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Section 3.1: Definition and Examples (Vector Spaces), Completed

1. Examples n Spaces: The set of n- vectors that we denoted by R is a . For simplicity, let’s consider n = 2.

T 2 A vector x = x1 x2 in R can be represented by a directed segment from the to the (x1, x2) or from any point (a, b) (same and direction).

We scale vectors by any α, multiplying each component by α.

We add and subtract vectors via components, with a nice geometrical interpretation. n The space of all m × n matrices: Just as R can be viewed as n × 1 matrices, we can generalize m×n n to R , the set of all m × n matrices. As we’ve seen, matrices behave similarly to vectors in R with / and scalar . n m×n Both of these vector spaces R and R possess desirable operations that we can perform on their elements. These algebraic rules form the axioms used to define a general vector space. Intuitively, a vector space is a set that has the essential operations of addition and defined, with some other well-desired properties satisfied. This concept, generalizing to defining abstract objects possessing certain properties, is one of the most powerful in all of mathematics. M309 Notes ©, R.G. Lynch, Texas A&M Section 3.1: Definition and Examples (Vector Spaces), Completed Page 2 of 9

Definition. An essential component of the definition are the closure properties: C1. If x ∈ V and α is a scalar, then αx ∈ V . C2. If x, y ∈ V , then x + y ∈ V . A vector space is closed under addition and scalar multiplication.

Theorem. If V is a vector space and x is any element of V , then (i)0 x = 0 (ii) x + y = 0 implies that y = −x (that is, the is unique) (iii)( −1)x = −x

Note. This is terrible notation and it is a shame that many authors introduce vector spaces as such. n Addition and scalar multiplication are NOT(!!!) necessarily what you are used to in R and this makes it look like it is. Usually, I will denote addition by ⊕ and scalar multiplication by ◦. While n most vector spaces we will deal with seem to behave very similarly to R , there are some weird ones (the last example). If I write + or ·, this will mean the usual addition and scalar multiplication.

m×n Example. When m and n are fixed, you can check that the R is a vector space. Is the set of all matrices (possibly different sizes) with the same addition and scalar multiplication a vector space? Solution. No, we cannot add two different sized matrices together. For example a 3 × 4 cannot be added to a 3 × 5, which we would need the to satisfy for it to make sense. M309 Notes ©, R.G. Lynch, Texas A&M Section 3.1: Definition and Examples (Vector Spaces), Completed Page 3 of 9

2 Example (Lines in R ). Let U := {(x, 1) : x ∈ R}, the set of all points lying on the horizontal line y = 1. Is U a vector space with usual addition and scalar multiplication? Solution. This set has a very undesired property. We can add any two vectors in the set together, say (3, 1) and (5, 1) to get (8, 2), which is no longer in U since it lies on the line y = 2! Thus, + is not really an operation on U. Similarly, neither is scalar multiplication because α(x, 1) = (αx, α) is only in U when α = 1. For example, 2(3, 1) = (6, 2) ∈/ W . Since it fails the closure properties, U is not a vector space.

Example. What about V := {(x, 0) : x ∈ R} with the same operations? Solution. This set no longer poses the problem we saw with U above. Adding (x1, 0) and (x2, 0) gives (x1 +x2, 0) which is still in V . Scalar multiplication also works because α(x, 0) = (αx, 0) ∈ V . It turns out that the axioms also hold for V and so it is a vector space. However, I won’t prove the axioms for V but instead prove them for W in the next example, since taking m = 0 will give V .

Example. What about W := {(x, mx): x ∈ R} with the same operations? Solution. Here we assume that m is any . Note that this set represents the set of all points on the line y = mx, so any nonvertical line through the origin. Showing closure: C1. α ∈ R, x = (x, mx) ∈ W implies that: αx = α(x, mx) = (αx, αmx) = (αx, m(αx)) ∈ W C2. x(x, mx), y = (y, my) ∈ W implies that: x + y = (x, mx) + (y, my) = (x + y, mx + my) = x + y, m(x + y) ∈ W

