hwapril12

Homework due Class 37

1. Let {v₁,...,v_n} be a basis of a vector space V. Let s₁,...,s_m be vectors of V and we have that

s₁ = a₁₁ v₁ +...+ a_1n v_n s₂ = a₂₁ v₁ +...+ a_2n v_n ........................ s_m = a_m1 v₁ +...+ a_mn v_n
for some numbers a_ij. Consider the following n-vectors:

t₁ = (a₁₁, ..., a_1n) t₂ = (a₂₁, ..., a_2n) ................... t_m = (a_m1, ..., a_mn)
Prove that the set {s₁,...,s_m} is linearly independent if and only if the set {t₁,...,t_m} is linearly independent.
Let S={s₁,...,s_m} and T={t₁,...,t_m}, and let * denote scalar multiplication.

To prove the statement "if S is linearly independent, then T is linearly independent," assume that S is linearly independent. Let x₁,...,x_m be scalars, and let the following be true:
x₁*s₁ + x₂*s₂ + ... + x_m*s_m = 0. (1)
Then, by the theorem about linearly independent sets of vectors, we know that x₁=x₂=...=x_m=0.
We may rewrite equation (1) by replacing each s_i with its equivalent expression a_i1*v₁ + a_i2*v₂ + ... + a_1n*v_n:
x_{1 (a₁₁*v₁ + a₁₂*v₂ + ... + a_1n*v_n)
+ x_{2 (a₂₁*v₁ + a₂₂*v₂ + ... + a_2n*v_n)
+ ... + x_{m (a_m1*v₁ + a_m2*v₂ + ... + a_mn*v_n) = 0 (2)}}}
Then we may collect the v_i terms:
(x₁*a₁₁ + x₂*a₂₁ + ... + x_m*a_m1) v₁ + (x₁*a₁₂ + x₂*a₂₂ + ... + x_m*a_m2) v₂ + ... + (x₁*a_1n + x₂*a_2n + ... + x_{m*a_mn) v_n = 0 (3)}
But {v₁,...,v_n} is a basis of V. So, from the definition of a basis, {v₁,...,v_n} is linearly independent. So the coefficient of each v_i in equation (3) must be identically zero; that is, the following system must be consistent:
x₁*a₁₁ + x₂*a₂₁ + ... + x_m*a_m1 = 0 x₁*a₁₂ + x₂*a₂₂ + ... + x_m*a_m2 = 0 ................................ x₁*a_1n + x₂*a_2n + ... + x_m*a_mn = 0.
We may rewrite this as one vector equation:
x_{1*(a₁₁,...,a_1n) + x_{2*(a₂₁,...,a_2n) + ... + x_m*(a_m1,..,a_mn) = 0}}
and, using the definitions of t₁,...,t_n given above, this is equivalent to the equation
x₁*t₁ + x₂*t₂ + ... + x_m*t_m = 0. (4)
By the theorem about linearly independent sets of vectors, {t₁,...,t_m} will be linearly independent if the above equation has only one solution, and if that one solution is x₁=x₂=...=x_m=0. At the beginning of this proof, we took a condition necessitating that, indeed, x₁=x₂=...=x_m=0. Since (4) is a homogeneous system of equations, it has either exactly one solution or infinitely many solutions; x₁=x₂=...=x_m=0 is certainly a solution, so it is thus the only solution.

So {t₁,...,t_m} is linearly independent.

To prove the statement "if T is linearly independent, then S is linearly independent," assume that T is linearly independent. Then we take x₁,...,x_m to be scalars such that equation (4) above is true. So x₁=x₂=...=x_m=0 by the theorem about linearly independent sets of vectors.

Now we transform equation (4) back into equation (1) by proceeding exactly backward through the steps laid out above. Equation (4) leads directly to the equation
x₁*(a₁₁,...,a_1n) + x₂*(a₂₁,...,a_2n) + ... + x_m*(a_m1,..,a_mn) = 0,
which in turn gives us the system
x₁*a₁₁ + x₂*a₂₁ + ... + x_m*a_m1 = 0 x₁*a₁₂ + x₂*a₂₂ + ... + x_m*a_m2 = 0 .............................. x₁*a_1n + x₂*a_2n + ... + x_m*a_mn = 0.
This guarantees that equation (3) will be consistent in all cases, which we may rewrite as equation (2):
x₁ (a₁₁*v₁ + a₁₂*v₂ + ... + a_1n*v_n) + x₂ (a₂₁*v₁ + a₂₂*v₂ + ... + a_2n*v_n) + ... + x_m (a_m1*v₁ + a_m2*v₂ + ... + a_mn*v_n) = 0, (2)
which is equivalent to equation (1):
x₁*s₁ + x₂*s₂ + ... + x_m*s_m = 0. (1)
But we know that x₁=x₂=...=x_m=0, so (1) has only the trivial solution, and thus the set {s₁,...,s_m} is linearly independent.

The proof is complete.