Dot product of two vectors squared identity

Question

Does the following equation hold true in general?

$$ (a^Tb )^ 2 = (a^T b b^T a) $$

Where $a, b \in R^n$ are vectors of the same size n=1, 2, ...

I know it's true in n=1 and n=2 because I manually expanded it and verified it but I am not sure how to verify this for all n.

The reason for confusion is following: I know I can manipulate the RHS into

$$ a^Tb (a^Tb)^T $$

And if we define $u=a^Tb$, then

$$ u^2 = uu^T $$

But why does $u^2$ have to be the outer product? (as opposed to, say, inner product?)

Answer 1

The inner product $a^T b$ is a scalar. It is always the case $a^T b = b^T a$, by the definition of inner product (and commutative law of multiplication).

$$a^T b = \sum_j a_j b_j = \sum_j b_j a_j = b^T a.$$

So yes,

$$(a^T b)^2 = (a^Tb)(b^T a) = a^Tb b^T a.$$

Since $u = a^Tb$ is a scalar, $u^2 = u^T u = uu^T$.

Answer 2

Since $a^Tb = a\cdot b = b^Ta$, the result is true.