Pick out the true statement(s):

  • (a) There exist $n \times n$ matrices with real entries such that $(I-(AB-BA))^n=0$.
  • (b) $A $ is a symmetric and positive definite $n\times n$ matrix then $(\operatorname{tr}(A))^n\ge n^n \det(A)$.
  • (c) Let $A$ be a $5 \times 5$ skew -symmetric matrix with real entries. Then $A$ is singular.

(a) If $C=AB-BA$ we see that $\operatorname{tr} C = 0$. If $(I-C)^n=0$, then $I-C$ is nilpotent and all eigenvalues are zero. Hence $\operatorname{tr} (I-C) = 0$, which implies that $\operatorname{tr}C = n$, a contradiction. Hence no such matrix can exist.

(b) The arithmetic and geometric means are related by ${x_1+...+x_n \over n} \ge \sqrt[n]{x_1\cdots x_n}$, which is equivalent to $(x_1+...+x_n)^n \ge n^n x_1\cdots x_n$. Since $\operatorname{tr} A = \lambda_1+\cdots + \lambda_n$ and $\det A = \lambda_1 \cdots \lambda_n$, we obtain the desired result.

(c) All eigenvalues of a skew symmetric matrix are imaginary. Since it is real and has odd dimension, it must have one real eigenvalue. The only number that is both real and imaginary is zero. Hence $A$ is singular.

  • $\begingroup$ Yuh dun good, +1! $\endgroup$ – Robert Lewis Jan 13 '14 at 4:30
  • $\begingroup$ b and c is clear . but for case a how Tr(C)=0? $\endgroup$ – nothingobvious Jan 13 '14 at 4:35
  • $\begingroup$ $\operatorname{tr}C = \operatorname{tr}(AB-BA) = \operatorname{tr}(AB) - \operatorname{tr}(BA) = \operatorname{tr}(AB)-\operatorname{tr}(AB) = 0$ (since $\operatorname{tr} (AB) = \operatorname{tr}(BA)$). $\endgroup$ – copper.hat Jan 13 '14 at 4:38

a) never true

Note that $AB - BA$ can never have $1$ as an eigenvalue. Hence $$ \text{If } \lambda \text{ is an eigenvalue of } I -(AB-BA) \text { then } \lambda \ne 0$$

c) If $\lambda$ is an eigenvalue, then so is $\lambda$. Any odd dimensional matrix must have a real eigenvalue. So zero is an eigenvalue. What can you conclude?

b) is easy

  • $\begingroup$ not getting. please elaborate $\endgroup$ – nothingobvious Jan 13 '14 at 3:29
  • $\begingroup$ $AB - BA$ can certainly have $1$ as an eigenvalue. Try e.g. $ A = \pmatrix{0 & 0\cr 1 & 0\cr}$, $B = \pmatrix{1 & 1\cr 0 & 0\cr}$. $\endgroup$ – Robert Israel Jan 13 '14 at 3:39
  • $\begingroup$ What is true is that $1$ is not the only eigenvalue. $\endgroup$ – Robert Israel Jan 13 '14 at 3:41
  • $\begingroup$ Sorry, my mistake $\endgroup$ – user44197 Jan 13 '14 at 3:47

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