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Pick out the true statements:

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

  2. If $A$ is symmetric and positive definite matrix then $tr(A)^n\geq n^n \det(A)$. :(

I am stucked, unable to solve this problem.

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For the second part, consider a scalar matrix. – EuYu May 10 '12 at 20:03
are you sure the second one is not $tr(A)^n \geq n^n det(A)$? – N. S. May 10 '12 at 20:04
do both $n$ in the first statement refer to the same natural number? is this first statement supposed to be true for all $n$ or for some $n$? – Xabier Domínguez May 10 '12 at 20:04
ya sir both refers to the same natural number – srijan May 10 '12 at 20:06
For (1), consider matrices $A$ and $B$ where $B=A^{-1}$. – Joel Cornett May 10 '12 at 20:30
up vote 1 down vote accepted

For the first statement, if $n=2$, let $A=\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & -\frac{1}{2}\end{pmatrix}$ and $B= \begin{pmatrix} 1 & 1 \\ -1 & -1\end{pmatrix}$


$$AB-BA= \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2}& \frac{1}{2}\end{pmatrix} - \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2}& +\frac{1}{2}\end{pmatrix}= \begin{pmatrix} 0 & 1 \\ 1& 0\end{pmatrix}$$

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:

$$\frac{tr(A)}{n}=\frac{\lambda_1+...+\lambda_n}{n} \geq \sqrt[n]{\lambda_1 \cdot ... \lambda_n}=\sqrt[n]{\det(A)}$$

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sir nick kidman has given counter example for second statement then how can it be true? – srijan May 10 '12 at 20:17
Well $A=I$ is a counterexample ;) – N. S. May 10 '12 at 20:24
For the first one, I think i am wrong. Actually, if $A,B$ are $2 \times 2$, then $(AB-BA)^2= [tr(AB-BA)] (AB-BA)-\det(AB-BA)I_2$. Thus all you need is to find two matrices $A,B$ so that $\det(AB-BA)=-1$. – N. S. May 10 '12 at 20:28
Here both statements are wrong? – srijan May 10 '12 at 20:30
Well not sure about the first one ;) – N. S. May 10 '12 at 20:32

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.)

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Yours first part is not clear to me. Can you explain me please? – srijan May 10 '12 at 20:20
I don't think it is in general true that $[A,B]^2=I\Rightarrow[A,B] = \pm I$, consider the matrix $\frac{1}{5}\begin{pmatrix} 3 & 4 \\ 4 & -3\end{pmatrix}$ – EuYu May 10 '12 at 20:22
Actually i was not clear about the notation. Now its clear to me – srijan May 10 '12 at 20:24
In your answer, you write that "If $[A,B]^2 = I$ then $[A,B] = \pm I$". I'm just seeking some clarification on that part since it's not true. – EuYu May 10 '12 at 20:29
@EuYu: Yeah, you're right, I don't know how to resove it. – NikolajK May 10 '12 at 20:31

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?

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For odd $n$, $(AB-BA)^n=I$ implies that $1$ is an eigenvalue for $AB-BA$ which might lead to issues or not :D – N. S. May 10 '12 at 21:03

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