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I'm looking for some help with the following practice question for an upcoming midterm:

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There is a simple proof that requires only linear algebra.

First notice that if you take $\beta = (\beta_0,\beta_1,...,\beta_{n-1})^T$ and the matrix $X = [\mathbb 1,x_1,x_2,...,x_{n-1}]$, where $\mathbb 1 = (1,1,...,1)^T$ and $x_i =(x_{i1},...,x_{i(n-1)})^T$, you can write the model equation as:

$$y = X\beta + \epsilon$$

You know that the least square estimate $\hat y$ is the orthogonal projection of $y$ over the linear space spawned by the columns of $X$. But if $p=n-1$ and the columns of $X$ are linearly independent you are projecting $y$ over $\mathbb R^n$ since $X$ has $n$ independent columns.

When you orthogonally projects $y\in\mathbb R^n$ over $\mathbb R^n$ you get $y$ again. For this reason you have $\hat y = y$ which clearly shows that the $RSS = ||y-\hat y||^2$ is equal to zero.

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