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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!

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

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

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