Subspaces
Lines and planes through the origin keep turning up, as spans and as solution sets of equations. What do they share that a line missing the origin, or a curve, does not?
Where you are 8 in Vectors in n dimensions
Why this exists
Carmela, Gennaro and Sokol run the vats at a dyehouse. When a batch comes out slightly off, they adjust the recipe for the next one. An adjustment is a vector (x, y, z) in \mathbb{R}^3: the change in grams per liter of red, yellow and blue dye, positive for more and negative for less. Their shade meter turns a recipe into one reading, and an adjustment (x, y, z) changes the reading by x - 2y + 3z.
An adjustment that leaves the reading unchanged is balanced. Two balanced adjustments made in turn are balanced, and so is one made twice as strongly, or reversed, or no change at all. Adding and scaling never lead out of the set. Sets like this are the subspaces: the flat pieces of \mathbb{R}^n through the origin, where the rest of linear algebra takes place.
Definition
A subspace of \mathbb{R}^n is a set W \subseteq \mathbb{R}^n with three properties.
- It contains zero: \mathbf{0} \in W.
- It is closed under addition: if \mathbf{u} \in W and \mathbf{v} \in W, then \mathbf{u} + \mathbf{v} \in W.
- It is closed under scalar multiplication: if \mathbf{u} \in W and c is any real number, then c\mathbf{u} \in W.
“Closed” means the operation never leads out of the set. \mathbb{R}^n is a subspace of itself, and so is \{\mathbf{0}\}, the zero subspace, since \mathbf{0} + \mathbf{0} = \mathbf{0} and c\mathbf{0} = \mathbf{0}.
The two closure conditions fold into one, the one-line test: W is a subspace exactly when \mathbf{0} \in W and
for all \mathbf{u}, \mathbf{v} \in W and all scalars c, d. So W is closed under linear combinations of two of its vectors, and so, adding one term at a time, of any number of them.
Intuition
A subspace contains everything that scaling and adding reach from its own vectors. One nonzero vector brings the whole line through the origin along it, in both directions and with no gaps; two vectors in different directions bring the whole plane through the origin that holds them. Nothing curved survives scaling, nothing bounded except \{\mathbf{0}\} survives it (doubling a nonzero vector again and again escapes any bound), and no nonempty set missing the origin survives scaling by 0.
The line x - 2y = 0 in \mathbb{R}^2 is a subspace. The parallel line x - 2y = 4 is not: it misses \mathbf{0}, and the sum of two of its vectors lands off it.
Results
Every span is a subspace. For any vectors \mathbf{v}_1, \dots, \mathbf{v}_k in \mathbb{R}^n, \operatorname{span}\{\mathbf{v}_1, \dots, \mathbf{v}_k\} is a subspace of \mathbb{R}^n. It is the smallest one containing them: every subspace that contains \mathbf{v}_1, \dots, \mathbf{v}_k contains their whole span.
Homogeneous systems. For an m \times n matrix A, the solution set of A\mathbf{x} = \mathbf{0} is a subspace of \mathbb{R}^n. When \mathbf{b} \neq \mathbf{0}, the solution set of A\mathbf{x} = \mathbf{b} is never a subspace.
Subspaces as spans. Every subspace in this lesson is written as a span by solving its equations for the free variables (Examples 1, 4 and 6). In fact every subspace of \mathbb{R}^n is the span of at most n vectors, proved in Basis and dimension.
The subspaces of \mathbb{R}^2 are exactly \{\mathbf{0}\}, the lines through the origin, and \mathbb{R}^2.
The subspaces of \mathbb{R}^3 are exactly \{\mathbf{0}\}, the lines through the origin, the planes through the origin, and \mathbb{R}^3. This is stated here; the full reason needs basis and dimension and is given in Basis and dimension.
Why it is true
The one-line test. If W is a subspace and \mathbf{u}, \mathbf{v} \in W, closure under scaling puts c\mathbf{u} and d\mathbf{v} in W, and closure under addition puts their sum there. Conversely, if W passes the test, c = d = 1 gives closure under addition, and \mathbf{v} = \mathbf{u}, d = 0 gives c\mathbf{u} + 0\mathbf{u} = c\mathbf{u} \in W.
