Linear transformations — definition, kernel, image: Visual walkthrough

This is the visual companion to the parent note. If a symbol here is new, it gets a picture before it gets used.

Step 0 — The words we are allowed to use

Before any derivation, let us pin down four plain-English ideas. We will need them constantly.

Keep the last one in mind: dimension = number of arrows in a basis. The whole theorem is a story about splitting one basis into two groups and counting them.

Our running machine will be $T : R^{3} \to R^{2}, T (x, y, z) = (x + y + z, x - z) .$ Here $V = R^{3}$ (input space, $3$ directions) and $W = R^{2}$ (output space, $2$ directions).

Step 1 — See the machine crush a direction

WHAT. We feed the machine every possible arrow in $R^{3}$ and watch where they land in $R^{2}$ .

WHY. The theorem is about what survives and what dies. So the first thing to look at is a direction that dies — an input the machine flattens to the zero output.

PICTURE. Below, the left box is the 3-D input space $V$ ; the right box is the 2-D output space $W$ . The magenta arrow $(1, - 2, 1)$ is fed in — and it lands exactly on the origin dot in $W$ . It got crushed.

Figure — Linear transformations — definition, kernel, image

Check it by hand: $T (1, - 2, 1) = (x 1 + y (- 2) + z 1, x 1 - z 1) = (0, 0) .$

The first output slot is $x + y + z = 1 - 2 + 1 = 0$ .
The second output slot is $x - z = 1 - 1 = 0$ .

So this arrow maps to the zero arrow. It is a member of the kernel.

Step 2 — Collect all the crushed arrows: the kernel

WHAT. Find every input arrow that lands on zero, not just one.

WHY. One crushed arrow is a fluke; the set of all crushed arrows is what carries information. That set is the kernel, $ker T = {v : T (v) = 0}$ .

PICTURE. In $V$ , all the crushed arrows form the magenta line through the origin. Every arrow on that line dies; nothing off it does.

Solve $T (x, y, z) = (0, 0)$ , i.e. two equations: $x + y + z = 0 x - z = 0 (top output = 0) (bottom output = 0)$

From the second equation $x = z$ (it is the simpler one, so use it to substitute).
Put $x = z$ into the first: $z + y + z = 0 \Rightarrow y = - 2 z$ .
So every kernel arrow is $(z, - 2 z, z) = z (1, - 2, 1)$ .

$ker T = span {(1, - 2, 1)} .$ This is one independent direction, so $dim (ker T) = 1$ . We call this number the nullity. Hold onto it: nullity = 1.

Step 3 — Watch the surviving directions come out: the image

WHAT. Look at where the machine can send things — the set of all reachable outputs.

WHY. Whatever isn't crushed has to go somewhere in $W$ . The collection of all outputs is the image, $im T = {T (v) : v \in V}$ . Its dimension is the rank.

PICTURE. In $W$ , we plot the outputs of the three standard input directions $e_{1}, e_{2}, e_{3}$ . Even two of them already fan out to cover the whole plane.

Because $T (x) = A x$ with $A = (1110 1 - 1)$ , feeding a basis vector picks out a column: $T (e_{1}) = col 1 (1, 1), T (e_{2}) = col 2 (1, 0), T (e_{3}) = col 3 (1, - 1) .$

$(1, 1)$ and $(1, 0)$ point in different directions → independent → together they span all of $R^{2}$ .
$(1, - 1)$ is then just a combination of those two — it adds no new direction.

So $im T = R^{2}$ , giving rank = 2.

Step 4 — The key idea: split a basis of $V$ into "crushed" + "extra"

WHAT. Build a basis of the whole input space $V$ by starting from a basis of the kernel and adding just enough extra arrows to fill up $V$ .

WHY. This is the engine of the proof. If we can group the input directions cleanly into crushed ones and extra ones, then counting each group separately gives us nullity + (something), and we will show that "something" is the rank.

PICTURE. In $V$ : the magenta arrow $k_{1} = (1, - 2, 1)$ spans the kernel. We add two violet arrows $w_{1}, w_{2}$ chosen so the three together are independent and span all of $R^{3}$ .

$basis of k e r T, p = 1 {k_{1}} \cup extra vectors, q = 2 {w_{1}, w_{2}} = p + q = 3 basis of V .$

$p = dim (ker T) = 1$ — the crushed group.
$q$ = however many extras we needed — the surviving group.
$p + q = dim V = 3$ because a basis of $R^{3}$ has exactly $3$ arrows.

We have not proved the theorem yet — we have only defined $q$ as "the leftover count". The next steps prove $q$ equals the rank.

Step 5 — The extras produce the whole image (they span it)

WHAT. Show that every output of $T$ is built from just $T (w_{1}), T (w_{2})$ — the images of the extras.

WHY. If those two vectors already generate every output, then they span the image. The crushed arrows contribute nothing, because they map to zero.

PICTURE. Feeding a general input on the left, the kernel part vanishes and only the violet extras' images survive on the right.

