Rank and collapse

Here are two matrices. They differ in a single entry.

$$A = \begin{bmatrix} 1 & 2 \ 2 & 4 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & 2 \ 2 & 5 \end{bmatrix}$$

One of them destroys a dimension. The other does not. Nothing about their appearance tells you which.

Read the columns

Use the column picture. For $A$, the columns are $(1,2)$ and $(2,4)$. Look closely: $(2,4)$ is just $2 \times (1,2)$. The second column points along the same line as the first.

So whatever vector $x = (x_1, x_2)$ you feed in, the output is

$$x_1\begin{bmatrix} 1 \ 2 \end{bmatrix} + x_2\begin{bmatrix} 2 \ 4 \end{bmatrix} = (x_1 + 2x_2)\begin{bmatrix} 1 \ 2 \end{bmatrix}$$

Every possible output is a multiple of $(1,2)$. The whole plane — every point in it — is squeezed onto that single line. There is no input you could choose that lands anywhere else.

Now $B$. Its columns are $(1,2)$ and $(2,5)$, and $(2,5)$ is not a multiple of $(1,2)$. Those two directions are independent, so combinations of them reach every point in the plane. Nothing is lost.

The word for this

The set of all reachable outputs is the column space, and its dimension is the rank.

• $A$ has rank 1. Its column space is a line. A dimension was crushed.
• $B$ has rank 2. Its column space is the whole plane. Full rank.

You can also see it in the determinant: $\det A = 1(4) - 2(2) = 0$, while $\det B = 1(5) - 2(2) = 1$. Zero determinant and rank below full are two descriptions of the same collapse — area is destroyed exactly when a dimension is.

Why it decides everything downstream

If a transformation is full rank, no information was lost, and you can run it backwards. If the rank dropped, many different inputs were mapped to the same output, and no reverse motion can possibly tell them apart.

That is not a rule to memorise. It is the reason the next two ideas — invertibility and the null space — exist at all.