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Lecture 11
Andrei Antonenko
February 26, 2003
1 Examples of bases
Last time we studied bases of vector spaces. Today we’re going to give some examples of bases.
Example 1.1. Consider the vector space P 2 — the space of polynomials with degree less than or equal to 2. Let’s consider the following 3 vectors in this vector space:
u 1 = t^2 + 1, u 2 = t + 1, u 3 = t − 1.
Let’s determine whether it is a basis or not. We have to check 2 conditions:
Spanning set To check that these vectors form a spanning set for P 2 we should take arbitrary vector from P 2 and try to express it as a linear combination of the vectors from the basis. Let’s take arbitrary polynomial at^2 + bt + c:
at^2 + bt + c = x(t^2 + 1) + y(t + 1) + z(t − 1) = xt^2 + (y + z)t + (x + y − z).
So, we can see that this is equivalent to the following system of linear equations, which we will try to solve:
x = a y + z = b x + y − z = c
subtract the 1st eq. from the 3rd one √
x = a y + z = b y − z = c − a
subtract the 2nd eq. from the 3rd one √
x = a y + z = b − 2 z = c − a − b
So, we see that z = 12 (a + b − c), y = 12 (b + c − a), and x = a. So, we got the expression for arbitrary polynomial as a linear combination of given:
at^2 + bt + c = a(t^2 + 1) +^12 (b + c − a)(t + 1) +^12 (a + b − c)(t − 1).
So, this system is a spanning set.
Linear independence To check that these vectors are linearly independent we form a linear combination which is equal to 0:
x(t^2 + 1) + y(t + 1) + z(t − 1) = 0 ⇔ xt^2 + (y + z)t + (x + y − z) = 0.
This is equivalent to the following linear system:
x = 0 y + z = 0 x + y − z = 0
subtract the 1st eq. from the 3rd one √
x = 0 y + z = 0 y − z = 0
subtract the 2nd eq. from the 3rd one √
x = 0 y + z = 0 − 2 z = 0
So, we see that the only solution for this system is x = 0, y = 0, and z = 0. Thus, these vectors are linearly independent.
So, since both properties hold for this system of vectors, we deduce that this system is a basis.
Example 1.2. Consider the vector space P 2 — the space of polynomials with degree less than or equal to 2. Let’s consider the following 3 vectors in this vector space:
u 1 = t^2 + t + 2, u 2 = t^2 + 1, u 3 = t + 1.
Let’s determine whether it is a basis or not. We have to check 2 conditions:
Spanning set To check that these vectors form a spanning set for P 2 we should take arbitrary vector from P 2 and try to express it as a linear combination of the vectors from the basis. Let’s take arbitrary polynomial at^2 + bt + c:
at^2 + bt + c = x(t^2 + t + 2) + y(t^2 + 1) + z(t + 1) = (x + y)t^2 + (x + z)t + (2x + y + z).
Thus these vectors are linearly dependent.
2 Dimension
Now we’ll state the following theorem about linear dependence.
Theorem 2.1 (Main lemma about linear dependence). Let u 1 , u 2 ,... , un is a basis for vector space V. Let m > n. Then any m vectors from V are linearly dependent.
Example 2.2. Vectors
and
form a basis for R^2. So, any 3 vectors from R^2 are
linearly dependent. For example we can say that
v 1 =
, v 2 =
, v 3 =
are linearly dependent without finding a nontrivial linear combination of them.
The following corollary is one of the main results in linear algebra.
Corollary 2.3. All bases of the given vector space V have the same number of vectors.
Definition 2.4. The number of vectors in basis of V is called the dimension of V. It is denoted by dim V.
Example 2.5. The space R^2 has 2 vectors in its basis, so dim R^2 = 2.
Example 2.6. The space P 2 of polynomials of degree less than or equal to 2 has dimension equal to 3, since it has a basis of 3 vectors: u 1 (t) = t^2 , u 2 (t) = t, and u 3 (t) = 1. So, dim P 2 = 3.
Now we are ready to give the proofs of these main results.
Proof of the Main Lemma about linear dependence. Let we have m vectors in the V : v 1 , v 2 ,
... , vm, and m > n, where n is dimension of V. Vectors u 1 , u 2 ,... , un form a basis for V , so we can express vectors vi’s as linear combinations of ui’s:
v 1 = a 11 u 1 + a 12 u 2 + · · · + a 1 nun v 2 = a 21 u 1 + a 22 u 2 + · · · + a 2 nun
... vm = am 1 u 1 + am 2 u 2 + · · · + amnun
Let’s form a linear combination of vi’s which is equal to zero, and prove that it may be non- trivial — then it will be proved that vi’s are linearly dependent.
λ 1 v 1 + λ 2 v 2 + · · · + λmvm = λ 1 (a 11 u 1 + a 12 u 2 + · · · + a 1 nun)
- λ 2 (a 21 u 1 + a 22 u 2 + · · · + a 2 nun)
- · · ·
- λm(am 1 u 1 + am 2 u 2 + · · · + amnun)
By rearranging terms, we write that the same linear combination is equal to
λ 1 v 1 + λ 2 v 2 + · · · + λmvm = u 1 (λ 1 a 11 + λ 2 a 21 + · · · + λmam 1 )
- u 2 (λ 1 a 12 + λ 2 a 22 + · · · + λmam 2 )
- · · ·
- un(λ 1 a 1 n + λ 2 a 2 n + · · · + λmamn)
In order for it to be equal to 0 , we will write that coefficients are equal to 0 (since ui’s are independent). We’ll have the system of linear equations:
λ 1 a 11 + λ 2 a 21 + · · · + λmam 1 = 0 λ 1 a 12 + λ 2 a 22 + · · · + λmam 2 = 0
........................................... λ 1 a 1 n + λ 2 a 2 n + · · · + λmamn = 0
This is a homogeneous system, and the number of equations is n, the number of variables is m, so the number of equations is less then the number of variables (since n < m). So, it has non-trivial solution — there exist λi’s not all equal to 0, such that linear combination of vi’s is equal to 0. So, vi’s are linearly dependent.
Proof of the Corollary 2.3. Let we have 2 bases with different numbers of vectors, say m in the first basis, and n in the second one. Let m > n. But by the previous theorem any m vectors are linearly dependent. But they are in basis, so they should be independent! Contradiction proves the corollary.
