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The trace and determinant of a 3x3 matrix satisfy Tr A=2 and det A=2. The sum of two eigenvalues of A is equal to the third eigenvalue. Then the trace and determinant of the matrix $A^2$ is equal to?

I know that the trace is equal to the sum of eigenvalues and determinant is equal to its products.

Let $\lambda_1,\lambda_2, \lambda_3$ be the eigenvalues

$\lambda_1+\lambda_2+\lambda_3= 2$

$\lambda_1\lambda_2\lambda_3=2$

$\lambda_1+\lambda_2=\lambda_3$

Even if i make some substitutions I do not know how to get it for $A^2$.

Please explain how to do this.

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  • $\begingroup$ If $A\vec v = \lambda \vec v$, then $A^2 \vec v = \lambda^2 \vec v$. What does that tell you about the trace of $A^2$? $\endgroup$ – DHMO Apr 18 '17 at 10:04
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From $\lambda_1+\lambda_2+\lambda_3=2$ and $\lambda_1+\lambda_2-\lambda_3=0$, we obtain $\lambda_1+\lambda_2=\lambda_3=1$.

From $\lambda_1\lambda_2\lambda_3=2$ we obtain $\lambda_1\lambda_2=2$.

Therefore, $\lambda_1^2+\lambda_2^2=(\lambda_1+\lambda_2)^2-2\lambda_1\lambda_2=-3$.

Therefore, $\lambda_1^2+\lambda_2^2+\lambda_3^2=-2$.


If $A\vec v = \lambda \vec v$, then $A^2 \vec v = \lambda^2 \vec v$. Therefore, $\lambda^2$ are the eigenvalues of $A^2$.

Therefore, $\operatorname{tr}(A^2) = \lambda_1^2+\lambda_2^2+\lambda_3^2=-2$.

$\det(A^2) = \det(A)^2 = 2^2 = 4$.

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