How can I check whether a vector is in the span of two given vectors?

Think of the two given vectors as building blocks. Their span is the set of every vector you can make by scaling each one and adding the results. [S1] So the direct test is to ask whether your candidate can be written as a*u + b*v for some numbers a and b. Set the candidate equal to that expression, expand it by components, and look for one pair of coefficients that works throughout.

For example, suppose the given vectors are u = (1, 2) and v = (2, 1), and the candidate is w = (5, 4). We’ll check whether there are numbers a and b such that a(1, 2) + b(2, 1) = (5, 4).

Scale each vector, then add matching coordinates:

a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b).

For this to equal (5, 4), the first coordinate must be 5 and the second must be 4. That gives a + 2b = 5 and 2a + b = 4. A pair that satisfies both is a = 1 and b = 2: the first expression gives 1 + 2(2) = 5, and the second gives 2(1) + 2 = 4.

So (5, 4) = 1(1, 2) + 2(2, 1). The same coefficients reproduce both coordinates, which confirms that this candidate is in the span.

No. The coefficients have to make the entire vector match, not just one coordinate. For vectors with two coordinates, check both component equations using the same a and b; for vectors with more coordinates, check every component. If no single pair works for all of them, then the candidate is not in the span. [S1]

For a quick contrast using the same given vectors, consider the candidate (1, 1). Its equations would be a + 2b = 1 and 2a + b = 1. They have the solution a = b = 1/3, so this candidate is also in the span. To conclude that a candidate is outside the span, you must establish that its full set of component equations has no solution—not merely stop after a failed guess.

Use this template, replacing the letters with your vectors:

Given u = (u1, u2, ...) and v = (v1, v2, ...), test w = (w1, w2, ...).

1. Write w = au + bv.
2. Expand: (w1, w2, ...) = (a*u1 + b*v1, a*u2 + b*v2, ...).
3. Match corresponding coordinates and solve for a and b.
4. Substitute the proposed coefficients back into every coordinate.
5. Conclude “in the span” if all coordinates match; conclude “not in the span” only if no coefficients satisfy all of them.

In the worked example, the check ends with (5, 4) = 1(1, 2) + 2(2, 1), so the conclusion follows directly from the combination. [S1]