# Understanding linear projection in “The Elements of Statistical Learning”

In the book "The Elements of Statistical Learning" in chapter 2 ("Linear models and least squares; page no: 12"), it is written that

In the (p+1)-dimensional input-output space, (X,Y) represent a hyperplane. If the constant is included in X, then the hyperplane includes the origin and is a subspace; if not, it is an affine set cutting the Y-axis at the point (0,$\beta$).

I don't get the sentence "if constant is ... (0,$\beta$)". Please help? I think the hyperplane would cut the Y-axis at (0,$\beta$)in both the cases, is that correct?

The answer below has helped somewhat, but I am looking for more specific answer. I understand that when $1$ is included in the $X$, it won't contain origin, but then how would the $(X,Y)$ would contain origin? Should not it depend on value of $\beta$? If intercept $\beta_0$ is not $0$, $(X,Y)$ should not contain origin, in my understanding?

• How much linear algebra have you done? Do you know what vectors are? What about vector spaces, subspaces, ... ? – Adrian Dec 8 '15 at 8:06
• I have basic understanding of linear algebra, vector and vector spaces. – Abhinav Gupta Dec 8 '15 at 8:09
• en.wikipedia.org/wiki/Hyperplane has a bit on affine hyperplanes and vector hyperplanes – Adrian Dec 8 '15 at 8:52
• Thnaks! just read this article. But I still don't understand how can one say that hyperplane includes the origin if contant is included in X. If this clear then I understand why hyperplane is a subspace. – Abhinav Gupta Dec 8 '15 at 9:06
• page no: 12. I have edited the question too. – Abhinav Gupta Dec 10 '15 at 9:53

Including the constant 1 in the input vector is a common trick to include a bias (think about Y-intercept) but keeping all the terms of the expression symmetrical: you can write $\beta X$ instead of $\beta_0 + \beta X$ everywhere.
If you do this, it is then correct that the hyperplane $Y = \beta X$ includes the origin, since the origin is a vector of $0$ values and multiplying it for $\beta$ gives the value $0$.
However, your input vectors will always have the first element equal to $1$; therefore they will never contain the origin, and will be place on an smaller hyperplane, which has one less dimension.
You can visualize this by thinking of a line $Y=mx+q$ on your sheet of paper (2 dimensions). The corresponding hyperplane if you include the bias $q$ your vector becomes $X = [x, x_0=1]$ and your coefficients $\beta = [m, q]$. In 3 dimensions this is a plane passing from the origin, that intercepts the plane $x_0=1$ producing the line where your inputs can be placed.