Showing the axioms: Let α, β ∈ R and x = (x, mx), y = (y, mx), z = (z, mz) ∈ W . A1. x + y = (x, mx) + (y, mx) = (x + y, mx + my) = (y + x, my + mx) = (y, my) + (x, mx) = y + x A2. On one hand we have (x + y) + z = (x, mx) + (y, mx) + (z, mz) = (x + y, mx + my) + (z, mz) = (x + y + z, mx + my + mz) and on the other x + (y + z) = (x, mx) + (y, mx) + (z, mz) = (x, mx) + (y + z, my + mz) = (x + y + z, mx + my + mz) from which we see that (x + y) + z = x + (y + z). A3. We need to find a c = (c, mc) ∈ W so that x + c = x. Note that I am using c for the Spanish “cero” since z is already being used AND this zero may not be the usual zero vector, (0, 0). We have to figure out if it is first! We have x + c = x ⇐⇒ (x, mx) + (c, mc) = (x, mx) ⇐⇒ (x + c, m(x + c)) = (x, mx) from which we can see we must take c = 0, so c = (0, 0). Again, we cannot technically subtract vectors yet since A4 has not been proven yet, so you can’t move (x, mx) over. A4. Again, we can’t be sure that −x is simply (−x, −mx) until we show that it is. So let n = (n, mn) and let’s check if n = −x. We have, recalling that c is the zero vector found in A3, x + n = c ⇐⇒ (x, mx) + (n, mn) = (0, 0) ⇐⇒ (x + n, m(x + n)) = (0, 0) from which we see that indeed n = −x, so −x = n = (−x, −nx). M309 Notes ©, R.G. Lynch, Texas A&M Section 3.1: Definition and Examples (Vector Spaces), Completed Page 4 of 9

A5. On one hand we have α(x + y) = α(x, mx) + (y, my) = α(x + y, mx + my) = (α(x + y), α(mx + my)) = (αx + αy, αmx + αmy) and on the other αx + αy = α(x, mx) + α(y, my) = (αx, αmx) + (αy, αmy) = (αx + αy, αmx + αmy) from which we see that α(x + y) = αx + αy. A6. On one hand we have (α + β)x = (α + β)(x, mx) = (α + β)x, (α + β)mx = (αx + βx, αmx + βmx) and on the other αx + βx = α(x, mx) + β(x, mx) = (αx, αmx) + (βx, βmx) = (αx + βx, αmx, βmx) from which we see that (α + β)x = αx + βx. A7. On one hand we have (αβ)x = (αβ)(x, mx) = (αβ)x, (αβ)mx = (αβx, αβmx) and on the other α(βx) = αβ(x, mx) = α(βx, βmx) = (αβx, αβmx) from which we see that (αβ)x = α(βx). A8. In class, we wrote u instead 1 and showed it was 1. I realize now that this is incorrect. The 1 is the unit number, that is, the number that 1 · α = α for any scalar α. Since the vector space is over the reals, that is, the scalars are real numbers, we know that 1 is the unit. Thus, we really do need to just check that 1 · x = x. Checking: 1 · x = 1 · (x, mx) = (1 · x, 1 · mx) = (x, mx) = x. Note that 1 · x = x and 1 · mx = mx since x and mx are just real numbers.

Try yourself. Technically we didn’t show that the set of all points on a vertical line that goes through the origin is a vector space. Try showing this yourself. M309 Notes ©, R.G. Lynch, Texas A&M Section 3.1: Definition and Examples (Vector Spaces), Completed Page 5 of 9

Example (Continuous real-valued functions on [a, b]). Let C[a, b] := {f :[a, b] → R : f is continuous} with addition and scalar multiplication defined by: (f ⊕ g)(x) = f(x) + g(x) (α ◦ f)(x) = α · f(x) Is C[a, b] a vector space? Solution. Yes, it is a vector space. Note that the vectors in this case are actually functions! This n m×n is already different than R or R . The closure properties follow because adding two continuous functions together still gives a continuous function and multiplying a continuous function by a scalar simply stretches or shrinks the function, but doesn’t affect continuity. You can rigorously check the closure properties (and it would be good practice) as we did in the last example, but I’m going to cheat a little and say that because we have basically defined the operations to be real number operations, adding real numbers at each value of x and multiplying. However, I want to point out what the zero vector in C[a, b] is, that is, show A3. A3. We need for any f ∈ C[a, b] to find a c ∈ C[a, b] so that (f ⊕ c)(x) = f(x) for all x ∈ [a, b]. Checking for any x ∈ [a, b]: (f ⊕ c)(x) = f(x) = f(x) ⇐⇒ f(x) + c(x) = f(x) ⇐⇒ c(x) = 0. So c(x) = 0 for any x. That is, c is the constant zero function! M309 Notes ©, R.G. Lynch, Texas A&M Section 3.1: Definition and Examples (Vector Spaces), Completed Page 6 of 9

