Linear algebra exam questions Pick out the true statements: 


*

*There exist $n\times n$ matrices $A$ and $B$ with real entries such that $(I-(AB-BA)^n) = 0$. 

*If $A$ is symmetric and positive definite matrix then $tr(A)^n\geq n^n \det(A)$. 
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I am stucked, unable to solve this problem. 
 A: For the second question: Consider the matrix $A=\text{diag}(a,b)$, then the relation is $a+b\ge 4 ab$. It's trivial to construct a counterexample, e.g. $a=b=1$. More concretely, all $a,b$ with $a>\frac{1}{4}$ and $b>\frac{a}{4a-1}$ violate the inequality.
(I cleared the answer for the first one, which was flawed.)
A: For the first statement, if $n=2$, let $A=\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & -\frac{1}{2}\end{pmatrix}$ and $B= \begin{pmatrix} 1 & 1 \\ -1 & -1\end{pmatrix}$
Then 
$$AB-BA= \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2}& \frac{1}{2}\end{pmatrix} - \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2}& +\frac{1}{2}\end{pmatrix}= \begin{pmatrix} 0 & 1 \\ 1& 0\end{pmatrix}$$
I think this can be generalized to $2m \times 2m$ matrices by using
$\begin{pmatrix} A & 0 & 0& ..&0 \\
0 & A & 0& ..&0 \\
. & .& . & ..& . \\
0 & 0 & 0& ..&A \\\end{pmatrix}$ and $\begin{pmatrix} B & 0 & 0& ..&0 \\
0 & B & 0& ..&0 \\
. & .& . & ..& . \\
0 & 0 & 0& ..&B \\\end{pmatrix}$
For the second statement $A=I$...
If the problem asks instead to prove that 
$$tr(A)^n \geq n^n \det(A)$$
then let $\lambda_1,..,\lambda_n$ be the eigenvalues. Then they are real and positive (WHY?), thus by $AM-GM$ we have:
$$\frac{tr(A)}{n}=\frac{\lambda_1+...+\lambda_n}{n} \geq \sqrt[n]{\lambda_1 \cdot ... \lambda_n}=\sqrt[n]{\det(A)}$$
A: Regarding the first question.
For $n$ even, this is certainly true. Every traceless matrix can be written as the commutator of two square matrices. That is, given $C$ such that $tr(C) = 0$, there exists $A,B$ such that $[A,B]=AB - BA=C$. Now for even $n$ we can take the diagonal matrix $$D = diag(1,-1,1,-1,\cdots,1,-1)$$ which can be written in terms of some $A,B$. Then we have $AB - BA = D$ which implies the equation.
For odd $n$, if we can show the existence of a traceless $n$th root of the identity, then we can certainly prove the statement.
Edit: I'm  not too familiar with real canonical forms, so someone might be able to correct me. Take the complex diagonal matrix with the diagonals given by the $n$th roots of unity. Would taking the real canonical form of such a matrix give us the desired $nth$ root traceless matrix?
