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?

Kade Reese 2022-07-23 Answered
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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Answers (1)

eri1ti0m
Answered 2022-07-24 Author has 11 answers
Let x = [ x 1 . . . x n ] and therefore x T = [ x 1 . . . x n ]
If you evaluate x 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 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.
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