Question

The equation Ax=b is solvable exactly when b is a linear combination of the columns of A. So Ax=b is solvable exactly when b lies in the column space, exactly when the column space doesn't get larger.

This is from one of the problem sets in MIT Opencourse. But I don't understand why b is only solvable when the column space doesn't get larger. Since b is already in the space of one of the columns of A, then if I happen to add a new column to matrix A that gets the column space to become larger, I can still solve Ax=b isn't it? A larger column space is definitely covers more, wouldn't it?

Answer 1

Oh and A is a matrix and x and b are vectors that satisfy the equation Ax=b.

Answer 2

And when I say I add a new column to matrix A that gets the column space larger, I mean I add a new column that is independent and therefore enlarges the span of the column space. As such, since the vector b, which was inside the "old" column space still exists inside this "new" column space, why can't Ax=b be solved when the column space gets larger?

Answer 3

I'd have to see the context for the statement. Certainly b would still lie in the span of the columns of A, it would just have 0 for the coefficient of the new column.

Answer 4

The question was like this:

Given an example where the column space gets larger and an example where it doesn't. Why is Ax=b solvable exactly when the column space doesn't get larger - is it the same for A and [A b]?

The given solution was:
The equation Ax=b is solvable exactly when b is a (nontrivial) linear combination of the columns of A (with the components of x as combing coefficients); so Ax=b is solvable exactly when b lies in the column space, so exactly when the column space doesn't get larger.

The last statement is what that I get confuse the most. Why must it exactly when the column space doesn't get larger? I thought if the columns space gets larger, it would just cover more and it would be merrier, wouldn't it? Since it covers more, then b would be inside the column space.