# Given the characteristic polynomial for the matrix, prove these statements about the trace of the matrix and the determinant of the matrix.

Let $$A\in M_{n\times n}(\mathbb{R})$$, and assume that $$det(A-\lambda I)$$ factors completely into $$(\lambda_1 - \lambda)...(\lambda_n - \lambda)$$, a product of linear factors.

I need to do two things:

1. Show that $$Tr(A)=\lambda_1+...+\lambda_n$$
2. Show that $$det(A)=\lambda_1\cdot \cdot \cdot \lambda_n$$.

How should I approach this? I know that the trace of the matrix is the sum of the diagonal entries of the matrix. Should I try to prove that all the diagonal entries are equal to $$\lambda_i$$ - I could try to prove that each entry of the matrix $$a_{ii} = \lambda_i$$. Even though I think that would pretty much be enough for both of these proofs, I do not know how to do it.

Also, a different approach would be trying to prove that our matrix has to be either a diagonal matrix or a lower (upper) triangular matrix to have this type of factorization for the characteristic polynomial.

Any type of help would be much appreciated.

• $\det$ is continuous, so setting $\lambda = 0$ gives 2. Commented Dec 16, 2019 at 17:56

$$det(A-\lambda I)= (\lambda_1 - \lambda)...(\lambda_n - \lambda)$$
Let $$\lambda =0$$ and you get $$det(A)=\lambda_1\cdot \cdot \cdot \lambda_n$$
For the trace identity you need to check the coefficient of $$-\lambda ^{n-1}$$ in the characteristic polynomial which is the sum of the eigenvalues and identify that with the coefficient of $$-\lambda ^{n-1}$$ in the expansion of the determinant $$det(A-\lambda I)$$ which is the trace of $$A$$
• @LukaDuranovic Start with a 2X2 matrix and see how trace appears as the coefficient of $\lambda$ in the charactristic polynomial. It is not quite straight forward but you will get it. Commented Dec 16, 2019 at 20:53