0
$\begingroup$

There is a matrix equation for solving a linear regression, $\vec{y}=X\vec{\beta}$ where $X$ is the matrix of features, $\vec{\beta}=[\beta_1,...,\beta_n]$ are the coefficients for each feature, and $\vec{y}$ is the measurements, or the true solutions you are trying to solve for.

The matrix equation I have seen to solve for $\beta$ is $$\vec{\beta}=(X^TX)^{-1}X^T\vec{y}$$

I am wondering, why isn't the solution just : $\vec{\beta}=X^{-1}\vec{y}$ ?

Is it because $X$ may be singular, and this is some kind of psuedo inverse?

$\endgroup$
0
$\begingroup$

Yes, that's exactly what you suggested.

Typically $X\in\mathbb{R}^{m\times n}$ is a tall matrix (m>=n) with $rank(X)=n$, because # of data $m$ should be at least greater than # of parameters $n$. Also, in practice most observed matrix attains its maximum rank.

In this situation, $X^\top X$ has full rank, and the pseudo-inverse $X^\dagger$ is exactly $(X^\top X)^{-1}X^\top$.

As a quick check, we have $$((X^\top X)^{-1}X^\top ) X=I$$ and $$X(X^\top X)^{-1}X^\top=\text{projection matrix to the column space of }X$$

To prove for $X\in\mathbb{R}^{m\times n}$, if $rank(X)=n$, then $X^\dagger=(X^\top X)^{-1}X^\top$, let the SVD of X be $U\Sigma V^\top$.

By definition of pseudo-inverse, $X^\dagger=V\Sigma^{-1}U^\top$, where

$$(\Sigma^{-1})_{ij}=\begin{cases} 0 & \text{ if }\Sigma_{ji}=0 \\ 1/\Sigma_{ji} & \text{ o.w. }\end{cases}$$

the (i,j) element of $\Sigma^{-1}$ is zero if the corresponding (j,i) element in $\Sigma$ is; otherwise, it is the reciprocal of it.

Also,

$$(X^\top X)^{-1}X^\top = (V \Sigma^\top \Sigma V^\top)^{-1} V\Sigma^\top U^\top= V (\Sigma^\top \Sigma)^{-1}\Sigma^\top U^\top$$

Finally, by definition of $\Sigma^{-1}$, we have $(\Sigma^\top \Sigma)^{-1}\Sigma^\top=\Sigma^{-1}$, so the proof is complete.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.