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In Elements of Statistical Learning, they state on p. 11 that all vectors are column vectors and start developing the least squares idea.

So if we have $$\mathbf{X} = \begin{bmatrix} 1 \\ X_1 \\ X_2 \\ \vdots \\ X_p\end{bmatrix}$$ and $$\hat{\boldsymbol{\beta}} = \begin{bmatrix} \hat{\beta}_0 \\ \hat{\beta}_1 \\ \vdots \\ \hat{\beta}_p \end{bmatrix}\text{,}$$ take $$\hat{Y} = \mathbf{X}^{T}\hat{\boldsymbol{\beta}} = \langle \mathbf{X}, \hat{\boldsymbol{\beta}} \rangle\text{.}$$ So $\hat{Y} \in \mathbb{R}^1$ because $\mathbf{X}^{T} \in M_{1 \times p}(\mathbb{R})$ and $\boldsymbol{\beta} \in M_{p \times 1}(\mathbb{R})$. Fine.

Now they say:

... in general $\hat{Y}$ can be a $K$-vector, in which case $\beta$ (I think this is a typo - they should probably have $\hat{\beta}$ instead) would be a $p \times K$ matrix of coefficients.

Well, here's the problem. If $\mathbf{X}^{T} \in M_{1 \times p}(\mathbb{R})$ and $\hat{\beta} \in M_{p \times K}(\mathbb{R})$, wouldn't $\mathbf{X}^{T}\hat{\beta}$ give a row vector for $\mathbf{Y}$ (dimensions $1 \times K$)?

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    $\begingroup$ Your $X$ makes no sense in this context unless its elements are interpreted as row vectors of dimension $K$. Both $\beta$ and its estimate $\hat\beta$ are column vectors of dimension $p+1$, as you show. Have you perhaps mixed up the roles of $X$ and $\beta$ in the model? $\endgroup$ – whuber Nov 13 '15 at 18:20
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I think you confuse yourself a bit. In page 10 the authors say:

"a set of $N$ input $p$-vectors $x_i$ , $i = 1, \dots, N$ would be represented by the $N \times p$ matrix $X$."

This means that the $p$ feature vectors/regressors they use will be represented as column vectors in the matrix $X$. For that matter and to quote from the text page 12 exactly :

"$X$ is an $N \times p$ matrix with each row an input vector".

So $X^T$ is not a row (or column) vector in any case. $X$ (and its transpose) will be (probably) of rank $p$. $\hat{Y}$ will be a matrix $N \times K$. For the case of OLS model $K$ is usually equal to unity (1) but this is not a strict prerequisite for its use (that's why usually we write $\hat{y}$).

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