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\(Author : \ Thomas\ Shores\), "\[IndentingNewLine]",
\(University\ of\ Nebraska\), "\[IndentingNewLine]",
\(Send\ comments\ \(to : \ tshores@math . unl . edu\)\)}], "Input",
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Let us recall the definition of an (abstract) vector space:
Definition: A VECTOR SPACE (v.s.) over a field of scalars is a set V of \
elements called vectors, together with two binary operations, scalar \
multiplication * (usually suppressed) and vector addition + , subject to the \
following vector space laws: For all u, v, w \[Element] V and scalars \
a and b,
(1) (Closure of vector additon) u + v \[Element] V.
(2) (Commutativity of addition) u + v = v + u.
(3) (Associativity of addition) u + (v + w) = (u + v) + w.
(4) (Additivive Identity) There exists an element 0 \[Element] V such that \
u + 0 = u = 0 + u.
(5) (Additive Inverse) There exists, for each u \[Element] V, an element -u \
\[Element] in V such that
\tu + (-u) = 0 = (-u) + u.
(6) (Closure of scalar multiplication) a*v \[Element] V.
(7) (Distributive Law) a*(u+v) = a*u + a*v.
(8) (Distributive Law) (a + b)*u = a*u + b*u.
(9) (Associative Law) (a b) * u = a*(b*u).
(10) (Monoidal Law) 1*u = u.
NOTATION: u + (-v) = u - v.
ALSO: we often refer to the \[OpenCurlyDoubleQuote]vector space V\
\[CloseCurlyDoubleQuote], without identifying explicitly the binary \
operations. They are taken for granted.
\t
SIMPLE FACTS: (deducible from the vector space properties alone):
\tFor all vectors u, 0*u = 0, where the latter 0 means the zero vector.
\tFor all vectors u, (-1)*u = -u.\
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EXAMPLES OF VECTOR SPACES:
(1) The spaces Rn and Cn, n >=1, over the fields R and C, \
respectively. Also, C is a vector space over R.
(2) The set M(2,2) of all 2x2 real matrices, together with the usual \
matrix addition and scalar multiplication.
(3) The set
\tV = C[0,1] = {f(x) | f(x) is a continuous function on the interval [0,1] }, \
together with the usual function addition and scalar multiplication:
\t\t(f+g)(x) = f(x) + g(x)
\t\t(a*f)(x) = a f(x).
\tThis is a very non-standard vector space. It seems kind of odd, but we are \
really thinking of functions f(x), 0 <= x <= 1, as \
\[OpenCurlyDoubleQuote]vectors\[CloseCurlyDoubleQuote]. Of course, one has \
to check that all the laws are satisfied. This is routine: they are. What \
plays the part of the zero vector is the zero constant function.
\t
\tNow, what is a subspace of a given vector space V?
DEFINITION: A subspace of the vector space V is a subset W of V such \
that W, together with the binary operations it inherits from V , forms a \
vector space (over the same field of scalars) in its own right.
EXAMPLES OF SUBSPACES:
\t(1) Let V = R3 and W = {[x,y,x] | x, y \:02db R}.
\t(2) Let V = M(2,2) and W = {{{a,b},{0,a}} | a , b \:02db R }
\t(3) Let V = C[0,1] and W = P_2 = {f(x) = a + b x + c x2 | a, \
b, c in R}
In every case we have to verify that laws (1) - (10) are valid for each \
subset W of the v.s. V. This is unpleasant!! Do we have to do all this \
work?? No, and here is why:
SUBSPACE TEST: Let W be a nonempty subset of the v.s. V. Then W is \
a subspace of V if and only if
\t(1) For all u, v \:02db W, u + v \:02db W (this is really closure of \
addition on elements of W).
\t(2) For all u \:02db W and scalars a, a*u \:02db W (this is really \
closure of scalar multiplication on elements of W).
\t
\tIt isn't too hard to see what the Subspace Test works: certainly, any \
subspace W of V must satisfy these conditions since it is a vector space \
in its own right with the operations inherited from V. Conversely, if the \
nonempty subset W of V satisfies these conditions, then vector space \
properties (1) and (6) follow. Properties (2)-(3) and (7)-(10) follow \
simply because elements of W are elements of V and they hold for elements of \
V. The only properties to be checked are (4) and (5). Since W is \
nonempty, there is an element v \:02db W. Now one of the Simple Facts above \
is that (-1)v = -v, the additive inverse of v as an element of V. By \
condition (2) of the Subspace Test, (-1)v \:02db W. Therefore -v \:02db W. \
By condition (1) of the Subspace Test, v + (-v) = 0 (the zero vector of \
the vector space V) also belongs to W. It follows that W contains an \
additive identity, namely, the very same one as V has, and that every \
element of W has an additive inverse in W. This disposes of vector space \
laws (4) and (5) and shows that W is a vector space in its own right, hence \
a subspace of V.
\t
APPLICATION: A very common occurrence is a linear combination (l.c.) of \
vectors v1, v2, ... , vn, by which we mean an expression of the form, for \
suitable scalars a1, a2, ...an,
\t
\t\ta1 v1 + a2 v2 + ... + an vn.
\t\t
The set of all possible linear combinations of the vectors v1, v2, ... ,vn \
has a name: the linear span of v1, v2, ..., vn.
DEFINITION: The linear span of vectors v1, v2, ...., vn is the set of all \
vectors
\tspan{v1, v2, ..., vn} = { a1 v1 + a2 v2 + ... + an vn | a1, a2, ..., an \
are scalars}.
\t
Note that the span depends on the field of scalars in use. Here is the \
application we referred to:
FACT: If v1, v2, ..., vn are vector in the v.s. V then lin{v1, v2, ..., \
vn} is a subspace of V.
This is a fact that you should be able to prove. It is also in your text.
ANOTHER IDEA: if V is a vector space such that V = lin{v1, v2, ..., vn}, \
then the vectors v1, v2, ..., vn are said to SPAN the vector space V.
EXAMPLE: {1,0,0}, {0,1,0}, and {0,0,1} span the vector space R3. As \
a matter of fact, so does the set of vectors {1,0,0}, {0, 1,0},{0,0,1}, \
{1,1,2}. Why do you think that one would prefer the first spanning set?
DEFINITION: A set of vectors v1, v2, ... , vn is a BASIS of the vector \
space V if this set spans V and no subset of v1, v2, ... vn spans \
V.\
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Suppose we are given a matrix A. For the sake of definiteness, we \
will work with the following matrix A:\
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"Here are three very important vector spaces that are associated with the \
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" A.v = 0} \n(Note: replace R by C in the case of complex scalars in all \
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C(A) are definitely vector spaces. As a matter of fact, R(A) is a subspace \
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". (How come?) \n\nEXAMPLE: Describe the row space, column space and null \
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Where do these subspaces live? For the null space, by the way, it might \
help to click on the cells after the definition of"
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Now let's take a look at a harder example, namely, the matrix A \
we defined above. How would you define the row space and column space of A? \
Write this out on a separate piece of paper. It might help to use the \
following comman\
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" Now for another idea about spanning sets. Remember the examples of the \
last section (look them up). There was something undesirable about one \
spanning set: it had unnecessarily many vectors in the list. A spanning set \
which does not have too many vectors is called a basis of the vector space \
it spans. Thus:\n\nDEFINITION: A ",
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Find the row space, the column space, and the null space of the \
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