Back to the on-screen lesson ·
The eight axioms, the spaces of polynomials, matrices and functions they cover, and the three-line test that decides whether a subset is a subspace.
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 state what a vector space is without mentioning coordinates, recognise polynomials, matrices and functions as vectors and name the zero vector of each of those spaces, apply the three-condition subspace test to a subset given by an equation or an inequality, say which single condition a failing subset fails and give the counterexample that shows it, and explain why the null space and the column space of a matrix are subspaces of different spaces.
Up to now a vector has been a column of numbers, and the two things you could do with one were add it to another and multiply it by a scalar. Every fact about solution sets in the last unit was really a fact about those two operations. This lesson takes that seriously: the operations are kept and the columns are thrown away, and everything proved afterwards holds for anything at all that has the two operations.
A vector space over the real numbers is a set $V$ with an addition and a scalar multiplication satisfying the eight axioms below. Its elements are vectors, whatever they happen to be made of. The zero vector $0$ is the additive identity, and it is unique. A subspace of $V$ is a subset that is itself a vector space under the same two operations. A set is closed under an operation when applying the operation to members never leaves the set.
A vector space is a set $V$ with an addition $u + v$ and a scalar multiplication $cv$ obeying eight rules: addition is commutative and associative; there is a zero vector; every vector has a negative; $1v = v$; $c(dv) = (cd)v$; and multiplication distributes over both kinds of sum, $c(u + v) = cu + cv$ and $(c + d)v = cv + dv$.
Not one of those rules mentions coordinates, and that is the point. The following are all vector spaces, and every theorem in this course applies to each of them without a word of change:
| The space | A vector in it | The zero vector |
|---|---|---|
| $\mathbb{R}^n$ | a column of $n$ numbers | the column of zeros |
| $P_n$, polynomials of degree at most $n$ | $3x^2 - x + 5$ | the zero polynomial |
| $M_{2 \times 2}$, the $2 \times 2$ matrices | a $2 \times 2$ array | the zero matrix |
| the real functions on $[0, 1]$ | $\sin$ | the function that is $0$ everywhere |
| the solutions of $y'' + y = 0$ | $\cos t$ | the zero function |
That last row is the return on the investment. The solutions of a homogeneous linear differential equation form a vector space, so a combination of solutions is a solution — the superposition principle — and it is a theorem about the axioms rather than a fact about differential equations.
The subspace test. Verifying eight axioms for a subset is wasted effort, because a subset of a vector space inherits every one of them that is an identity. What it does not inherit is that the answers stay inside. So a non-empty subset $W \subseteq V$ is a subspace exactly when:
Three lines, and they are the only test this course ever uses.
Another way: picture
In $\mathbb{R}^3$ the subspaces are exactly four kinds of thing: the single point at the origin, every line through the origin, every plane through the origin, and the whole space — dimensions $0$, $1$, $2$ and $3$. Nothing else qualifies. A plane that misses the origin is not a subspace; neither is a half-plane, a disc, a cone or a pair of crossing lines. Every subspace is flat, unbounded in the directions it contains, and nailed to the origin, and those three words are the geometric content of the three conditions.
Another way: steps
Condition 1 looks redundant. If $W$ is non-empty and closed under scalars then $0 = 0u$ is in it already, so why state it separately?
Because it is the cheapest test in the subject and it disqualifies almost every set that fails. Checking closure means reasoning about two arbitrary members; checking the origin means one substitution with zeros. In practice the sets that come up are defined by equations, and moving the right-hand side off zero is the commonest way a candidate goes wrong:
That last pair is worth dwelling on. Passing the origin test proves nothing on its own; it only ever refutes. $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ are both singular and their sum is the identity, so the singular matrices fail condition 2.
*The pattern to remember: a homogeneous linear condition gives a subspace. Anything inhomogeneous, non-linear or one-sided does not.*
Two subspaces come free with every $m \times n$ matrix $A$, and the rest of this unit is about them.
The null space $N(A) = \{x \in \mathbb{R}^n : Ax = 0\}$ is a subspace of $\mathbb{R}^n$, and the three conditions are one line each: $A0 = 0$; if $Ax = 0$ and $Ay = 0$ then $A(x + y) = Ax + Ay = 0$; and $A(cx) = c(Ax) = 0$. Every step used only that $A$ distributes over sums and pulls scalars out — which is the definition of a linear map, not a fact about this particular $A$.
The column space $C(A)$, the set of all $Ax$ as $x$ ranges over $\mathbb{R}^n$, is a subspace of $\mathbb{R}^m$, by the same three lines read in the other direction.
This is why the solution set of $Ax = b$ has the shape it does. When $b = 0$ the solutions are a subspace. When $b \ne 0$ they are not — the zero vector is missing — but they are a single solution plus the whole null space, a flat set of the same dimension pushed off the origin. Consistency asks whether $b$ lies in the column space; uniqueness asks whether the null space is just $\{0\}$. Two questions, two subspaces, and they are independent of each other.
