# Finding matrix determinant based on Trace, Spectrum and Eigenspace

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!

• what is $E$ ? Is it the identity ? – zwim Jan 20 at 20:52
• Yep. E is the Identity Matrix – Wolf Jan 20 at 22:13

## 1 Answer

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.

• Really neat and concise answer! Thank you! – Wolf Jan 20 at 22:12