The nullspace: finding the inputs that give zero

An input that gives zero

Take a matrix whose second column is twice the first: A = 1224.

Try the input x = 2−1: two of column 1, minus one of column 2.

The two columns cancel exactly. So 2−1 is in the nullspace.

You can also check it row by row, the way lesson 1 taught Ax. Each row of A, dotted with x, must give 0:

The nullspace is a subspace

If Ax = 0 and Ay = 0, then A(x + y) = Ax + Ay = 0 + 0 = 0, and A(cx) = c·Ax = 0. Sums and multiples stay in. So every multiple c·2−1 also gives zero. For example, c = 3:

Here N(A) is a whole line through the origin, and nothing else: Ax = 0 says x₁ + 2x₂ = 0 (row 2 says the same thing, times 2), which is exactly this line. Any nonzero vector on a line can name it: multiples of 2−1 and multiples of −21 are the same line. Later slides use −21.

Look at the row 12 of A. Its dot product with 2−1 is 0, and lesson 1 said that means a right angle. So the nullspace line is perpendicular to the row. Slide 6 shows that this always happens.

What a nullspace vector tells us

Ax = 0 says x₁·(column 1) + x₂·(column 2) = 0. So a nonzero x in N(A) is a recipe that mixes the columns into zero. That means the columns are dependent: one is a combination of the others (slide 5 says more). Here 2·12 − 1·24 = 0 tells us column 2 = 2·column 1.

When only zero gets through

If A is invertible (lesson 3), multiply Ax = 0 by A⁻¹: x = A⁻¹0 = 0. The nullspace is just the zero vector. For a 2×2 matrix, ad − bc ≠ 0 tells us that.

Then every output comes from exactly one input. A nullspace with more than 0 in it means some different inputs share an output: A40 and A21 are both 48 for our first matrix, because 40 − 21 = 2−1 is in N(A).