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I am reading about Schubert calculus and have come across this definition of a linear space:

A linear space $L$ in $mathbb{P}^{n}$ is defined as the set of points $P = (p(0), p(1), ldots, p(n))$ of $mathbb{P}^{n}$ whose coordinates $p(j)$ satisfy a system of linear equations $sum_{j=0}^{n} b_{alpha j}p(j) = 0$, where $alpha = 1, ldots, (n-d)$. [Here, I get confused: ] We say that $L$ is $d$-dimensional if these $(n-d)$ equations are independent, that is if the $(n-d)times (n+1)$ matrix of coefficients $[b_{alpha j}]$ has a nonzero $(n-d) times (n-d)$ minor.

I would have thought that this space is of dimension $(n+1)$. Maybe because each equation is zero and the $p(j)$ satisfy such an equation, in each combination there is an "extra" vector by dependence. Thus, this leaves $(d+1)$ independent vectors, but not $d$ (this may have to do with the fact that we are in $mathbb{P}^{n}$).

Any help is appreciated.

Think of how $mathbb P^n$ is defined:

$$mathbb P^n = mathbb K^{n+1}setminus {0} /sim, $$

where $vec x sim vec y$ if they are proportional. Now the matrix $B$ is nondegenerate, so the linear map $B : mathbb K^{n+1} to mathbb K^{n+1}$ has a $d+1$ dimensional Kernel $tilde L$.

However, when you take quotient $sim$, then $L = tilde Lsetminus{0} /sim$ has one less dimension (The reason is the same as why $mathbb P^n$ is $n$-dimensional instead of $n+1$-dimensional).

Note that you might think that it is an (established) abuse of languages to call that $L$ linear: $mathbb P^n$ is not a vector space, after all.

The reason for the "off-by-one" error you are seeing is that in projective space, points are really given by equivalence classes of coordinates. Therefore the point
$$
(p_0,p_1,p_2,ldots,p_n)
$$
is in the same equivalence class as
$$
(lambda p_0,lambda p_1,lambda p_2,ldots,lambda p_n),
$$
for any $lambdanot=0$.

I figure I'll add my two cents as well, even though the other answers are also good. I agree that it is confusing that projective dimension is always one less than the vector space dimension. You have to be careful to specify which you mean. However there is a good reason for this double usage of the term, using geometric intuition.

We can view a vector space and its subspaces geometrically; the zero vector is included in every subspace. So one nonzero vector determines a line ($1$-dimensional), and two linearly independent vectors form a plane ($2$-dimensional). Moving to a projective setting, one nonzero vector now determines only a point ($0$-dimensional) while two linearly independent vectors form a line ($1$-dimensional). We need three linearly independent vectors to give a triangle/plane ($2$-dimensional) in this setting. So really the dimension can be interpreted in a geometric way.

Of course when we deal with a projective space in analytic geometry, we have the complication that we need to keep track of both systems simultaneously: we are representing geometric objects in the projective setting with geometric objects in a vector space setting. The same object has a different geometric dimension, depending on the point of view (i.e. a $(d+1)$-dimensional vector subspace corresponds to a $d$-dimensional projective subspace).