Spans. \mathbf{0} = 0\mathbf{v}_1 + \dots + 0\mathbf{v}_k is in the span. If \mathbf{x} = a_1\mathbf{v}_1 + \dots + a_k\mathbf{v}_k and \mathbf{y} = b_1\mathbf{v}_1 + \dots + b_k\mathbf{v}_k, distributing c and d and collecting the terms in each \mathbf{v}_j gives
another combination of the \mathbf{v}_j, so the span passes the one-line test. A subspace containing each \mathbf{v}_j contains each c_j\mathbf{v}_j by scaling, and then their sum by adding one term at a time, so it contains the whole span.
Homogeneous systems. Let N be the solution set of A\mathbf{x} = \mathbf{0}. Since A\mathbf{0} = \mathbf{0}, \mathbf{0} \in N. A matrix times a vector distributes over sums and lets scalars out (Multiplying a matrix by a vector), so for \mathbf{u}, \mathbf{v} \in N,
and c\mathbf{u} + d\mathbf{v} \in N. If \mathbf{b} \neq \mathbf{0}, then A\mathbf{0} = \mathbf{0} \neq \mathbf{b}, so \mathbf{0} is not a solution and the first condition fails.
Subspaces of \mathbb{R}^2. The three kinds are subspaces: a line through the origin is \operatorname{span}\{\mathbf{u}\} for any nonzero \mathbf{u} on it. Conversely, let W \neq \{\mathbf{0}\} be a subspace of \mathbb{R}^2. It holds some \mathbf{u} \neq \mathbf{0}, and so the line \operatorname{span}\{\mathbf{u}\}. If W is more than that line, it holds some \mathbf{v} off the line. Then \mathbf{v} is not a multiple of \mathbf{u}, and \mathbf{u} is not a multiple of \mathbf{v} either: if \mathbf{u} = t\mathbf{v}, then t \neq 0 because \mathbf{u} \neq \mathbf{0}, and \displaystyle \mathbf{v} = \frac{1}{t}\mathbf{u}. Two vectors in \mathbb{R}^2, neither a multiple of the other, span \mathbb{R}^2 (Span). W contains their span, so W = \mathbb{R}^2 = \operatorname{span}\{\mathbf{u}, \mathbf{v}\}.
Subspaces of \mathbb{R}^3. The same argument runs one step further. A nonzero W holds a line; if it holds a vector off that line, it holds a plane through the origin; if it also holds a vector off that plane, it holds three vectors, each outside the span of the ones before it. That such three vectors span \mathbb{R}^3 is the step that needs Linear independence and Basis and dimension.
Worked examples
The lesson is taught in two parts. Examples 1 to 3 belong to the first: proving from the definition that a set is a subspace, or that it is not. Examples 4 to 6 belong to the second: spans and solution sets as subspaces, and the subspaces of \mathbb{R}^2.
A plane through the origin
Show from the three conditions that the plane P given by x - 2y + 3z = 0 is a subspace of \mathbb{R}^3, and write P as a span.
SolutionZero. 0 - 2(0) + 3(0) = 0, so \mathbf{0} \in P.
Addition. Let \mathbf{u} = (u_1, u_2, u_3) and \mathbf{v} = (v_1, v_2, v_3) be in P, so u_1 - 2u_2 + 3u_3 = 0 and v_1 - 2v_2 + 3v_3 = 0. The entries of \mathbf{u} + \mathbf{v} are u_i + v_i. Distributing, then reordering and regrouping the terms (the commutative and associative properties of addition),
So \mathbf{u} + \mathbf{v} \in P.
Scaling. Factoring out c,
so c\mathbf{u} \in P.
All three hold, so P is a subspace. To write it as a span, solve the equation for x: x = 2y - 3z, with y and z free. Every vector of P is
and every such combination satisfies the equation, so P = \operatorname{span}\{(2, 1, 0), (-3, 0, 1)\}. Check: 2 - 2 + 0 = 0 and -3 - 0 + 3 = 0.
Third entry the sum of the first two
Let W be the set of all vectors (a, b, a + b) in \mathbb{R}^3, for real a and b. Show with the one-line test that W is a subspace. Decide whether (3, -5, -2) and (2, 2, 3) are in W, and write W as a span and by an equation.