Example (Polynomials). Let

Pn[a, b] := {p :[a, b] → R : p is a polynomial of degree less than n} with addition and scalar multiplication defined as on C[a, b]. Is Pn[a, b] a vector space? Solution. Note that the zero constant function is a 0-degree polynomial, so it is in Pn[a, b], showing that Pn[a, b] is nonempty. The closure properties hold since adding any polynomials of degree less than n leaves us with a polynomial of degree less than n and so does multiplying by a scalar. I claim that this is all we need to check. Since polynomials are continuous, this set is a subset of C[a, b], which we know is already a vector space. Notice that the axioms for a vector space have nothing to do with the closure properties and only rely on the vectors to lie in the set. For example, if p, q ∈ Pn[a, b], then we also have p, q ∈ C[a, b] and since we already know C[a, b] is a vector space, we can apply A1 for C[a, b] to obtain that p ⊕ q = q ⊕ p. All axioms are shown in this way. This leads us to the following definition.

Definition. A subset S of a vector space V is a subspace of V if S is nonempty and it satisfies the closure properties (i) αx ∈ S whenever x ∈ S and α is a scalar (ii) x + y ∈ S whenever x, y ∈ S, ALWAYS check that S is nonempty first. It’s possible for the closure properties to hold, but for S to be empty! Actually, if we suspect that S is a subspace of V , then if the closure properties hold, the same zero vector for V must for S, so you can always just check that the zero vector is S, which I did above for Pn[a, b].

Think about it. What happens if we require that the polynomials must be of degree exactly n? Solution. This is too restrictive because adding two polynomials of degree exactly n might give a polynomial with degree less than n. For example, in p(x) = x2 and q(x) = −x2 are two polynomials of degree 2, however, (p ⊕ q)(x) = p(x) + q(x) = x2 − x2 = 0. That is, p ⊕ q is the constant zero function, which is not exactly of degree 2, but instead degree zero. This is why we must make the relaxation to polynomials less than a positive integer and not exactly it. M309 Notes ©, R.G. Lynch, Texas A&M Section 3.1: Definition and Examples (Vector Spaces), Completed Page 7 of 9

Example (Shifted Reals).1 Let S = {x : x ∈ R} be the set of real numbers with addition defined by x ⊕ y = x + y + 7 and scalar multiplication defined by α ◦ x = α · x + 7(α − 1). Is S a vector space? Solution. I’ve decided to drop the bold letters because it makes it more apparent that these are still real numbers, just with different operations defined. Using the different notation ⊕ and ◦ will hopefully make it clear as to when I mean vector addition and scalar multiplication S versus just the usual real number operations. Now to see why this space is weird, note that 3 ⊕ 8 = 3 + 8 + 7 = 18 which is different than 11! This hints that the zero vector in S will be different than 0. Indeed, 3 ⊕ 0 = 3 + 0 + 7 = 10 6= 3. Multiplying a vector by a scalar is also weird: 5 ◦ (−2) = 5 · (−2) + 7(5 − 1) = −10 + 28 = 18 6= −10. Nevertheless, S is still a vector space under these operations as we now check. Showing closure: Even though the operations are unusual, we still obtain a real number as the output and so S is closed under these operations. Careful though, the outputs are real numbers, but they are also vectors in S.

