3.3 One-to-One and Onto

A transformation can fail its job in two different ways, and it is worth having names for both before meeting a matrix that manages both at once. $T$ is onto when every vector of the target space is $T (𝐱)$ for some input — nothing is missed. $T$ is one-to-one when different inputs always produce different outputs — nothing is confused. Neither property implies the other in general, and neither is guaranteed.

Take the matrix Chapter 2 finished with, now read as a motion of $ℝ^{3}$ :

$A = (\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{matrix}), T (𝐱) = A 𝐱$

Its rank was $2$ and its nullity $1$ , and those two numbers are about to be re-read as answers to the two questions above.

A semi-transparent tilted plane, everything the transformation can produce. A blue straight line runs through space at an angle to the plane, and a short thick green marker slides along it as a slider is dragged. From the green marker a dashed amber arrow always points to one and the same amber dot lying on the plane: wherever the input slides to, the output does not move. A separate violet dot floats clearly above the plane, marking a target the transformation can never produce, because every output lies in the sheet. — The plane $z = x - y$ is everything $T$ can produce. The blue line is a whole line of inputs, and the green marker slides along it as $a$ is dragged; the dashed amber arrow shows where $T$ sends the marker, and it always arrives at the same amber point $open parenthesis 1, 0, 1 close parenthesis$ . The violet point $open parenthesis 0, 0, 1 close parenthesis$ floats off the plane — a target $T$ cannot produce at all.

Explore the figure

Look at the violet point first. It sits clear of the sheet, and since every output of $T$ lies in the sheet, no input whatsoever is sent there. That single missed target is enough to say $T$ is not onto.
Drag $a$ from $minus 2$ to $2$ and watch the two ends of the dashed arrow. The green marker travels a long way; the amber point it lands on does not move at all. Every input on the blue line has the same image.
Stop at $a equals 0$ . The input is $open parenthesis 1, 0, 0 close parenthesis$ and the output is $open parenthesis 1, 0, 1 close parenthesis$ . Now stop at $a equals 1$ : the input is $open parenthesis 0, minus 1, 1 close parenthesis$ , a genuinely different vector, and the output is $open parenthesis 1, 0, 1 close parenthesis$ again. Two different inputs, one output — $T$ is not one-to-one.
Notice the blue line is not the amber line of the previous chapter but a parallel copy of it, shifted off the origin. Sliding along it changes the input by multiples of $(- 1, - 1, 1)$ , and that is exactly the direction $T$ annihilates.

Both failures are one failure

Start with onto. The outputs of $T$ are precisely the vectors $A 𝐱$ , which is precisely the column space — the same set section 2.4 called the reach. Here that is a plane, a two-dimensional slice of a three-dimensional target, so almost every vector of $ℝ^{3}$ is missed. In general, $T$ is onto exactly when the column space is everything, which is exactly when $rank A$ equals the number of rows.

Now one-to-one, which has a shortcut worth knowing. Suppose two inputs share an image: $A 𝐱 sub 1 equals A 𝐱 sub 2$ . Subtracting and using linearity gives $A (𝐱_{1} - 𝐱_{2}) = 0$ , so their difference lies in the null space. Run it backwards too: if $𝐧 not equal to 0$ is in the null space then $𝐱$ and $𝐱 + 𝐧$ are different inputs with the same image. So

$T is one-to-one ⟺ the null space is only {0} ⟺ nullity A = 0$

You never have to compare pairs of inputs; you only have to ask what gets sent to zero. Our $A$ has nullity $1$ , so it is not one-to-one, and the blue line in the figure is a picture of the failure: it is the set $𝐱 sub 0 plus null space$ , and every such shifted copy of the null space is a batch of inputs that $T$ cannot tell apart.

Here is the part that makes square matrices special. The rank theorem says $rank + nullity = 3$ , the number of columns, and for a square matrix the number of rows is also $3$ . Onto wants rank $3$ ; one-to-one wants nullity $0$ ; and the theorem says those are the same condition. A square matrix is onto if and only if it is one-to-one — not because the two properties are similar, but because there is only one budget and both are asking for all of it.

0

One number, both failures ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The determinant of $A$ vanishes, and for a square matrix that single number decides both questions at once. A nonzero determinant would mean three independent columns, hence rank $3$ and nullity $0$ , hence onto and one-to-one. Zero means rank at most $2$ , so something is missed and something is confused — which is what the violet point and the blue line respectively show you.

Example 1 What has to change to fix it

Replace the third column of $A$ by $open parenthesis 3, 1, 3 close parenthesis$ instead of $open parenthesis 3, 1, 2 close parenthesis$ , leaving the first two columns alone. Decide whether the new transformation is one-to-one, whether it is onto, and what has happened to the blue line of the figure.

The new matrix has columns $open parenthesis 1, 0, 1 close parenthesis$ , $open parenthesis 2, 1, 1 close parenthesis$ and $open parenthesis 3, 1, 3 close parenthesis$ . The first two still span the plane $z = x - y$ ; the question is whether the third escapes it. Test it: $x - y = 3 - 1 = 2$ , but the third coordinate is $3$ , so the column is off the plane by one unit and the three columns are independent.

Independent columns means rank $3$ , so the column space is all of $ℝ^{3}$ and the map is onto — the violet point $open parenthesis 0, 0, 1 close parenthesis$ now has a preimage. By the rank theorem, nullity $= 3 - 3 = 0$ , so the null space collapses to $open brace 0 close brace$ and the map is one-to-one as well. Both failures were repaired by the same single change, exactly as the equivalence promises.

The blue line is gone as a picture of anything. It was a shifted copy of the null space, and the null space is now a single point; the inputs that used to share an image have been separated, and each output has exactly one source.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.