GO Classes DPP | GATE CS | Linear Algebra | Matrix Algebra

Answer 1

Statement A is true. For compatible matrices, $\operatorname{tr}(AB)=\operatorname{tr}(BA)$.

Statement B is true. By the cyclic property of trace, $\operatorname{tr}(CAB)=\operatorname{tr}(ABC)=\operatorname{tr}(BCA)$.

Statement C is false. A counterexample is

$A=\begin{bmatrix}1&0\end{bmatrix}$ and $B=\begin{bmatrix}0\\1\end{bmatrix}$.

Then

$AB=\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}=0$,

but

$BA=\begin{bmatrix}0\\1\end{bmatrix}\begin{bmatrix}1&0\end{bmatrix}=\begin{bmatrix}0&0\\1&0\end{bmatrix}\ne 0$.

So statement C is false.

Statement D is true. If $AB=I_m$, then using statement A,

$\operatorname{tr}(BA)=\operatorname{tr}(AB)=\operatorname{tr}(I_m)=m$.

Hence, statements A, B, and D are true, and statement C is false.

Answer 2

If A is m×n  and B is n×m, then AB and BA have the same non-zero eigenvalues with the same algebraic multiplicities. If m≠n, the matrix with larger dimension has additional zero eigenvalues.

 

Even if AB and BA have different dimensions, their traces are equal because they share the same non-zero eigenvalues and any extra eigenvalues are zero, which do not affect the sum.