Let V be a vector space over the body K, In order to have something concrete in mind, imagine K to be the body of all real and complex numbers or also the body of all real numbers. When we speak in connection with V of numbers, we will always mean numbers out of K.
You say that the l vectors a1, ··· , al of K are linearly dependent when there exist l numbers a1,···,al, not all of which vanish, such that
a1a1 + ··· + anan = 0.*
* Obviously, this equation can always be satisfied in a trivial manner with a1 = ··· = al = 0. However, the essential aspect of linear dependence is that you are not dealing with this trivial condition, but that at least one of the numbers a1,···,an is non-zero.
In contrast, m vectors b1, ··· , bm of V are linearly independent when it follows from each equation
b1b1 + ··· + bmbm = 0,
that all of the numbers b1, ··· , bm are zero.
The vectors e1, ··· , en of any base of the vector space V are always linearly independent, because, according to the basic property (19.9) of every base, every vector of V can be expressed in one and only one way as the sum of the products of the base vectors multiplied by certain numbers. As a consequence,
0·e1 + ··· + 0·en = 0
is such a presentation for 0, and indeed the only one, that is, it follows from
b1e1 + ··· + bnen = 0
that b1 = ··· = bn = 0.
For example, in the vector space of all two-rows, square, real matrices
are linearly dependent, because 2A + B - C = 0.
If the vectors a1,···,al include the null-vector, they are certainly linearly dependent, for if, for example, a1 = 0, the relation
1·a1 + 0·a2 + ··· + 0·al = 0
is one of the kind required, in which not all coefficients vanish. If already a partial set of the vectors a1,···,al is linearly dependent, then all of them are linearly dependent, because, say, a1,···,ak are linearly dependent, then you have a relation
a1a1 + ··· + alal = 0,
in which not all of the a1,···,ak vanish, so that
a1a1 + ··· + akak + 0·ak+1+ ··· + 0·al = 0,
that is, a1,···,al are linearly dependent.
You say that the vector a depends linearly on the vectors c1,···,cr, if it can be expressed as a sum of the vectors c1,···,cr with suitable coefficients, that is, if
a = g1c1 + ··· + grcr.
If A depends linearly on c1,···,cr , then a, c1,···,cr together are linearly dependent, because in the relation
(-1)·a + g1c1 + ··· + grcr= 0
not all the coefficients on the left hand side vanish.
The vector space V has a base consisting of n vectors
e1, ··· , en . (20.1)
The number n is the dimension of V. As has already been point out above, the base vectors are linearly independent. We will now prove that there does not exist in V a system of more than n linearly independent vectors. We only have to show that any n + 1 vectors a0, a1,···,an of V are linearly dependent. Certainly, this is so when already a1,···,an are linearly dependent, whence we need only consider the case that a1,···,an are linearly independent. In particular, none of these vectors is the null-vector. Since e1, ··· , en form a base, there exists one and only one representation
a1 = a1c1 + ··· + ancn. (20.2)
Since a1
0,
at least one of the numbers a1,···,an is non-zero.
Generality will not be restricted by assuming that a1 0,
as if necessary the ei
can be renumbered, which obviously is without
consequence for the property of the base. Hence (20.2) yields an
equation of the form
e1 = b1a1 + b2e2 + ··· + bnen. (20.3)
We can now conclude that every vector of V can be represented as a sum of a1, e1, ··· , en, for , first of all, every vector of V can be represented as a sum of the base vectors (20.1), in which one replaces e1 by the right hand side of (20.3) and thus obtains a sum of a1, e2, ··· , en. In particular, you can represent a2 by
a2 = g1a1 + g2e2 + ··· + gnen. (20.4)
Since a1,···,an
together are linearly independent, at least one of the numbers g2,···,gn must be non-zero.
Again generality is not affected by setting g2 0. Thus (20.4) yields
e2 = d1a1 + d2e2 + ··· + dnen. (20.5)
You conclude from this equation that every vector of V can be represented as a linear combination of a1, a2, e3,.··· , en, because, first of all, , as we have seen already, there exists for every vector of V a representation by a sum of a1,e2,···,en; replace now in this representation e2 by the right hand side of (20.5) and obtain thus a sum of a1, a2, e3,.··· , en. This argument can be continued and arrives after n steps at the result that every vector of V can be represented as a sum of the vectors a1, a2, ··· , an. In particular, there exists such a representation for a0:
a0 = r1a1 + ··· + rnan.
Hence a0, a1, ··· , an. are linearly dependent.
Moreover, every base of V
consists of n vectors. In any case, there exists no base
consisting of more than n vectors; as we have just seen,
more than n vectors are always linearly dependent, while
the vectors of a base must be linearly independent. On the other
hand, a base can not comprise less than n vectors, for
if there existed a base of m vectors with m
< n, we could conclude, as above, that all m +
1 vectors of V are linearly dependent, which due
to m + 1 
n contradicts the fact that the n
vectors (20.1) are linearly independent.
The last proved statement that every base consists of n linearly independent vectors has also an inverse: Every system f1, ··· , fn of linearly independent vectors of V forms a base of V. In fact, is a is an arbitrary vector, then the n+1 vectors a, f1, ··· , fn are, as has been shown above, linearly dependent, whence there exists the non-trivial relation
la + l1f1 + ··· + lnfn. (20.6)
Since the f1, ···
, fn are linearly
independent, you must have l 0,
whence
a = m1f1 + ··· + mnfn. (mi = - li/l). (20.7)
Consequently, every vector of V can be represented as a linear sum of the vectors f1, ··· , fn. In order to be sure that the f1, ··· , fn form a base , you still must make sure that the representation (20.7) of a is unique. If
a = m'1f1 + ··· + m'nfn
is any representation of a, you obtain by subtraction from (20.7)
0 = (m1- m')f1 + ··· + (mn - m'n)fn.
From the linear independence of f1, ··· , fn follows that mi - m'i = 0 for i=1,···,n, that is, apart from (20.7), that there does not exist another representation.
We summarize the last results in
There corresponds to every vector space a natural number n - its dimension. In V, there exists at least one base of n vectors. Every base of V consists of n vectors and every system of n linearly independent vectors of V forms a base. Any n + 1 vectors of V are linearly dependent.
It follows from the fact that any n +1 vectors of V are linearly dependent that every system of more vectors is linearly dependent, because every such system contains partial systems of n + 1 vectors; from the linear dependence of such a partial system follows, according to an earlier comment, the linear dependence of the entire system.
The following information proves frequently to be useful: Let there be given in a vector space V of dimension n a system of r linearly independent vectors v1,···,vr in terms of which every vector can be expressed. Then r = n and v1,···,vr form a base of V, because the uniqueness of the representation of a vector as a linear combination of the v1,···,vr follows from their linear independence, so that the v1,···,vr form a base of V and, by Theorem 8, r must be equal to n.