Suppose A has eigenvalues 1,2, 4.

Question

a) What is the trace of $A^2$ b) What is the determinant of $(A^{-1})^T$

I need someone to check my answers and correct me, am especially not sure about part a), help me me out;

for a), I did--- Trace $A = 1+2+4 = 7$. So trace $A^2 = 14$

for b) $det(A^{-1})^T = 1/(1\times2\times4) = 1/8$

Answer 1

Yes this is correct as long as you assume that $A$ is a $3 \times 3$ matrix, except $tr(A^2) = 1^2+ 2^2 + 4^2 = 21$, since $Av = \lambda v \implies A^2v = \lambda^2v$

Answer 2

Multiplying a diagonal matrix by itself, $A^2$, will result in each of the diagonal entries being squared. For example, your matrix $A$ has diagonal entries 1, 2, and 4. $A^2$ has entries 1, 4, and 16. The trace of this matrix is 21.

Answer 3

Part a) seems solved.

Here is part b) solution

$$|(A^{-1})^T|=|(A^{-1})|=1/|A|.$$