$\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$

### Linear Approximation

Linear approximation is a fundamental problem in machine learning, and one that has a surprising amount of mathematical structure built around it for such a seemingly simple problem. Consider the following problem: We have a Hilbert space $\mathbf{S}$ and a subspace $\mathbf{T} \subseteq \mathbf{S}$. We also have an element $\mathbf{x} \in \mathbf{S}$. What is the closest element $\mathbf{\hat{x}} \in \mathbf{T}$ to $\mathbf{x}$?

This Hilbert space $\mathbf{S}$ has an inner product $\langle\cdot, \cdot\rangle$ and induced norm $\norm{\cdot}$. So we can frame the problem as finding the point $\mathbf{\hat{x}} \in \mathbf{T}$ such that $\norm{\mathbf{\hat{x} - x}}$ is minimized.

We can find a unique minimizer by exploiting orthogonality. In fact, $\mathbf{\hat{x} \in T}$ is the closest point to $\mathbf{x \in S}$ if $\mathbf{\hat{x} - x}$ is orthogonal to all other points $\mathbf{y \in T}$. This means that $\langle \mathbf{\hat{x} - x}, y\rangle = 0$ for all $\mathbf{y \in T}$

Lets show that if $\langle\mathbf{\hat{x} - x}, \mathbf{y}\rangle = 0$ for all $\mathbf{y \neq \hat{x} \in T}$ then $\mathbf{\hat{x}}$ is minimizer of $(1)$.

The last equality follows from the Pythagorean theorem. This is valid because we required that $\mathbf{x - \hat{x}}$ was orthogonal to all points in $\mathbf{T}$, and $\mathbf{y} - \mathbf{\hat{x}}$ is certainly in $\mathbf{T}$!

Therefore, if $\norm{\mathbf{y} - \mathbf{\hat{x}}}^{2} \neq 0$ (i.e. $\mathbf{y} \neq \mathbf{\hat{x}}$), then

where equality is achievd only when $\mathbf{y} = \mathbf{\hat{x}}$. This implies that $\mathbf{\hat{x}}$ is a unique minimizer of $(1)$. This is a pretty intuitive result: If $\mathbf{x - y}$ is not orthogonal to $\mathbf{T}$, then there is some other point $\mathbf{\hat{x}}$ that comes closer to $\mathbf{x}$ while still remaining inside $\mathbf{T}$. This can be seen visually in the image above.

#### Computing the closest point

So we know that $\mathbf{\hat{x}}$ is a unique minimizer of $(1)$ if $\langle\mathbf{x - \hat{x}}, y\rangle = 0$ for all $\mathbf{y} \neq \mathbf{\hat{x}}$ in $\mathbf{T}$, but how do we actually compute $\mathbf{\hat{x}}$? If $\mathbf{T}$ is an $N$-dimensional subspace, that means we can represent any point in the space by a linear combination of $N$ basis vectors - call them $\mathbf{v_{1}}, \mathbf{v_{2}}, …, \mathbf{v_{N}}$.

for some constants $\{ \alpha \}_{1}^{N}$. Orthogonality also tells us

If we take the inner product of $\mathbf{x - \hat{x}}$ with one of the basis vectors we generate a linear equation.

In fact, we can generate $N$ different linear equations by taking the inner product with each of the basis vector separately. That means we can solve this linear system of equations for $\mathbf{\alpha}$, the vector of coefficients!

Where $\mathbf{G}$ is the matrix if inner products and is called the Gram Matrix or Grammian of the basis $\{\mathbf{v}\}_{n=1}^{N}$. After we solve for our coeeficientls, we can easily reconstruct the closest point in $\mathbf{T}$ to $\mathbf{x}$ by

Take a second to appreciate what we did. We took a minimization problem, converted it to a finite dimensional linear algebra problem by exploiting our basis to ask the question “what basis coefficients will create a $\mathbf{\hat{x}}$ that minimizes the objective?”. This idea is central to many more topics we will cover.

$\mathbf{G}$ is invertible because the basis vectors are linearly independent. Also, since the inner product is a symmetric function, the Gram Matrix is also symmetric. Because the Gram matrix is square and invertible, $\mathbf{b} = \mathbf{G\alpha}$ always has a solution. Further, if we have an orthogonal basis, then the Gram Matrix is exactly the Identity transformation, and the coefficients can be calculated by simply taking inner products of $\mathbf{x}$ with each basis vector.

##### Example

We will close with an example to drive this idea home. Let our Hilbert space $\mathbf{S} = \mathbb{R}^{3}$ with the standard inner product and

The vectors we defined $\mathbf{T}$ with form a basis for the subspace. What is the closest point in $\mathbf{T}$ to $\mathbf{x}$? We can write $\mathbf{\hat{x}}$ as

and our Gram Matrix and $\mathbf{b}$ are

The inverse Gram Matrix is

Finally, $\mathbf{\alpha} = \begin{bmatrix}1 & -1\end{bmatrix}^{T}$. We reconstruct our solution using the coefficients: $\mathbf{\hat{x}} = \mathbf{v_{1}} - \mathbf{v_{2}} = \begin{bmatrix}2 & 0 & 0\end{bmatrix}^{T}$