Linear Algebra, Theory And Applications, 2012a

8.4. EXERCISES 219
8.3.4 The Lindemannn Weierstrass Theorem And Vector Spaces
As another application of the abstract concept of vector spaces, there is an amazing theorem
due to Weierstrass and Lindemannn.
Theorem 8.3.34 Suppose a 1 , ··· ,a n are algebraic numbers and suppose α 1 , ··· ,α n are
distinct algebraic numbers. Then
n∑
a i e αi
≠0
i=1
In other words, the {e α1 , ··· ,e αn } are independent as vectors with field of scalars equal to
the algebraic numbers.
There is a proof of this in the appendix. It is long and hard but only depends on
elementary considerations other than some algebra involving symmetric polynomials. See
Theorem F.3.5.
A number is transcendental if it is not a root of a polynomial which has integer coefficients.
Most numbers are this way but it is hard to verify that specific numbers are
transcendental. That π is transcendental follows from
e 0 + e iπ =0.
By the above theorem, this could not happen if π were algebraic because then iπ would also
be algebraic. Recall these algebraic numbers form a field and i is clearly algebraic, being
arootofx 2 +1. This fact about π was first proved by Lindemannn in 1882 and then the
general theorem above was proved by Weierstrass in 1885. This fact that π is transcendental
solved an old problem called squaring the circle which was to construct a square with the
same area as a circle using a straight edge and compass. It can be shown that the fact π is
transcendental implies this problem is impossible. 1
8.4 Exercises
⎛⎛
1. Let H denote span ⎝⎝ 1 2
0
and determine a basis.
⎞ ⎛
⎠ , ⎝ 1 ⎞ ⎛
4 ⎠ , ⎝ 1 ⎞ ⎛
3 ⎠ , ⎝ 0 ⎞⎞
1 ⎠⎠ . Find the dimension of H
0 1 1
2. Let M = { u =(u 1 ,u 2 ,u 3 ,u 4 ) ∈ R 4 : u 3 = u 1 =0 } . Is M a subspace? Explain.
3. Let M = { u =(u 1 ,u 2 ,u 3 ,u 4 ) ∈ R 4 : u 3 ≥ u 1
}
. Is M a subspace? Explain.
4. Let w ∈ R 4 and let M = { u =(u 1 ,u 2 ,u 3 ,u 4 ) ∈ R 4 : w · u =0 } . Is M a subspace?
Explain.
5. Let M = { u =(u 1 ,u 2 ,u 3 ,u 4 ) ∈ R 4 : u i ≥ 0foreachi =1, 2, 3, 4 } . Is M a subspace?
Explain.
6. Let w, w 1 be given vectors in R 4 and define
M = { u =(u 1 ,u 2 ,u 3 ,u 4 ) ∈ R 4 : w · u = 0 and w 1 · u =0 } .
Is M a subspace? Explain.
1 Gilbert, the librettist of the Savoy operas, may have heard about this great achievement. In Princess
Ida which opened in 1884 he has the following lines. “As for fashion they forswear it, so the say - so they
say; and the circle - they will square it some fine day some fine day.” Of course it had been proved impossible
to do this a couple of years before.