Disclaimer: these are my personal solutions; they have not been reviewed, so be careful relying on them as there might be errors. Some competitions may have official solutions from the contest publishers which you may wish to refer to instead if you’re looking for verified solutions.
Problem 3
Let $A$ be an $n \times n$ nonsingular matrix with complex elements, and let $\bar{A}$ be its complex conjugate. Let $B = A\bar{A} + I$, where $I$ is the $n \times n$ identity matrix.
Prove or disprove:
- $A^{-1}BA = \bar{B}$
- The determinant of $A \bar{A} + I$ is real.
Hints
Hint #1: general approach for this problem
If you want to prove that they are equal, but you can’t directly convert $A^{-1}BA$ into $\bar{B}$ or vice versa, maybe try simplifying each of these expressions independently, and get them to equal to the same expression. For example, if $A^{-1}BA = X$ and $\bar{B} = X$, then $A^{-1}BA = \bar{B}$.
If you want to prove that they are not equal, try to reduce each side to an expression that is clearly unequal, e.g., if $A^{-1}BA = X$ and $\bar{B} = Y$, yet $X \ne Y$, then $A^{-1}BA \ne \bar{B}$.
Hint #2: should we aim to prove or disprove the equality?
Try to prove the equality. If you need a general hint for how to approach proving the equality in this case, see hint #1 above.
Solution
(a) Prove or disprove: $A^{-1}BA = \bar{B}$
Open to see the solution, but try solving it first or see the hints above.
Starting from the left side of the equation, we have:
$$ \begin{align*} A^{-1}BA & = A^{-1} (A \bar{A} + I) A & \quad \text{expand $B$} \\ & = (A^{-1} A \bar{A} + A^{-1} I) A & \quad \text{distribute $A^{-1}$} \\ & = (\bar{A} + A^{-1}) A & \quad \text{simplify} \\ & = \bar{A} A + A^{-1} A & \quad \text{distribute $A$} \\ & = \bar{A} A + I & \quad \text{simplify (1)} \\ \end{align*} $$Starting from the right side, we have:
$$ \begin{align*} \bar{B} & = \overline{A \bar{A} + I} & \quad \text{expand $B$} \\ & = \overline{A \bar{A}} + \bar{I} & \quad \text{since $\overline{A+B} = \bar{A} + \bar{B}$} \\ & = \bar{A} A + \bar{I} & \quad \text{since $\overline{AB} = \bar{A} \bar{B}$ and $\overline{\bar{A}} = A$} \\ & = \bar{A} A + I & \quad \text{since $\bar{I} = I$ (2)} \\ \end{align*} $$Using both $(1)$ and $(2)$, we find that:
$$ A^{-1}BA = \bar{A} A + I = \bar{B} $$
and hence,
$$ \boxed{ A^{-1}BA = \bar{B} } \quad \blacksquare $$
(B) Prove or disprove: $\det (A\bar{A} + I)$ is real
Still working on it.