Suppose that V is a vector space over C, and that U is a subset of V. If U is a vector space over C using the same definition of addition and scalar multiplication as used in V, then U is called a subspace of V.
Now, as we found in our studies of the real numbers, any subset (but not necessarily a vector space) W of a vector space V will automatically satisfy properties V2, V5, V7, V8, V9, and V10. So if we want to prove that W is itself a vector space, we only need to look at properties V1,
V3, V4, and V6. Now, we can easily use the same proof as we used in the real numbers to show that 0v = 0 vector and that (-1)v will equal the inverse of V in our complex numbers as well. So this means that properties V3 and V4 will follow from property V6, the closure under scalar multiplication property. And so, just as before, we get the following alternate definition of a subspace.
Suppose that V is a vector space over C. Then U is a subspace of V if it satisfies the following three properties.
S0) U is a non-empty subset of V.
S1) That w + z is an element of U for all w and z in U—that is, that U is closed under addition.
S2) (alpha)z is in U for all z in U and all alpha in C—that is, that U is closed under scalar multiplication.
So let’s see how this really works. Let’s show that the set U equal to the set of all vectors [z; 2z] where z is a complex number is a subspace of C-squared. We need to verify the three defining properties. S0: Well, we note that [z; 2z] is, in fact, an element of C2 for all z in C, so U is a subset of C2. To see that it is non-empty, we’ll note that 2(0) = 0, so the 0 vector is an element of U.
S1: Let’s let w and z be elements of U. Well, then we must have that w =[ w; 2w] and that z =
[z; 2z] for some complex numbers w and z. Well, then we have that w + z = the vector [w; 2w] + the vector [z; 2z], which will equal the vector [w + z; 2w + 2z], which equals the vector [w + z; 2-quantity-(w + z)]. And since w + z is a complex number, we see now that the vector sum w + z satisfies the definition of being an element of U.
S2: Let’s let z be an element of U, so that z equals the vector [z; 2z], and we’ll let alpha be a complex number. Then (alpha)z will equal alpha times the vector [z; 2z]. Well, this equals [(alpha)z; (alpha)(2z)], but since complex multiplication is commutative, we can rewrite this as [(alpha)z; 2((alpha)z)], and since, of course, (alpha)z is a complex number, we see that alpha times the vector z satisfies the definition of being in U. And since S0, S1, and S2 all hold, U is a subspace of C2.
Let’s look at another example. Let’s let C(2, 2) be the set of all 2-by-2 matrices with entries from the complex numbers, and we’ll let the set A be the set of all matrices [z1, z2; z1, z2] where z1 and z2 are in the complex numbers. Then A is a subspace of C(2, 2). Again, to prove this, we need to check the three properties.
S0: We first note that the matrix [z1, z2; z1, z2] is an element of C(2, 2) for any complex numbers z1 and z2, and so we have defined A as a subset of C(2, 2). To see that A is non-empty, we can simply set z1 and z2 equal to 0 and see that the 0 matrix is an element of A.
S1: So let’s let A and B be two matrices from A. We’ll say that A = [a1, a2; a1, a2] and B = [b1, b2; b1, b2]. Well, then we have that the sum A + B will be the matrix [a1 + b1, a2 + b2; a1 + b1, a2 + b2]. And since our a1 + b1 and our a2 + b2 are both complex numbers, we see that the matrix sum A + B satisfies the definition of being an element of A.
S2: Let’s let A be an element of A, so we’ll say that A is the matrix [a1, a2; a1, a2], and we’ll let alpha be an element of C. Then (alpha)A = [(alpha)a1, (alpha)a2; (alpha)a1, (alpha)a2], and since (alpha)a1 and (alpha)a2 are both complex numbers, we see that (alpha)A satisfies the definition of being a member of A. And since properties S0, S1, and S2 all hold, this means that A is a subspace of C(2, 2).
Now let’s look at an example of something that would not be a subspace. So we’ll define the set W equal to vectors of the form [z; z-squared] where z is a complex number. To show that it is not a subspace of C2, consider that the vector [i; -1] is an element of W, and [-i; -1] is also an element of W, but if we were to sum these vectors, we get the vector [0; -2], but [0; -2] does not satisfy the definition of being in W. So as such, our property S1 (closure under addition) fails, and this means that W is not a subspace.