Practice: Vector Spaces and Subspaces

Recognition · Interpretation

Let $W = {(x, y) \in R^{2} : x y = 0}$ , the union of the two coordinate axes. $W$ contains $(0, 0)$ , and for any $a$ and any $(x, y) \in W$ the point $(a x, a y)$ has $(a x) (a y) = a^{2} x y = 0$ , so $W$ is closed under scaling. Is $W$ a subspace?

2 hints available, least help first.

Hint 1: Retrieval cue

How many conditions does the subspace test have, and which has not been checked here?

Hint 2: Concept cue

Pick one point on each axis and add them. Where does the sum land?

Direct application

Let $W = {(x, y, z) \in R^{3} : x + 2 y - z = 5}$ .

Evaluate the left-hand side $x + 2 y - z$ at the zero vector $(0, 0, 0)$ . What value do you get?

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

Put zero in place of each of $x$ , $y$ and $z$ in $x + 2 y - z$ .

Hint 2: Concept cue

Compare the value you get with the 5 the set demands. What does a mismatch tell you?

Recognition · Error diagnosis

Let $W \subset R^{2}$ be the union of the line $y = x$ and the line $y = - x$ .

A learner argues: " $W$ contains $(0, 0)$ , and scaling any point of $W$ by $a$ keeps it on the same line, so $W$ is closed under scalar multiplication. Closure under addition follows, because a sum is built out of scalings. So $W$ is a subspace."

Which response identifies the error?

2 hints available, least help first.

Hint 1: Retrieval cue

The subspace test has three conditions. Which one has been asserted rather than checked?

Hint 2: Concept cue

Take one point from each line and add them. Does the sum satisfy $y = x$ or $y = - x$ ?

Classification · Construction · Explanation

For each set below, decide whether it is a subspace of the stated space. Where it is, prove both closure conditions for arbitrary elements. Where it is not, give explicit elements witnessing the failure and name the condition they break.

(a) $A = {(x, y, z) \in R^{3} : x = 3 z}$ .

(b) $B = {(x, y) \in R^{2} : x \geq 0 and y \geq 0}$ , the first quadrant.

(c) $C = {p \in P_{3} : p (2) = 0}$ , the polynomials of degree at most 3 vanishing at 2.

(d) $D = {M \in M_{2 \times 2} (R) : det M = 0}$ , the singular $2 \times 2$ matrices.

(e) One of the failing sets fails a condition the others pass. Say which, and why that makes it a different kind of failure from the rest.

Write your answer

Write your answer, then compare it with the worked solution.

3 hints available, least help first.

Hint 1: Retrieval cue

Start each part with the zero test. It is the cheapest, and it settles some cases outright.

Hint 2: Concept cue

For a positive verdict, write two arbitrary members with symbols and show the defining condition still holds for their sum and for a multiple.

Hint 3: Strategy cue

For part (e), look at what kind of condition defines each failing set: an inequality, or an equation that is not linear.

Compare with the worked solution

Comparing does not record a result. Judging your own written answer cannot show that you can do this without help.

(a) $A$ is a subspace. Zero. $0 = 3 (0)$ ✓. Addition. Let $u = (x_{1}, y_{1}, z_{1})$ and $v = (x_{2}, y_{2}, z_{2})$ lie in $A$ , so $x_{1} = 3 z_{1}$ and $x_{2} = 3 z_{2}$ . The first coordinate of $u + v$ is $x_{1} + x_{2} = 3 z_{1} + 3 z_{2} = 3 (z_{1} + z_{2})$ , which is three times the third coordinate of $u + v$ . So $u + v \in A$ . Scaling. For $a \in R$ , the first coordinate of $a u$ is $a x_{1} = a (3 z_{1}) = 3 (a z_{1})$ , three times its third coordinate, so $a u \in A$ . Both arguments used arbitrary members. ( $A$ is the plane spanned by $(3, 0, 1)$ and $(0, 1, 0)$ .) (b) $B$ is not a subspace. It contains $(0, 0)$ and is closed under addition. Two points with nonnegative coordinates sum to one with nonnegative coordinates. It fails closure under scaling: $(1, 1) \in B$ , but

(- 1) (1, 1) = (- 1, - 1) \notin B,

since $- 1 < 0$ . Any negative scalar applied to any nonzero member of $B$ witnesses the same failure. (c) $C$ is a subspace. Zero. The zero polynomial satisfies $0 (2) = 0$ ✓. Addition. If $p (2) = 0$ and $q (2) = 0$ then $(p + q) (2) = p (2) + q (2) = 0 + 0 = 0$ , and $p + q$ still has degree at most 3, so $p + q \in C$ . Scaling. $(a p) (2) = a \cdot p (2) = a \cdot 0 = 0$ , and $a p$ has degree at most 3, so $a p \in C$ . The condition is linear and homogeneous, evaluation at a point is a linear map, and the requirement is that it return zero, which is what makes it work. (d) $D$ is not a subspace. It contains the zero matrix, since $det 0 = 0$ , and it is closed under scaling, because $det (a M) = a^{2} det M = 0$ for $2 \times 2$ matrices. It fails closure under addition:

M = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), N = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}),

both with determinant $0$ , while

M + N = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), det (M + N) = 1 \neq 0 .

(e) The odd one out is $B$ . It is the only set here that fails closure under scaling while passing the zero test and closure under addition. $D$ fails the other closure condition, and $A$ and $C$ fail nothing. The difference is structural. $B$ 's defining condition is an inequality, and inequalities are reversed by negative scalars, so the failure is guaranteed for every nonzero member, and it arises at the moment a negative scalar is allowed. $D$ 's condition is an equation, but a nonlinear one: $det$ is quadratic in the entries, which is precisely why $det (a M) = a^{2} det M$ preserves the condition while addition does not. So the two failures come from opposite sources. An inequality survives addition and dies under negative scaling; a homogeneous nonlinear equation survives scaling and dies under addition. Only a linear homogeneous condition, as in $A$ and $C$ , survives both.

A complete answer does each of these:

applies the zero test
tests both closures
argues for arbitrary elements
exhibits a counterexample

Practice: Vector Spaces and Subspaces

Recognition · Interpretation

Direct application

Recognition · Error diagnosis

Classification · Construction · Explanation

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