SolutionWith a = b = 0, \mathbf{0} = (0, 0, 0 + 0) \in W. Take \mathbf{u} = (a, b, a + b) and \mathbf{v} = (a', b', a' + b') in W, and scalars c, d:
The third entry regroups to (ca + da') + (cb + db'), the sum of the first two. So c\mathbf{u} + d\mathbf{v} has the form (s, t, s + t) with s = ca + da' and t = cb + db', and lies in W. W is a subspace.
A vector (x, y, z) is in W exactly when z = x + y, taking a = x and b = y. For (3, -5, -2): 3 + (-5) = -2, so it is in W. For (2, 2, 3): 2 + 2 = 4 \neq 3, so it is not.
Splitting by the free letters, (a, b, a + b) = a(1, 0, 1) + b(0, 1, 1), so W = \operatorname{span}\{(1, 0, 1), (0, 1, 1)\}. The condition z = x + y says W is the plane x + y - z = 0.
Four sets that are not subspaces
Prove that none of these is a subspace of \mathbb{R}^2: (a) the line L given by x - 2y = 4; (b) the first quadrant Q, the vectors (x, y) with x \geq 0 and y \geq 0; (c) the union U of the lines y = x and y = -2x; (d) the curve C given by y = x^2.
SolutionOne condition failing for specific vectors is enough.
(a) 0 - 2(0) = 0 \neq 4, so \mathbf{0} \notin L. Addition fails too: \mathbf{p} = (4, 0) and \mathbf{q} = (0, -2) are on L, but \mathbf{p} + \mathbf{q} = (4, -2) gives 4 - 2(-2) = 8 \neq 4, as the first figure shows.
(b) Q contains \mathbf{0}, and adding vectors with nonnegative entries gives nonnegative entries, so Q is closed under addition. Scaling fails: \mathbf{u} = (2, 1) \in Q, but (-1)\mathbf{u} = (-2, -1) has negative entries, so it is not in Q.
(c) U contains \mathbf{0}, and a multiple of a vector on either line stays on that line, so U is closed under scaling. Addition fails: (1, 1) is on y = x and (1, -2) is on y = -2x, but their sum (2, -1) is on neither, since -1 \neq 2 and -1 \neq -4.
(d) \mathbf{0} \in C, since 0 = 0^2. But (1, 1) \in C while 2(1, 1) = (2, 2) is not, since 2 \neq 2^2.
The solutions of a homogeneous system
Let
(a) Write the solution set of A\mathbf{x} = \mathbf{0} as a span. (b) Show that the solution set of A\mathbf{x} = \mathbf{b} with \mathbf{b} = (1, 3) is not a subspace, and describe it.
Solution(a) By the result on homogeneous systems, the solution set N is a subspace of \mathbb{R}^4; a span shows what it looks like. Subtracting twice row 1 from row 2 gives
The pivots are in columns 1 and 2, so x_3 and x_4 are free. Row 2 gives x_2 = -3x_3 + 2x_4, and row 1 then gives x_1 = -2x_2 + x_3 - 3x_4 = 7x_3 - 7x_4. So
and N is the span of (7, -3, 1, 0) and (-7, 2, 0, 1), a plane through the origin in \mathbb{R}^4 (neither vector is a multiple of the other). Check: A(7, -3, 1, 0) = (7 - 6 - 1, 14 - 15 + 1) ={} (0, 0), and A(-7, 2, 0, 1) = (-7 + 4 + 3, -14 + 10 + 4) ={} (0, 0).
(b) A\mathbf{0} = (0, 0) \neq (1, 3), so \mathbf{0} is not a solution and the solution set is not a subspace. The same row operation turns the right side into (1, 3 - 2) = (1, 1), so x_2 = 1 - 3x_3 + 2x_4 and x_1 = 1 - 2x_2 + x_3 - 3x_4 = -1 + 7x_3 - 7x_4. Setting x_3 = x_4 = 0 gives the solution \mathbf{p} = (-1, 1, 0, 0), and in general
The solution set is the plane N slid off the origin by \mathbf{p}: flat, but not through \mathbf{0}. That shape is studied in Homogeneous systems and solutions in parametric vector form. Check: A\mathbf{p} = (-1 + 2, -2 + 5) = (1, 3).
What a subspace of the plane can be
(a) A subspace W of \mathbb{R}^2 contains (5, -2) and (1, 3). Show that W = \mathbb{R}^2 by writing a general (x, y) as a combination of the two. (b) A subspace V of \mathbb{R}^2 is known to contain (5, -2) and (-15, 6). What can V be?
