# $2$ question on matrix with some condition

we need to tell whether $a,b$ true or false,

I know that there does not exist $n\times n$ $A,B$ such that $(AB-BA)=I$,as if we take trace of both side they are not equal, so $a$ is false?

$b$ I have no idea, could any one give me hint?

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if $(a)$ must be true for all $n$, the $(a)$ is false (for $n=1$, it's impossible). – Xoff Aug 14 '12 at 20:47

(a) Let $C = AB - BA$. Then we have that the eigenvalues of $(I-C)^n$ are all $0$, and hence the eigenvalues of $I - C$ are also all zero. (Here we use the fact that if $\lambda$ is an eigenvalue of $A$ then $\lambda^n$ is an eigenvalue of $A^n$).

Hence, the eigenvalues of $C$ are all $1$, which implies that the trace of $C$ is $n$. On the other hand, $tr(AB) = tr(BA)$ implies that $tr(C) = tr(AB - BA) = 0$, a contradiction.

So, no such matrices exist.

(b) Take $a_1,a_2, \cdots, a_n$ to be the eigenvalues of $A$, and note that $\displaystyle tr(A) = \sum_{i=1}^n a_i, \ \ \ \det(A) = \prod_{i=1}^n a_i$.

As positive definiteness of $A$ gives us that $a_1, a_2, \cdots, a_n$ are all positive, therefore we apply the AM -GM Inequality which immediately gives us what is required.

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As $A$ is symetric and positive, $A$ is diagonalizable and there exist $a_1,a_2 \cdots a_n$ are the eigenvalues of A nonnegatives. Then, $$\dfrac{a_1+a_2 \cdots +a_n}{n} \ge (a_1\cdot a_2 \cdots a_n)^{1/n},$$ and follows $$\dfrac{tr(A)}{n}\ge det(A))^{1/n}.$$

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We don't exactly need A to be diagonalizable actually, just that every eigenvalue is a positive real suffices. – Rijul Saini Aug 14 '12 at 20:52