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Lecture 13
Andrei Antonenko
March 3, 2003
1 Dimension and basis of the span
Last lecture we formulated the problem of finding the basis and the dimension of the span of given vectors. This lecture we will give the algorithm to determine these characteristics of the span.
1.1 Algorithm
Step 1. Write the given vectors as a rows of a matrix.
Step 2. Reduce this matrix to REF keeping track which row in the matrix corresponds to which vector, i.e. initially, i-th row corresponds to ui – so, label i-th row as i, and after Type 1 elementary operation we interchange row labels.
Step 3. The number of nonzero rows is a dimension of a span. The labels of nonzero rows are subscripts of vectors in basis. Moreover, nonzero rows form a basis for a span as well.
Example 1.1. Consider the following 4 vectors in R^4 : u 1 = (2, 3 , 1 , 1), u 2 = (1, 1 , 0 , −1), u 3 = (3, 4 , 1 , 0), and u 4 = (1, 2 , 1 , 3). Let’s find the basis and a dimension of their span. To do
this we’ll form a matrix with labels and reduce it to REF.
1 2 3 4
^ interchange 1st and 2nd √
^ subtract 1st from others √
^ subtract 2nd from others^ √
^ interchange 3rd and 4th^ √
So, the number of nonzero rows is 3, so, the dimension of a span is 3. Moreover, we see that vectors u 1 , u 2 and u 4 form a basis of the span, and moreover vectors (1, 1 , 0 , −1), (0, 1 , 1 , 3) and (0, 0 , 0 , 1) form another basis of a span.
Using this last basis formed by nonzero rows of a matrix in REF we can determine whether a given vector belongs to the span or not.
Example 1.2. Consider a vector (1, 2 , 3 , 4) and let’s find whether this vector belongs to the span from the previous example. We have a basis with 3 vectors: (1, 1 , 0 , −1), (0, 1 , 1 , 3) and (0, 0 , 0 , 1). Let’s try to express our given vector as a linear combination of these 3 basis vectors:
(1, 2 , 3 , 4) = a(1, 1 , 0 , −1) + b(0, 1 , 1 , 3) + c(0, 0 , 0 , 1) = (a, a + b, b, −a + 3b + c).
So, a = 1, then a + b = 2, so b = 2. Now when we’re considering the third component, we see that b = 3. So, we got a contradiction. So, vector (1, 2 , 3 , 4) cannot be expressed as a linear combination of 3 vectors from the basis, and so it is not in basis. Now consider vector (5, 7 , 2 , 0). Let’s try to express our given vector as a linear combination of 3 basis vectors:
(5, 7 , 2 , 0) = a(1, 1 , 0 , −1) + b(0, 1 , 1 , 3) + c(0, 0 , 0 , 1) = (a, a + b, b, −a + 3b + c).
So, a = 5, then a + b = 7, so b = 2. Now when we’re considering the third component, we see that b = 2 – that’s ok, no contradiction, and then −a + 3b + c = 0 ⇔ −5 + 6 + c = 0 ⇔ c = − 1. So, (5, 7 , 2 , 0) = 5(1, 1 , 0 , −1) + 2(0, 1 , 1 , 3) − (0, 0 , 0 , 1),
and thus this vector belongs to the span of initial vectors.
of nonzero rows was equal to the dimension of a span of these vectors. So, we see that for any given matrix
A =
a 11 a 12... a 1 n a 21 a 22... a 2 n
................... am 1 am 2... amn
its rank rk A = dim span(a 1 , a 2 ,... , am),
where
a 1 = (a 11 , a 12 ,... , a 1 n); a 2 = (a 21 , a 22 ,... , a 2 n);
... ; am = (am 1 , am 2 ,... , amn).
