The trace and determinant of a 3×3 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.
Best Answer
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, $\operatorname{tr}(A^2) = \lambda_1^2+\lambda_2^2+\lambda_3^2=-2$.
$\det(A^2) = \det(A)^2 = 2^2 = 4$.