# Suppose A has eigenvalues 1,2, 4.

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

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"So trace A²=14": beware that there is no logical connection with the previous statement "Trace A=7". What is the relation between 7 and 14 ? – Yves Daoust Jul 23 '14 at 8:32
You're wrong about $a)$. If $\lambda_{1},\lambda_{2},\lambda_{3}$ are the eigenvalues of $A$, then : $$\mathrm{Tr}(A) = \lambda_{1}+\lambda_{2}+\lambda_{3}$$ and $$\mathrm{Tr}(A^{2}) = \lambda_{1}^{2} + \lambda_{2}^{2} + \lambda_{3}^{2}$$ – jibounet Jul 23 '14 at 8:33
This question relies on the property that the Eigenvalues of matrix powers are the powers of the Eigenvalues. In other words, you can reason on the diagonalized matrix instead of the original, and perform the computations explicitly on the diagonal matrix. – Yves Daoust Jul 23 '14 at 8:35
@Lost1 : No, for a matrix $A$ in $\mathcal{M}_{n}(\mathbb{C})$, it always work. One just need to triangularize $A$. By squaring $A$, one squares the eigenvalues of $A$ on the diagonal of the [upper] triangular matrix. For your example, $\displaystyle A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, $\mathrm{Sp}(A) = \left\{ -1,1 \right\}$ and $\mathrm{Sp}(A^{2}) = \left\{ 1,1 \right\}$. As a consequence, $\mathrm{Tr}(A^{2}) = 2$, which is what we expected since $\displaystyle A^{2} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. – jibounet Jul 23 '14 at 9:14

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 $Av = \lambda v \implies A^2v = \lambda^2v$

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@jibounet Thanks, I am not very good at squaring numbers=) – flawr Jul 23 '14 at 8:37

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.

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Part a) seems solved.

Here is part b) solution

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

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