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Linear Algebra
Step 15 of 22

The null space

Understand first: rank and collapse · inverse

When a transformation collapses space, something specific gets crushed. Find it.

A=[1122]A = \begin{bmatrix} 1 & 1 \ 2 & 2 \end{bmatrix}

Its determinant is 1(2)−1(2)=01(2) - 1(2) = 0, so you already know the plane is being flattened onto a line. The question now is which vectors get annihilated.

Hunting for the destroyed direction

Try (1,−1)(1, -1), using the column picture:

1[12]+(−1)[12]=[00]1\begin{bmatrix} 1 \ 2 \end{bmatrix} + (-1)\begin{bmatrix} 1 \ 2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}

A non-zero vector was sent to the origin. Any multiple works the same way: (2,−2)(2,-2), (−5,5)(-5,5), all crushed to nothing.

That whole line is the null space: every vector satisfying Ax=0Ax = 0. In the visual, it is the glowing line whose arrows all collapse to the centre.

The null space is always a subspace — it contains the origin, and sums and scalings of crushed vectors stay crushed. Act I gave you the vocabulary for exactly this.

Why solutions arrive in families

Here is the payoff, and it explains something that confuses almost everyone about systems of equations.

Suppose x0x_0 solves Ax=vAx = v, and nn is any vector in the null space. Then

A(x0+n)=Ax0+An=v+0=vA(x_0 + n) = Ax_0 + An = v + 0 = v

x0+nx_0 + n is also a solution. So if the null space is a line, there is not one solution — there is an infinite line of them, all landing on the same target.

This is why linear systems have exactly three outcomes and never anything else:

  • One solution. Null space is only the origin, nothing was destroyed.
  • Infinitely many. Null space is bigger, and solutions come as a family.
  • None. The target vv lies outside the column space, unreachable.

Act III is done. You can see what a motion does, what it destroys, and whether it can be undone. Act IV turns to measurement, and then to the single best question you can ask a matrix.