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Find characteristic Polynomial of a matrix?

Posted 9 years ago

I'm trying to find the Characteristic Polynomial of a matrix. I have the worked solution to the problem, but all my polynomials seem to be the solutions I find, but multiplied by -1. I don't know what I'm doing incorrectly.

For example, I run the line"

CharacteristicPolynomial[{{-6, 4, -8}, {28, -15, 32}, {21, -12, 25}}, x]

and get

2 - 5 x + 4 x^2 - x^3

where as my book's solution is: -2 + 5 x - 4 x^2 + x^3

This is making my graphs completely upside-down. Here is the full problem:

[M] Let A = [ {-6, 4, -8}, {28, -15, a}, {21, -12, 25}] where each {} contains a column of matrix A

For each value of a in the set {32, 31.9, 31.8, 32.1, 32.2}, compute the characteristic polynomial of A and the eigenvalues. In each case, create a graph of the characteristic polynomial p(t) = det(A-tI) for 0<=t<=3. If possible, construct all graphs on one coordinate system. Describe how the graphs reveal the changes in the eigenvalues as a changes.

Any help would be greatly appreciated!

POSTED BY: Joseph Savin
3 Replies
Posted 9 years ago

Hi,

I agree with others that the sign does not matter here. The roots are the only values of interest. It's worthwhile -- though somewhat circular -- to calculate the characteristic polynomial as follows:

Poly =  Times @@ ((x - #) & /@ Eigenvalues[{{-6, 4, -8}, {28, -15, 32}, {21, -12, 25}}])
Expand@Poly

Out[]= (-2 + x) (-1 + x)^2
Out[]= -2 + 5 x - 4 x^2 + x^3

Then we see that, possibly, your book's convention involves choice between either

$$P_{+}(x) = \prod (x-x_{i})$$

$$P_{-}(x) = - \prod (x-x_{i})$$

with $x_i$ the $i$ eigenvalue. I think this is probably the clearest way to state the sign convention, but it is also equivalent to saying ''highest power of $x$ positive''.

POSTED BY: Brad Klee
Posted 9 years ago
POSTED BY: David Keith

It is a matter of definition: as documented, Mathematica computes the characteristic polynomial of A as Det[A - x I] while some authors may use Det[x I - A]. Of course going from one definition to the other is trivial, just multiply by (-1)^First[Dimensions[A]].

POSTED BY: Ilian Gachevski
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