Solution(a) W contains the span of any of its vectors, so it contains every c_1(5, -2) + c_2(1, 3). Matching entries, 5c_1 + c_2 = x and -2c_1 + 3c_2 = y. Three times the first minus the second gives 17c_1 = 3x - y, and then c_2 = x - 5c_1:
Weights exist for every (x, y), so W = \mathbb{R}^2.
(b) (-15, 6) = -3(5, -2), so V contains the line \operatorname{span}\{(5, -2)\}, which is 2x + 5y = 0. By the classification, V is that line or \mathbb{R}^2 (not \{\mathbf{0}\}, since it holds a nonzero vector). Both contain the two vectors, so the information given does not decide.
Balanced adjustments at the dyehouse
At the dyehouse, an adjustment (x, y, z) changes the shade reading by x - 2y + 3z and the total dye per liter by x + y + z. (a) Carmela wants the adjustments that change neither. Show that they form a subspace, write it as a span, and decide whether Sokol’s adjustment (10, -4, -6) is one of them. (b) Gennaro wants the adjustments that keep the total but raise the reading by 6. Show that they do not form a subspace, and describe them.
Solution(a) Carmela’s adjustments are the solutions of the homogeneous system
so they form a subspace by the result on homogeneous systems. Subtracting the first equation from the second gives 3y - 2z = 0, so \displaystyle z = \frac{3}{2}y, and then \displaystyle x = -y - z = -\frac{5}{2}y. Writing y = 2t, every solution is t(-5, 2, 3): the subspace is \operatorname{span}\{(-5, 2, 3)\}, a line through the origin. Check: -5 - 4 + 9 = 0 and -5 + 2 + 3 = 0. Sokol’s (10, -4, -6) = -2(-5, 2, 3) is on the line, so it qualifies.
(b) Gennaro’s system is x - 2y + 3z = 6, x + y + z = 0. The zero adjustment raises the reading by 0, not 6, so the set is not a subspace. Subtracting the first equation from the second gives 3y - 2z = -6; y = 0 gives z = 3 and x = -3. So (-3, 0, 3) is one of his adjustments: -3 + 9 = 6 and -3 + 3 = 0. All of them are
a line parallel to Carmela’s that misses the origin.
Boundaries
- Through the origin, always. A line or plane that misses \mathbf{0} is never a subspace, however flat. It is a subspace slid off the origin, the shape of the solutions of A\mathbf{x} = \mathbf{b} with \mathbf{b} \neq \mathbf{0} (Example 4).
- Every real scalar. Closure under scaling is for all real c. A set closed under positive scalars only, like the quadrant in Example 3, is not a subspace.
- Not empty. The empty set passes both closure conditions trivially but lacks \mathbf{0}. The smallest subspace is \{\mathbf{0}\}.
- Intersections, not unions. The intersection of two subspaces is a subspace (Exercise 20); their union usually is not (Example 3, Exercise 21).
Where this goes next
A spanning list with nothing redundant in it, a linearly independent one, is a basis, and the number of vectors in a basis is the dimension of the subspace: 1 for a line, 2 for a plane. That is what finishes the list of subspaces of \mathbb{R}^3. Every matrix carries two subspaces: the solution set of A\mathbf{x} = \mathbf{0} is its kernel, and the span of its columns is its image. Projecting a vector onto a subspace is the start of Orthogonality and least squares. The area is mapped in Vectors in n dimensions.
Exercises
Exercises 1 to 9 go with the first part (Examples 1 to 3), Exercises 10 to 18 with the second (Examples 4 to 6), and the reasoning group draws on both.
Part 1: proving a set is or is not a subspace
- Show from the three conditions that the plane 2x + y - 4z = 0 is a subspace of \mathbb{R}^3, and write it as a span.
- Let W be the set of vectors (a - b, b, 2a) for real a, b. Show with the one-line test that W is a subspace of \mathbb{R}^3, decide whether (1, 2, 6) is in W, and write W as a span and by an equation.
In Exercises 3 to 9, decide whether the set is a subspace. Prove a yes; for a no, show one condition failing with specific vectors.
- The line x + y = 1 in \mathbb{R}^2.
- The vectors (x, y) in \mathbb{R}^2 with xy \geq 0.
