Why is $(\|x\|_2)^2$ given by $x^T\!x$ where x is a vector?

Question

Suppose that $x$ is a real number, then

$$(\|x\|_2)^2 = x \cdot x = x^2$$

Now suppose $x$ is a real vector, then

$$(\|x\|_2)^2 = x^T\! x $$

Why should it be obvious that the multiplication sign turns into the transpose?

Is it by definition?

Answer 1

The definition of the norm $\|x\|$ is $(x,x)$, where the latter is the standard inner product. The inner product $(x,y)$ is given by $\sum_i x_iy_i$, i.e., the sum of the products of the components. If you treat $x$ and $y$ as $n\times 1$ matrices, then this expression is simply $x^Ty$. From all of this you get $\|x\|=x^Tx$. (By the way, the reason for defining the standard inner product this way is coming from the law of cosines.)

Answer 2

The multiplication didn't turn to a transpose - you are still multiplying $x^T$ and $x$. The value is $(x^T)\cdot x$.

If you had assumed that $x$ was a row vector, then you'd have $\|x\|^2 = xx^T = x\cdot (x^T)$. So you are treating that internal $T$ as a binary operation of some sort, when it is not - it is a unary operation.

If $z$ is a complex number, then $|z|^2 = \overline{z}\cdot z$.

So the transpose is acting like complex conjugate (and, indeed, in the standard $2\times2$ matrix representation of the complex numbers, conjugation of complex numbers is transposition of matrices.)