Checking only closure under addition. The first quadrant of the plane contains the origin and is closed under addition, and it is not a subspace: $-1$ times $(1, 1)$ leaves it. Scalars include the negative ones and zero, and a set that forgets this is the commonest false positive there is.
Testing with particular vectors instead of general ones. Finding two members whose sum is in the set proves nothing; closure is a claim about every pair. One counterexample refutes, but no number of examples confirms. To prove closure you must argue about arbitrary members.
Assuming the origin test is enough. It refutes and never confirms. The singular matrices and the union of the two axes both contain $0$ and both fail closure under addition.
Thinking a subspace must be smaller than the space. $V$ is a subspace of itself, and $\{0\}$ is a subspace of everything. Both are legitimate answers, and the second is the reason the phrase trivial subspace exists.
Forgetting that the vectors need not be arrows. A question about polynomials or matrices is not a different kind of question. Write down what addition, scaling and zero mean in that space, and then run the same three lines.
Is $W = \{(x, y, z) : x + 2y - z = 0\}$ a subspace of $\mathbb{R}^3$? Substitute $(0, 0, 0)$: $0 = 0$, so it survives.
The origin test first.
Take $u, v \in W$. Then $(u_1 + v_1) + 2(u_2 + v_2) - (u_3 + v_3) = 0 + 0 = 0$, and $cu_1 + 2cu_2 - cu_3 = c \cdot 0 = 0$.
Both closures, argued generally.
So $W$ is a subspace — a plane through the origin. Change the condition to $x + 2y - z = 5$ and the origin test fails at the first line, so that set is not.
Homogeneous is what matters.
Let $U$ be the union of the two coordinate axes in $\mathbb{R}^2$, that is $\{(x, y) : xy = 0\}$. It contains $(0, 0)$, and every multiple of a member stays on the same axis.
Two of the three conditions hold.
But $(1, 0)$ and $(0, 1)$ are both in $U$, and their sum $(1, 1)$ has $xy = 1$.
One counterexample settles closure.
Not a subspace. Note what went wrong: the defining condition $xy = 0$ is homogeneous but not linear, and it is linearity that makes the closure arguments work.
Homogeneous is necessary, not sufficient.
The zero matrix has trace $0$, so the first condition holds.
Substitute the zero vector of this space.
The trace adds the diagonal entries, so $\operatorname{tr}(A + B) = \operatorname{tr} A + \operatorname{tr} B = 0$ and $\operatorname{tr}(cA) = c \operatorname{tr} A = 0$.
Both closures, in one line each.
It is a subspace. The condition is a single homogeneous linear equation on the four entries, which is the pattern to look for.
Match each subset of $\mathbb{R}^{2}$ to its verdict.
| A subspace | Not a subspace: the zero vector is missing | Not a subspace: closed under addition, but not under multiplying by $-1$ | |
|---|---|---|---|
| The set of $(x, y)$ with $5x + 5y = 0$ | |||
| The set of $(x, y)$ with $5x + 5y = 4$ | |||
| The set of $(x, y)$ with $x \ge 0$ and $y \ge 0$ |
The $2 \times 2$ matrices of trace $0$ form a subspace of all $2 \times 2$ matrices. Write the member of that subspace whose first row is $3 \; 4$ and whose lower-left entry is $3$.
This task has no paper form; do it on a device.
Which of these subsets of $\mathbb{R}^{3}$ is a subspace?
$W$ is the set of $(x, y)$ in $\mathbb{R}^{2}$ with $5x + 6y = 0$. The vector $(6, k)$ lies in $W$. What is $k$?
Answer:
You are asked whether the set of $(x, y)$ with $4x + 6y = 0$ is a subspace. Put the steps of the test in order.
Number the steps in order (write the number in the box):
Inside the vector space of all real functions, select every set below that is a subspace.
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.
Match each subset of $\mathbb{R}^{2}$ to its verdict.
| A subspace | Not a subspace: the zero vector is missing | Not a subspace: closed under addition, but not under multiplying by $-1$ | |
|---|---|---|---|
| The set of $(x, y)$ with $6x + 5y = 0$ | |||
| The set of $(x, y)$ with $6x + 5y = 5$ | |||
| The set of $(x, y)$ with $x \ge 0$ and $y \ge 0$ |
You can apply the subspace test to a set of columns, of matrices or of functions, and name the condition that fails when one does. Say in your own words why checking the origin refutes but never confirms. Next: what a set of vectors reaches, and when one of them is redundant.
10. Your turn: is the set of $2 \times 2$ matrices with $\operatorname{tr} A = 0$ a subspace?, step 3
Name the pattern, not just the verdict.