# Finding eigenvalues of a matrix given its characteristic polynomial and the trace and determinant

I am told a matrix A has characteristic polynomial: $$(\lambda−1)^3(a\lambda+\lambda^2+b),$$ and that $$\text {tr}(A)=12,$$ and $$\det(A) =14.$$

I am asked to find the eigenvalues.

Is the only to do this by simply using the fact that the sum of the eigenvalues is the trace, and the product is the determinant?

I had done so, and had gotten that the eigenvalues are 1, 2 and 7, (1 with multiplicity 3), which was correct.

I am asking because won't the number of eigenvalues have to be less than the number of rows and columns of the matrix? What if, if given different trace and determinant values, I somehow found eigenvalues which satisfied the trace and determinant given, but had a number of eigenvalues that exceeded the number of columns and rows of the matrix?

Also, I was not given the dimensions of the matrix A.

• The characteristic polynomial has degree equal the dimension of the matrix, and the eigenvalues are exactly the root of the characteristic polynomial. So you can't have more eigenvalues than the dimension of the matrix. – AlexL Oct 13 '18 at 0:15

## 2 Answers

I hope I understand everything you said.

1. The degree of your characteristic polynomial is 5. Therefore, the matrix it is coming from is $$5 \times 5$$.
2. An $$n \times n$$ matrix has $$n$$ (complex) eigenvalues (counting multiplicities). That's called the fundamental theorem of algebra.
3. Should the matrix be real and should you be only interested in real eigenvalues, you would see that some of the eigenvalues are conjugate from one another.
4. Given that the eigenvalues are the zeros of the characteristic polynomial, and given that it is factored, you directly notice that $$\lambda = 1$$ is an eigenvalue with algebraic multiplicity $$3$$. Therefore, you only have $$5-3 = 2$$ remaining eigenvalues to find.
5. You have two conditions (trace and determinant), i.e. two equations, and solve for two eigenvalues. This shouldn't be an issue and solvable for (almost) every pair (trace,det).

For an $$n\times n$$ matrix, the characteristic polynomial is an $$n^{th}$$ degree polynomial so the number of eigenvalues counting the multiplicities, is the same as number rows and which is also the same as the number of columns.

You are correct about the trace being the sum of eigenvalues but the determinant is $$(-1)^n$$ times the product of the eigenvalues.

Thus they did not have to give you the dimension of the matrix, because it was understood from the degree of the characteristic polynomial.