Any $v \in V$ is written in our basis as $v = kernel part a_{1} k_{1} + extra part b_{1} w_{1} + b_{2} w_{2} .$ Apply $T$ and use linearity ( $T$ splits over sums and pulls scalars out): $T (v) = a_{1} = 0 T (k_{1}) + b_{1} T (w_{1}) + b_{2} T (w_{2}) = b_{1} T (w_{1}) + b_{2} T (w_{2}) .$

$T (k_{1}) = 0$ because $k_{1}$ is in the kernel — that is what kernel means.
So every output is a combination of just $T (w_{1}), T (w_{2})$ .

Therefore ${T (w_{1}), T (w_{2})}$ spans $im T$ . That is half of "it's a basis".

Step 6 — The extras' images don't collapse (they're independent)

WHAT. Show $T (w_{1})$ and $T (w_{2})$ point in genuinely different directions — no accidental overlap.

WHY. To be a basis of the image, spanning is not enough; they must also be independent. If they secretly collapsed, the rank would be smaller than $q$ and the count would break.

PICTURE. We suppose they collapse (a hypothetical dashed collapse in $W$ ) and trace the contradiction back into $V$ : it would force a surviving direction to secretly live in the kernel.

Suppose some combination gives zero: $b_{1} T (w_{1}) + b_{2} T (w_{2}) = 0.$ Pull $T$ back outside (linearity, in reverse): $T (call this u b_{1} w_{1} + b_{2} w_{2}) = 0 \Rightarrow u \in ker T .$

$u$ is made only of the extra arrows.
But $ker T$ is spanned by $k_{1}$ alone, so $u = c k_{1}$ for some scalar $c$ .

Now we have $b_{1} w_{1} + b_{2} w_{2} - c k_{1} = 0.$ But ${k_{1}, w_{1}, w_{2}}$ was chosen to be a basis — independent! The only way a combination of independent vectors is zero is if every coefficient is zero: $b_{1} = 0, b_{2} = 0, c = 0.$ So the only way $b_{1} T (w_{1}) + b_{2} T (w_{2}) = 0$ is the trivial one → $T (w_{1}), T (w_{2})$ are independent. Combined with Step 5, they form a basis of the image, so $rank = dim (im T) = q .$

Step 7 — Put the counts together

WHAT. Add up the two groups.

WHY. We now know each group's size: crushed group $= p =$ nullity, extra group $= q =$ rank, and together they tile $V$ .

PICTURE. The single balance scale: the $3$ input directions of $V$ split exactly into $1$ crushed (magenta) $+$ $2$ surviving (violet).

$dim V = Step 4: kernel basis size p + Step 4: extras q = nullity dim (ker T) + rank dim (im T) .$ For our machine: $3 = 1 + 2$ . ✓ The theorem holds. $■$

See Rank–Nullity Theorem for the abstract statement, and Column space and null space for the language "column space = image, null space = kernel".

Step 8 — Edge and degenerate cases (so nothing surprises you)

WHAT. Push the counting to its extremes. WHY. A visual proof must survive every corner, not just the tidy example.

PICTURE. Three mini-machines on the same axes: the zero map, an injective map, and a bijective map.

The one-picture summary

Everything above compressed into a single flow: the $3$ input directions of $V$ split, one dies in the kernel, two survive into the image, and $1 + 2 = 3$ .

Recall Feynman retelling (plain words, no symbols)

Picture a machine with three input dials (that's $V$ being 3-dimensional) and a two-light output panel (that's $W$ ). You wiggle the dials and watch the lights. One special combination of dials does nothing to the lights — that's a direction the machine ignores completely; it's "crushed". That's one dead direction (nullity = 1). The other two directions actually move the lights, and between them they can make the panel show any pattern — the lights are fully controllable. That's two live directions (rank = 2). Now the punchline: three dials total = one dead + two live. You never lose or gain directions out of nowhere — every input direction is either crushed or survives, never both, never neither. That accounting, forced to always balance, is the Rank–Nullity Theorem: $dim V = nullity + rank$ .

Recall Self-test

Which step proves the extras span the image? ::: Step 5 (kernel terms vanish, so every output is a combo of $T (w_{i})$ ). Which step proves they're independent? ::: Step 6 (a collapse forces a basis combination to be zero, impossible). Why is nullity $= 1$ for our $T$ ? ::: The kernel is a single line $span {(1, - 2, 1)}$ . Why is rank $= 2$ ? ::: The image fills $R^{2}$ ; two columns already span it. In the zero map, what are rank and nullity? ::: rank $= 0$ , nullity $= dim V$ .

Connections

Rank–Nullity Theorem — this page is its visual derivation.
Column space and null space — image = column space (Step 3), kernel = null space (Step 2).
Matrix multiplication — Step 3 reads off columns because $T (e_{i})$ = column $i$ .
Injective surjective bijective — Cases B and C of Step 8.
Change of basis — the invariance in the Step 8 mistake box.
Eigenvalues and eigenvectors — bijective (Case C) means no zero eigenvalue.