- The vectors (x, y, z) in \mathbb{R}^3 with x^2 + y^2 = z^2.
- The vectors (x, y, z) in \mathbb{R}^3 with x + y + z = 0 and x = 2y. If it is a subspace, write it as a span.
- The vectors (x, y) in \mathbb{R}^2 with \lvert x \rvert = \lvert y \rvert.
- The vectors (x, y, z, w) in \mathbb{R}^4 with x = w and y = 0. If it is a subspace, write it as a span.
- The vectors in \mathbb{R}^2 whose two entries are both integers.
Part 2: spans, solution sets and the subspaces of the plane
- Write the solution set of x_1 - x_2 + 2x_3 = 0, 2x_1 + x_2 + x_3 = 0 as a span, and say what shape it is.
- Explain why the solution set of x_1 - x_2 + 2x_3 = 3, 2x_1 + x_2 + x_3 = 3 is not a subspace, and describe it using Exercise 10.
- Write the solution set of x_1 + x_2 - x_3 + 2x_4 = 0 in \mathbb{R}^4 as a span.
- Say whether each is \{\mathbf{0}\}, a line through the origin or all of \mathbb{R}^2: (a) \operatorname{span}\{(6, -4), (-9, 6)\}; (b) \operatorname{span}\{(1, 3), (2, 5)\}; (c) the solution set of 2x - y = 0, 4x + y = 0.
- A subspace W of \mathbb{R}^2 contains (2, -5) and (1, 4). Show that W = \mathbb{R}^2 by writing a general (x, y) as a combination of the two.
- A subspace V of \mathbb{R}^2 contains (2, -5) and (-4, 10). What can V be?
- Gennaro adds black dye, so an adjustment is (x, y, z, w), with w the change in black. The meter reads hue, which black leaves alone, so the reading still changes by x - 2y + 3z, and the total changes by x + y + z + w. Write the adjustments that change neither as a span.
- For which numbers k is the plane 3x - y + 2z = k a subspace of \mathbb{R}^3?
- Write the set of vectors in \mathbb{R}^3 orthogonal to (1, -1, 4) as a span, and explain why it is a subspace.
Reasoning
- Prove that if W is a subspace and \mathbf{u}, \mathbf{v} \in W, then \mathbf{u} - \mathbf{v} \in W.
- Prove that the intersection of two subspaces of \mathbb{R}^n is a subspace.
- Let V and W be subspaces of \mathbb{R}^n, neither contained in the other. Prove that their union is not a subspace.
- Explain why a subspace other than \{\mathbf{0}\} contains infinitely many vectors.
- Prove that a subspace of \mathbb{R}^3 containing (1, 0, 0), (0, 1, 0) and (0, 0, 1) is all of \mathbb{R}^3.
Answers
- It is the span of (1, -2, 0) and (0, 4, 1). Zero: 0 + 0 - 0 = 0. If \mathbf{u} and \mathbf{v} satisfy the equation, the left side for \mathbf{u} + \mathbf{v} regroups to (2u_1 + u_2 - 4u_3) + (2v_1 + v_2 - 4v_3) = 0, and for c\mathbf{u} it is c(2u_1 + u_2 - 4u_3) = 0. Solving y = -2x + 4z gives (x, -2x + 4z, z) = x(1, -2, 0) + z(0, 4, 1).
- Yes, (1, 2, 6) \in W, with a = 3, b = 2. W = \operatorname{span}\{(1, 0, 2), (-1, 1, 0)\}, the plane 2x + 2y - z = 0, since a = x + y and z = 2a. \mathbf{0} comes from a = b = 0, and c(a - b, b, 2a) + d(a' - b', b', 2a') = (s - t, t, 2s) with s = ca + da', t = cb + db'.
- No: 0 + 0 \neq 1, so \mathbf{0} is not on the line.
- No. It holds \mathbf{0}, and (cx)(cy) = c^2xy \geq 0, so scaling is fine, but (2, 1) + (-1, -3) = (1, -2), and 1 \cdot (-2) = -2 < 0.
- No. It holds \mathbf{0} and is closed under scaling, but (1, 0, 1) and (0, 1, 1) are in it while their sum (1, 1, 2) is not: 1 + 1 \neq 4.
