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Vector Spaces), Completed Section 3.1: Definition and Examples (Vector Spaces), Completed 1. Examples n Euclidean Vector Spaces: The set of n-length vectors that we denoted by R is a vector space. For simplicity, let's consider n = 2. T 2 A vector x = x1 x2 in R can be represented by a directed line segment from the origin to the point (x1; x2) or from any point (a; b) (same magnitude and direction). We scale vectors by any scalar 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 addition/subtraction and scalar multiplication. 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 scalar multiplication 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 2 V and α is a scalar, then αx 2 V . C2. If x; y 2 V , then x + y 2 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 additive inverse 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 operation 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 := f(x; 1) : x 2 Rg, 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) 2= W . Since it fails the closure properties, U is not a vector space. Example. What about V := f(x; 0) : x 2 Rg 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) 2 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 := f(x; mx): x 2 Rg with the same operations? Solution. Here we assume that m is any real number. 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 2 R; x = (x; mx) 2 W implies that: αx = α(x; mx) = (αx; αmx) = (αx; m(αx)) 2 W C2. x(x; mx); y = (y; my) 2 W implies that: x + y = (x; mx) + (y; my) = (x + y; mx + my) = x + y; m(x + y) 2 W Showing the axioms: Let α; β 2 R and x = (x; mx); y = (y; mx); z = (z; mz) 2 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) 2 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] := ff :[a; b] ! R : f is continuousg 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.
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