Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
"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 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 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 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 at 9:14

3 Answers 3

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

share|improve this answer
@jibounet Thanks, I am not very good at squaring numbers=) –  flawr Jul 23 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.

share|improve this answer

Part a) seems solved.

Here is part b) solution


share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.