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The subspace a linear map sends to zero and the subspace it reaches, the two questions they settle, and the solution set of an inhomogeneous equation as a translate of the kernel.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute the kernel and the image of a linear map from one row reduction, decide injectivity by asking whether the nullity is zero and surjectivity by comparing the rank with the dimension of the codomain, apply the rank-nullity theorem over the domain rather than the codomain, rule out injectivity or surjectivity from the shape of a map before computing anything, explain why the rank of a composition can never exceed either factor's, and describe the solution set of $T(x) = b$ as one solution plus the whole kernel.
You have met both of these already. The null space of $A$ is the solution set of $Ax = 0$; the column space is the span of the columns. What changes here is only the point of view: they are attached to the map rather than to the matrix, they get the names kernel and image, and they keep their meaning when the basis changes and the matrix does not stay the same.
For a linear map $T : V \to W$, the kernel is $\ker T = \{v \in V : T(v) = 0\}$ and the image is $\operatorname{im} T = \{T(v) : v \in V\}$. The dimension of the image is the rank; the dimension of the kernel is the nullity. $T$ is injective, or one-to-one, when distinct vectors have distinct images; surjective, or onto, when the image is all of $W$; and an isomorphism when it is both.
Two subspaces describe a linear map completely enough to answer the questions anyone asks of it.
The kernel lives in the domain and collects the vectors sent to zero. It is a subspace because $T(0) = 0$, and because $T(cu + dv) = cT(u) + dT(v) = 0$ whenever $T(u)$ and $T(v)$ are zero. The image lives in the codomain and collects everything the map actually reaches; it is a subspace for the same one-line reason.
$T$ is injective exactly when $\ker T = \{0\}$. This is the fact worth understanding rather than memorising. If $T(u) = T(v)$ then $T(u - v) = 0$, so $u - v$ is in the kernel; if the kernel is trivial that forces $u = v$. Conversely a non-zero kernel vector $z$ gives $T(z) = T(0)$ with $z \ne 0$. So for a linear map, checking injectivity means solving one homogeneous system — you never have to compare pairs of vectors.
$T$ is surjective exactly when $\dim \operatorname{im} T = \dim W$, since a subspace of $W$ of full dimension is $W$.
And the two are tied together: $\dim \ker T + \dim \operatorname{im} T = \dim V$. Every direction of the domain is either flattened or survives, and each does exactly one of the two.
Another way: picture
Picture the domain as a bundle of directions arriving at the map. Some are crushed flat against zero — those span the kernel — and the rest come out the other side and span the image. Nothing is crushed and survives, and nothing is neither, which is the whole content of rank-nullity. The codomain is simply the room the survivors arrive in; it may be much larger than the image, and the theorem does not care how large it is.
Another way: steps
Before any arithmetic, the dimensions of the domain and codomain rule some answers out.
$T : \mathbb{R}^n \to \mathbb{R}^m$ has rank at most $\min(m, n)$, because the image sits inside $\mathbb{R}^m$ and is spanned by $n$ columns. So:
None of these needs the entries. They follow from the shape, and a first check of any claim about a map should be whether its shape permits the claim at all.
$\operatorname{rank}(ST) \le \min(\operatorname{rank} S, \operatorname{rank} T)$. Both halves are easy to see once the subspaces are in view.
The image of $ST$ is $S$ applied to the image of $T$, which is $S$ applied to something at most $\operatorname{rank} T$ dimensional — and a linear map never raises dimension. That gives $\operatorname{rank}(ST) \le \operatorname{rank} T$. And the image of $ST$ sits inside the image of $S$, which gives the other half.
The consequence is worth stating plainly: a direction a map has collapsed is gone for good. Project the plane onto the $x$-axis and then rotate, and the result still has rank $1$; no later map can tell apart two vectors that an earlier one has already made equal. Information is destroyed once and cannot be recovered downstream, which is why an invertible map must be built from invertible pieces.
The kernel moves the other way: $\ker T \subseteq \ker(ST)$, because anything $T$ kills stays killed.
Using the codomain in rank-nullity. The sum is the dimension of the domain. For $T : \mathbb{R}^5 \to \mathbb{R}^2$ with rank $2$, the nullity is $5 - 2 = 3$, not $2 - 2 = 0$. Writing the theorem as $\text{rank} + \text{nullity} = n$ and then forgetting which letter $n$ was is the single commonest slip here.
Calling the kernel a vector. The kernel is a subspace. "The kernel is $(-2, 1)$" should be "the kernel is spanned by $(-2, 1)$", and the difference matters the moment a question asks for its dimension.
