Is $r.b*=b*A$ true?

Question

Let $a, b \in \Bbb C^n$ and $A\in \Bbb C^{n \times n}$. If $b^*\cdot a=1$ and $r=b^*\cdot A\cdot a$, is it true that: $r\cdot b^*=b^*A$?

Answer 1

The answer to your question is no. For an example where this doesn't work, take $$ a = \pmatrix{1\\0}, \quad b = \pmatrix{1\\0}, \quad A = \pmatrix{0&1\\1&0} $$

Answer 2

Taking the conjugate transpose, we get $$(rb^*)^* = b(b^*Aa)^* = ba^*A^*b$$ Which we need to be equal to $$(b^*A)^* = A^*b$$ Since matrix multiplication is associative, we can explicitly write $$ ba^*A^*b = ((ba^*)A^*)b$$ Then, we have the desired equality for all $b$ if and only if $$ba^* = I$$ for all $b$, which is not possible. Thus, the equality doesn't hold for all $b$.