Mathematica does. The coefficient of $\lambda^1 = \lambda^{n-2}$ isn't described by your formulas, and I don't understand how you computed it.

share|improve this answer
1  
@Christian: That was exactly the point I was trying to explain to Yuval: that you get your solution by computing $c_{1}$ and $c_{2}$ using the formula for $c_{n-1}$. Aren't you given an $n \times n$-matrix and you have to prove the formulas for $c_{k}$ the three values $k = 0$, $k = n-1$ and $k=n$? –  t.b. Jul 16 '11 at 14:35
    
@Theo Buehler: I am not given any specific matrix. The exercise is to prove the correctness of the relations... might it be true that $c_{n-1}$ can be computed this way and the other parts like $c_1$ not? Therefore i could proove the correctness of this without havong a look at $c_1$ if e.g. i let $n=3$. –  Christian Ivicevic Jul 16 '11 at 14:39
1  
@Christian: Yes, that was my point. You're given $n$ and you can ignore $c_1, c_2, \ldots, c_{n-2}$. In general, the coefficients in the characteristic polynomial are given by the elementary symmetric polynomials in the eigenvalues (up to a sign): $c_{k} = (-1)^k e_{k}(\lambda_1,\lambda_2,\ldots,\lambda_n)$. –  t.b. Jul 16 '11 at 14:48
    
@Theo Buehler: Thanks for the help - und danke sehr ;) –  Christian Ivicevic Jul 16 '11 at 14:49
    
@Christian: my pleasure - gern geschehen. Yuval: sorry about the pings. –  t.b. Jul 16 '11 at 14:51
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.