Let A=pmatrix0&2&-3-2&0&13&-1&0pmatrix and B be a matrix such that B(I-A)=I+A.

Question

Let $A=\begin{pmatrix}0&2&-3\-2&0&1\3&-1&0\end{pmatrix}$ and $B$ be a matrix such that $B(I-A)=I+A$. Then the sum of the diagonal elements of $B^TB$ is equal to \_\_\_\_.

Answer: 3

Accepted Answer

Note $A^T=-A$ (skew-symmetric). From $B(I-A)=I+A$, $B=(I+A)(I-A)^{-1}$.

Then $B^T=\big[(I-A)^{-1}\big]^T(I+A)^T=(I-A^T)^{-1}(I+A^T)=(I+A)^{-1}(I-A)$ (using $A^T=-A$).

So
$$B^TB=(I+A)^{-1}(I-A)\cdot(I+A)(I-A)^{-1}=(I+A)^{-1}(I+A)(I-A)(I-A)^{-1}=I.$$

Hence $B^TB=I_3$, and the sum of its diagonal elements is $1+1+1=3$.

Solutions · 1