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.

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

1 Answer 1


$$det(A-\lambda I)= (\lambda_1 - \lambda)...(\lambda_n - \lambda)$$

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 $A$

  • $\begingroup$ Thank you for your answer. I think that the first part makes complete sense, but I am still confused as to how to actually do the second part in practice. $\endgroup$ Commented Dec 16, 2019 at 19:09
  • $\begingroup$ @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. $\endgroup$ Commented Dec 16, 2019 at 20:53

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