Showing the axioms: be very careful of which addition and multiplication you’re doing! Let α, β ∈ R and x, y, z ∈ S. A1. This is obvious as x ⊕ y = x + y + 7 = y + x + 7 = y ⊕ x. A2. On one hand we have (x ⊕ y) ⊕ z = (x + y + 7) ⊕ z = (x + y + 7) + z + 7 = x + y + z + 14 and on the other x ⊕ (y ⊕ z) = x ⊕ (y + z + 7) = x + (y + z + 7) + 7 = x + y + z + 14 from which we see that (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z). A3. We need to find a c ∈ S so that x ⊕ c = x. We already saw from above that c 6= 0! So, what is it? Let’s find out: x ⊕ c = x ⇐⇒ x + c + 7 = x from which we see that we must take c = −7. So that −7 is the zero vector! A4. Here we see that, for example, −5 as a vector is not equal to −5 as a number! Really weird!! For each real number x ∈ S, we need to find an n ∈ S so that x ⊕ n = c. We have, using that c = −7 from A3, x ⊕ n = c ⇐⇒ x + n + 7 = −7 from which we see that we must take n = −x − 14. That is, −x as a vector in S is actually the vector −x − 14!

1Dan Kalman’s Notes, Example 2. M309 Notes ©, R.G. Lynch, Texas A&M Section 3.1: Definition and Examples (Vector Spaces), Completed Page 8 of 9

A5. On one hand we have α ◦ (x ⊕ y) = α ◦ (x + y + 7) = α(x + y + 7) + 7(α − 1) = αx + αy + 14α − 7 and on the other (α ◦ x) ⊕ (α ◦ y) = αx + 7(α − 1) ⊕ αy + 7(α − 1) = αx + 7(α − 1) + αy + 7(α − 1) + 7 = αx + αy + 14α − 7 from which we see that α ◦ (x ⊕ y) = (α ◦ x) ⊕ (α ◦ y). A6. On one hand we have (α + β) ◦ x = (α + β)x + 7(α + β) − 1 = αx + βx + 7α + 7β − 7 and on the other (α ◦ x) ⊕ (β ◦ x) = αx + 7(α − 1) ⊕ βx + 7(β − 1) = αx + 7(α − 1) + βx + 7(β − 1) + 7 = αx + βx + 7α + 7β − 7 from which we see that (α + β) ◦ x = (α ◦ x) ⊕ (β ◦ x). A7. On one hand we have (αβ) ◦ x = (αβ)x + 7((αβ) − 1) = αβx + 7αβ − 7 and on the other α ◦ (β ◦ x) = α ◦ (βx + 7(β − 1)) = α(βx + 7(β − 1)) + 7(α − 1) = α(βx + 7β − 7) + 7α − 7 = αβx + 7αβ − 7 from which we see that (αβ) ◦ x = α ◦ (β ◦ x). A8. Lastly, again, 1 is the unit number of the underlying scalar field. Thus, it is just the usual 1 that we all know and love in the real numbers. Checking: 1 ◦ x = 1 · x + 7(1 − 1) = x. M309 Notes ©, R.G. Lynch, Texas A&M Section 3.1: Definition and Examples (Vector Spaces), Completed Page 9 of 9

2 2 Example (R Shifted in a Single Coordinate). Let S be the set S = {x = (x1, x2): x1, x2 ∈ R} scalar multiplication defined as usual, but with addition defined by

x ⊕ y = (x1 + y1, x2 + y2 + 1). Is S a vector space? Solution. We will see that this is not a vector space. The closure properties are fine as you can check. Note that the 1 is a consequence of how we define ⊕ and not how S is defined, so we don’t run into the issue that we did with the points lying on the horizontal line y = 1. Actually, all of the axioms hold except for A5 and A6. They’re straightforward to verify, except for maybe A3 and A4 where you must realize that the zero vector is c = (0, −1). Let’s see why A5 fails. Note that 7((1, 0) ⊕ (1, 2)) = 7(1 + 1, 0 + 2 + 1) = 7(2, 3) = (14, 21), but (7(1, 0)) ⊕ (7(1, 2)) = (7, 0) ⊕ (7, 14) = (7 + 7, 0 + 14 + 1) = (14, 15) 6= (14, 21). You can actually check for any choice of x, y ∈ S that as long as α 6= 1, this axiom will fail. Also, even though it is now unnecessary because we already showed A5 fails, you can show A6 fails in a very similar fashion. I leave this for you to check yourself. Note that once you show one thing fails, whether it’s one of the closure properties (always check these first) or an axiom, for a specific choice of α, β or x, y, z ∈ S, then you’re done and can conclude that S is not a vector space.

2Thanks to Tom Vogel, also in the Texas A&M Math Department.