Reading a zero kernel as an empty kernel. The kernel always contains $0$; it is never empty. "Trivial kernel" means it contains nothing else.
Treating the image as the codomain. $T : \mathbb{R}^2 \to \mathbb{R}^5$ has an image of dimension at most $2$, sitting inside $\mathbb{R}^5$ as a plane. The codomain is where the answers are allowed to live; the image is where they actually do.
Expecting the solution set of $T(x) = b$ to be a subspace. It is a translate of the kernel, and translates of subspaces are not subspaces unless the translation is by zero.
$A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 7 \end{pmatrix}$, a map from $\mathbb{R}^3$ to $\mathbb{R}^2$. Row-reduce: $\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 1\end{pmatrix}$.
One reduction serves both questions.
Two pivots, in columns $1$ and $3$: rank $2$, so the image is all of $\mathbb{R}^2$ and the map is onto.
Rank equals the codomain's dimension.
Column $2$ is free, so the nullity is $3 - 2 = 1$: the kernel is the line spanned by $(-2, 1, 0)$, and the map is not injective — as the shape already promised, since $3 > 2$.
Onto but not one-to-one.
$T(x, y) = (x, y, x + y)$, from $\mathbb{R}^2$ to $\mathbb{R}^3$, with matrix having columns $(1, 0, 1)$ and $(0, 1, 1)$.
Two columns, three rows.
$T(x, y) = 0$ forces $x = 0$ and $y = 0$ from the first two coordinates alone, so the kernel is trivial and $T$ is injective.
Trivial kernel, one-to-one.
The rank is therefore $2$, and the image is the plane $z = x + y$ inside $\mathbb{R}^3$ — a proper subspace, so $T$ is not onto and no map from $\mathbb{R}^2$ could be.
The shape forbade it from the start.
Rank-nullity over the domain: $4 + \text{nullity} = 4$.
The domain has dimension $4$.
So the nullity is $0$ and the kernel is $\{0\}$.
Nothing is flattened.
A trivial kernel is exactly injectivity, so yes — and this is the square case, where onto and one-to-one arrive together.
Give the dimension of the kernel and of the image of each map.
| Dimension of the kernel | Dimension of the image | |
|---|---|---|
| $S(x, y) = (4x + 4y,\; 0)$, from $\mathbb{R}^{2}$ to $\mathbb{R}^{2}$ | ||
| $T(x, y, z) = (4x,\; 4y,\; 2z)$, from $\mathbb{R}^{3}$ to $\mathbb{R}^{3}$ | ||
| $U(x, y, z) = 4x + 4y + 2z$, from $\mathbb{R}^{3}$ to $\mathbb{R}$ |
$A = \begin{pmatrix} 1 & 3 \\ 4 & 12 \end{pmatrix}$. Enter, as a column, the vector in the kernel of $A$ whose second entry is $1$.
This task has no paper form; do it on a device.
$T$ is a map from $\mathbb{R}^{5}$ to $\mathbb{R}^{2}$ whose image is all of $\mathbb{R}^{2}$. Which is it?
Each map goes from $\mathbb{R}^{8}$ to $\mathbb{R}^{8}$. Put them in order of increasing kernel dimension, smallest first.
Number the steps in order (write the number in the box):
$T : \mathbb{R}^{7} \to \mathbb{R}^{8}$ has a kernel of dimension $3$. What is the dimension of its image?
Answer:
$T : \mathbb{R}^{3} \to \mathbb{R}^{2}$ has a kernel that is a line, $b$ is not the zero vector, and $u = (8, 4, 4)$ satisfies $T(u) = b$. Select every true statement about the set of solutions of $T(x) = b$.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Give the dimension of the kernel and of the image of each map.
| Dimension of the kernel | Dimension of the image | |
|---|---|---|
| $S(x, y) = (5x + 3y,\; 0)$, from $\mathbb{R}^{2}$ to $\mathbb{R}^{2}$ | ||
| $T(x, y, z) = (5x,\; 3y,\; 2z)$, from $\mathbb{R}^{3}$ to $\mathbb{R}^{3}$ | ||
| $U(x, y, z) = 5x + 3y + 2z$, from $\mathbb{R}^{3}$ to $\mathbb{R}$ |
You can find a kernel and an image, decide one-to-one and onto from their dimensions, and say what the solution set of an inhomogeneous equation looks like. Say in your own words why a trivial kernel is the same thing as being one-to-one. Next: the numbers a vector has once a basis is chosen.
10. Your turn: $T : \mathbb{R}^4 \to \mathbb{R}^4$ has an image of dimension $4$. Is it injective?, step 3
Half the work did all of it.