Is it possible for $A^T\vec{b} \bullet \vec{a} = A\vec{b} \bullet \vec{a}$ without $A=A^T$?

Question

Is it possible for $A^T\vec{b} \bullet \vec{a} = A\vec{b} \bullet \vec{a}$ without $A=A^T$?

Geometrically speaking, $A^T\vec{b} \bullet \vec{a}$ is the magnitude of the projection of $A^T\vec{b}$ onto $\vec{a}$ multiplied by the magnitude of $\vec{a}$

$A^T\vec{b} \bullet \vec{a}=|proj(A^T\vec{b}\rightarrow\vec{a})|*|\vec{a}|$

And, geometrically speaking, $A\vec{b} \bullet \vec{a}$ is the magnitude of the projection of $A\vec{b}$ onto $\vec{a}$ multiplied by the magnitude of $\vec{a}$

$A\vec{b} \bullet \vec{a}=|proj(A\vec{b}\rightarrow\vec{a})|*|\vec{a}|$

So, is there a way for...

$$\|proj(A\vec{b}\rightarrow\vec{a})\| = \|proj(A^T\vec{b}\rightarrow\vec{a})\|$$

...without $A$ being symmetric? Without $A=A^T?$

Thanks! The more geometric the answer, the better!

Answer 1

The identity is equivalent to $b^TAa=b^TA^Ta$ for all $a,b$. This implies $b^T(A-A^T)a=0$. Take $a=e_i$ and $b=e_j$, standard basis vectors, which implies $[(A-A^T)]_{ij}=0$. So $A=A^T$ is necessary.

Answer 2

If $A^T\vec{b}\bullet\vec{a} = A\vec{b}\bullet\vec{a}$ all $a$ and $b$, it means that if you choose a specific orthonormal basis $\vec{e_i}$ and a specific vector $\vec b$ , then the projections of $A\vec{b}$ on the vectors of this basis are the same as projections of $A^T\vec{b}$ on the vectors of this basis. This means that $$ A^T\vec b = \sum_i {\rm proj}(A^T\vec{b}\rightarrow\vec e_i)e_i = \sum_i {\rm proj}(A\vec{b}\rightarrow\vec e_i)e_i = A\vec{b}$$ Since for all $b$ we have $A^T\vec{b} = A\vec b$, then by the definition of equality of operators it means that $A^T=A$.

If the condition $A^T\vec{b}\bullet\vec{a} = A\vec{b}\bullet\vec{a}$ is not supposed to be satisfied by all $a$ and $b$, but only for some specific ones, then it's possible that $A\neq A^T$. For example for any symmetric matrix $B$ and any antisymmetric matrix $C$ such that $b\in \ker C$, $C\neq 0$ you can take $A=B+C$ and you have $$ A^T\vec{b}\bullet\vec{a} = (B^T + C^T)\vec{b}\bullet\vec{a} = (B - C)\vec{b}\bullet\vec{a} = B\vec{b}\bullet\vec{a} = (B +C)\vec{b}\bullet\vec{a}= A\vec{b}\bullet\vec{a}$$ but $$A^T = B - C \neq B+C = A$$