Finding matrix determinant based on Trace, Spectrum and Eigenspace

Question

Given the Matrix $A$ $\in$ $\mathbb {R}^{4\times4}$ with the following known properties:

1. $|Spec(A)| = 3$

2. $Tr(A) = 92$

3. $Tr(A^2) = 4060$

4. $Rank(A - E)= 2$

I need to determine the determinants of this matrix. So far I'm gathering that since $|Spec(A)|$ = 3 I have 3 eigenvalues and because the determinant is the product of all the eigenvalues I have $ det(A) =\lambda_1\lambda_2\lambda_3 $

From 4. I gather that $Rank(A-E) = 2 < 4 \iff \lambda = 1$ is an eigenvalue of $A$

The relationship between determinant and trace as well as the trace of $A^2$ is not clear to me aside from one of the coefficients of the characteristic of the equation being written in the form of the trace but not for an $n$ this high.

Thank you all in advance!

Answer 1

The fact that $\operatorname{rank}(A - E) = 2$ tells you not only that $1$ is an eigenvalue but also that its algebraic multiplicity is at least $2$. Because $A$ has $3$ distinct eigenvalues, the eigenvalues must be $1,1,\lambda_1,\lambda_2$ for some distinct $\lambda_1,\lambda_2$ not equal to $1$.

The trace of a matrix is the sum of its eigenvalues. Thus, we have $$ \operatorname{tr}(A) = 2 \cdot 1 + \lambda_1 + \lambda_2 = 92,\\ \operatorname{tr}(A^2) = 2 \cdot 1^2 + \lambda_1^2 + \lambda_2^2 = 4060. $$ The rest is algebra.