Function taking vectors of Rn as inputs

[gursimran]

Hello friends,

Is it possible that a function take inputs as vectors in Rⁿ and return one real value R. One such example is dot product. But keeping mathematical rigor in mind what should I call this function.

Problem 1.
Its a function from R²ⁿ to R
or
a function such that f(Rⁿ,Rⁿ)->R

Problem 2.
I want to solve this equation using gradient descent
|c - x.y|

here
c - scalar (R)
x and y are vectors in Rn

How should I visualize its graph?

This problem is inspired by the mathematical treatment used to calculated SVD here.

As proved by Ecart and young there could only be one minimum of the equation and the minimum should be SVD. So it means it should be a sort of parabolloid. How to visualize it.

 

[fresh_42]

Hello friends,

Is it possible that a function take inputs as vectors in Rⁿ and return one real value R. One such example is dot product. But keeping mathematical rigor in mind what should I call this function.

Problem 1.
Its a function from R²ⁿ to R
or
a function such that f(Rⁿ,Rⁿ)->R

If you want to consider it a bilinear function as usual, then it is the latter. If you are only interested in analytic properties, then you can use the former, but then we no longer have linearity.

Problem 2.
I want to solve this equation using gradient descent
|c - x.y|

here
c - scalar (R)
x and y are vectors in Rn

You can analytically consider $\left| \sum_k (c_k - x_ky_k) \right|= n \circ \sum_k (c_k - x_ky_k)$ and use the definition of your norm $n$.

How should I visualize its graph?

The graph of a function $f\, : \,\mathbb{R}^{2n} \longrightarrow \mathbb{R}$ is an object in $\mathbb{R}^{2n+1}$. This constraints visualizations to $n = 1$. Probably not very informative but at least it displays the principle.

This problem is inspired by the mathematical treatment used to calculated SVD here.

As proved by Ecart and young there could only be one minimum of the equation and the minimum should be SVD. So it means it should be a sort of parabolloid. How to visualize it.