Why does x^Tx=norm(x)^2? The way I am thinking that while x^T is the transpose of x, then we cross product with itself using x^T, which results in a symmetric matrix, R. So R is a square matrix. How can we jump from a square matrix to norm(x)^2, where norm(x) supposed to mean the normal of x, then we square it with 2?

How can we prove that "We know that for any vector x, ${x}^{T}x=||x|{|}^{2}$. Thus , ..... "
The way I am thinking that while ${x}^{T}$ is the transpose of x, then we cross product with itself using ${x}^{T}$, which results in a symmetric matrix, R. So R is a square matrix. How can we jump from a square matrix to $||x|{|}^{2},$, where ||x|| supposed to mean the normal of x, then we square it with 2?
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eri1ti0m
Let $x=\left[\begin{array}{c}{x}_{1}\\ ...\\ {x}_{n}\end{array}\right]$ and therefore ${x}^{T}=\left[\begin{array}{ccc}{x}_{1}& ...& {x}_{n}\end{array}\right]$
If you evaluate $x\cdot {x}^{T}$ you multiply the rows of x (just one entry) with the columns of ${x}^{T}$ with also just one entry. This gives you the matrix. This is also called a dyadic product.
If you evaluate ${x}^{T}\cdot x$ you multiply the one single row of ${x}^{T}$ with the single column of x to give you just a single number. And by some kind of definition, this equals the squared length of the vector in a euclidian sense.
So: Just write x and ${x}^{T}$ as matrices and use the usual way of multiplying them.