Given $X\in\mathbb R^{n\times p}$ and $y\in \mathbb R^n$, the least square coefficients are: $\hat{\beta} = \text{argmin} \| X\beta - y\|^2_2$.

Is $\hat{\beta}$ unique in the case $\text{rank}(X)=p$?

  • $\begingroup$ Yes, providing that $n \ge p$. The key is that $\beta$ is unique if the columns of $X$ are linearly independent, which $rank(X) = p$ combined with $n \ge p$ ensures. $\endgroup$ Apr 4 '16 at 0:22
  • $\begingroup$ can you write a proof ? $\endgroup$
    – Donbeo
    Apr 4 '16 at 0:23
  • 6
    $\begingroup$ I could, but I leave that to you as an exercise. $\endgroup$ Apr 4 '16 at 0:24
  • $\begingroup$ There's a algebraic-geometric proof if you know the theory of vector spaces. Are you familiar with vector spaces ? $\endgroup$ Apr 4 '16 at 10:40
  • $\begingroup$ Also see stats.stackexchange.com/questions/194460 $\endgroup$
    – whuber
    Apr 4 '16 at 16:09

Assume $\text{rank}(X) = p$ and $n \geq p$ (in fact $\text{rank}(X) = p$ implies that $n \geq p$), the least square estimator has an explicit expression: $$\hat{\beta} = (X^TX)^{-1}X^Ty,$$ in which $X^TX$ is non-singular. Since the inverse is unique, $\hat{\beta}$ is unique.

So probably the only difficult part is to show that $X^TX$ is non-singular provided $\text{rank}(X) = p$. There are multiple ways to show that, perhaps the easiest way is to use $$\text{rank}(X^TX) = \text{rank}(X) \tag{$*$}$$ for any matrix $X$, also note that $X^TX$ is a $p \times p$ matrix, thus $X^TX$ has to be non-singular. Let me know if you need a proof for $(*)$.


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