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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How can we prove that &quot;We know that for any vector x,       x    T    x  =      |        |    x      |              |        2  . Thus , ..... &quot;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?

eri1ti0m · Accepted Answer

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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