(NBHM_PhD_2010_Screening Test_Algebra)
Pick out the true statement(s):
(a) There exist $n×n$ matrices $A$ and $B$ with real entries such that $(I −(AB−BA))^n = 0$.
(b) If $A$ is a symmetric and positive definite n×n matrix, then $(tr(A))^n ≥ n^ndet(A)$ where $‘tr’$ denotes the trace and $‘det’$ denotes the determinant of a matrix.
(c) Let A be a $5×5$ skew-symmetric matrix with real entries. Then $A$ is singular.
(a) I was searching for $A$ and $B$ such that $I −(AB−BA)$. I didn't get. I took $A= \left[ {\begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{array} } \right]$ and $B= \left[ {\begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{array} } \right]$. then applied to the equation and tried to solve. For $n \times n$, this work is tedious and time consuming.
(b)I don't know how to prove it. Using induction, I tried. but $n=1$ it is true. I assumed $n=k$, the statement is true. I don't know how to prove for $n=k+1$?
(c)I could judge the statement without any confusion. The statement is true.
But this problem appeared in an entrance examination, time is limited. Which elementary linear algebra theorem should I use to judge? How to prove/disprove (a) and (b)? Please help me.
Problem $a$) the matrices $AB$ and $BA$ have the same trace. We conclude $AB-BA$ has trace equal to $0$, it follows from here that at least one eigenvalue of $AB-BA$ is different from $1$. This implies $I-(AB-BA)$ has at least one non-zero eigenvalue, and thus $I-(AB-BA)$ is not nilpotent.
Problem $b$). Let $0<\lambda_1,\dots,\lambda_n$ be the eigenvalues of $A$.
we have $tr(A^n)=(\lambda_1+\dots+ \lambda_n) ^n\geq n^n(\lambda_1\lambda_2\dots \lambda_n)=n^n\det(A)$
To prove the inequality notice it is equivalent to
$\frac{\lambda_1+\dots+\lambda_n}{n}\geq \sqrt[n]{\lambda_1\dots\lambda_n}$ which is AM-GM
problem $c$) since $A$ and $A^T=-A$ are similar we conclude the only complex eigenvalue with odd multiplicity can be $0$. So the matrix must be singular since $5$ is odd.