- Yes: \operatorname{span}\{(2, 1, -3)\}, a line. Both equations hold at \mathbf{0} and are kept by sums and multiples, as in Example 1. With y free, x = 2y and z = -x - y = -3y.
- No. It holds \mathbf{0} and is closed under scaling, but (3, 3) + (2, -2) = (5, 1), and 5 \neq 1.
- Yes: the span of (1, 0, 0, 1) and (0, 0, 1, 0). \mathbf{0} satisfies both equations; if x = w, y = 0 and x' = w', y' = 0, then x + x' = w + w', y + y' = 0, cx = cw and cy = 0. Every vector is (x, 0, z, x) = x(1, 0, 0, 1) + z(0, 0, 1, 0).
- No. It holds \mathbf{0} and is closed under addition, but (1, 0) is in it and \displaystyle \frac{1}{2}(1, 0) = \left(\frac{1}{2}, 0\right) is not.
- \operatorname{span}\{(-1, 1, 1)\}, a line through the origin. Row 2 minus twice row 1 gives 3x_2 - 3x_3 = 0, so x_2 = x_3, and x_1 = x_2 - 2x_3 = -x_3.
- \mathbf{0} is not a solution, since 0 \neq 3. The same row operation gives 3x_2 - 3x_3 = -3; x_3 = 0 gives the solution (2, -1, 0), and the solution set is (2, -1, 0) + t(-1, 1, 1), the line of Exercise 10 slid off the origin.
- The span of (-1, 1, 0, 0), (1, 0, 1, 0) and (-2, 0, 0, 1), from x_1 = -x_2 + x_3 - 2x_4 with x_2, x_3, x_4 free.
- (a) A line: \displaystyle (-9, 6) = -\frac{3}{2}(6, -4), so the span is the line 2x + 3y = 0. (b) All of \mathbb{R}^2: (2, 5) is not a multiple of (1, 3), since 2 \cdot 3 \neq 5. (c) \{\mathbf{0}\}: the first equation gives y = 2x, and then 4x + 2x = 0 forces x = 0 and y = 0.
- \displaystyle c_1 = \frac{4x - y}{13} and \displaystyle c_2 = \frac{5x + 2y}{13}, from 2c_1 + c_2 = x and -5c_1 + 4c_2 = y (four times the first minus the second gives 13c_1 = 4x - y). Weights exist for every (x, y), so W, which contains their span, is \mathbb{R}^2.
- The line 5x + 2y = 0 or all of \mathbb{R}^2. (-4, 10) = -2(2, -5), so V contains the line \operatorname{span}\{(2, -5)\}, and by the classification it is that line or \mathbb{R}^2.
- The span of (2, 1, 0, -3) and (-3, 0, 1, 2). With y, z free, x = 2y - 3z and w = -x - y - z = -3y + 2z.
- Only k = 0. \mathbf{0} is on the plane exactly when 0 = k, and for k = 0 the plane is the solution set of a homogeneous equation.
- \operatorname{span}\{(1, 1, 0), (-4, 0, 1)\}. Orthogonal to (1, -1, 4) means x - y + 4z = 0 (Dot product, length and orthogonality in n dimensions), a homogeneous equation, so the set is a subspace; with y, z free, x = y - 4z.
- \mathbf{u} - \mathbf{v} = 1\mathbf{u} + (-1)\mathbf{v}, a linear combination of vectors of W, so it is in W by the one-line test.
- \mathbf{0} is in both, so in the intersection. If \mathbf{u} and \mathbf{v} are in both, then c\mathbf{u} + d\mathbf{v} is in each by the one-line test, so in the intersection.
- Pick \mathbf{v} \in V not in W and \mathbf{w} \in W not in V. If \mathbf{v} + \mathbf{w} were in V, then \mathbf{w} = (\mathbf{v} + \mathbf{w}) - \mathbf{v} would be in V by Exercise 19, which it is not. In the same way \mathbf{v} + \mathbf{w} \notin W. So the union is not closed under addition.
- It holds some \mathbf{u} \neq \mathbf{0}, and so every c\mathbf{u}. These are all different: c\mathbf{u} = d\mathbf{u} gives (c - d)\mathbf{u} = \mathbf{0}, and since \mathbf{u} has a nonzero entry, c = d.
- Every (x, y, z) = x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1) is a linear combination of the three, and a subspace contains every linear combination of its vectors.