My Math Forum  

Go Back   My Math Forum > College Math Forum > Linear Algebra

Linear Algebra Linear Algebra Math Forum

LinkBack Thread Tools Display Modes
December 22nd, 2018, 02:19 AM   #1
Joined: Dec 2018
From: Tel Aviv

Posts: 4
Thanks: 0

Question Feedback for an answer involving Span, Dimension and Linear Independence

Hi guys,

I have worked out an answer to the following question, but it's stretching my understanding of Linear Algebra to the limit, so… if anybody could have a look at it, and just tell me whether my reasoning is sound, it would be a great help...

Let v_1,…,v_k,u,w be vectors in the linear space V.
It is given that the equation x_1·v_1+…x_k·v_k=u has a single solution and that the equation x_1·v_1+…+x_k·v_k=w has no solution.

a. Prove that the group {v_1,…,v_k} is linearly independent.
b. Find the dimension of Sp{v_1,…,v_k,w}.

My answer:
According to the information provided, the equation
x_1·v_1+…x_k·v_k=u has a single solution.
That means that for the solution to hold, all variables in this equation must have a fixed value, with no free variables. If u is the zero vector, each one of the variables must take on the value 0 for the single solution to hold: it is a given that there is a single solution only, so in this case it must be the trivial solution, as the trivial solution always exists in a homogeneous system. If the trivial solution is the only solution, then v_1 to v_k is linearly independent.

per definition {v_1,…,v_k} spans Sp{v_1,…,v_k}, and it was shown that {v_1,…,v_k} is linearly independent, so {v_1,…,v_k} is a basis with dimension k. Sp{v_1,…,v_k,w} has an additional member, w, which according to the information given is not a linear combination of {v_1,…,v_k} (given that the equation x_1·v_1+…+x_k·v_k=w has no solution). Therefore {v_1,…,v_k,w} is a basis of Sp{v_1,…,v_k,w}, which dimension k+1.

Last edited by skipjack; December 22nd, 2018 at 05:12 PM.
pregunto is offline  
January 27th, 2019, 06:52 PM   #2
Joined: Jan 2019
From: Adelaide Australia

Posts: 3
Thanks: 0

Hi pregunto,

Here's how I would approach the first problem:

We are given that x_1 * v_1 + ... + x_n * v_n = u has a unique solution.

Let's call this unique solution b = (b_1, ..., b_n).

Therefore, b_1 * v_1 + ... + b_n * v_n = u. .... (1)

Now we'll prove that {v_1, ..., v_n} is linearly independent using proof by contradiction. That is, we'll assume that it is NOT linearly independent and from that derive something that's impossible.

So let's assume that it is NOT linearly independent. In other words, it is a linearly dependent set of vectors. By definition of linear dependence, this means that there exists a vector a = (a_1, ..., a_n) such that

a_1 * v_1 + ... + a_n * v_n = 0 (the zero vector) .... (2)

Adding equations (1) and (2), we get:

(a_1 + b_1) * v_1 + ... + (a_n + b_n) * v_n = u + 0 = u.

Therefore, the vector a+b = (a_1 + b_1, ..., a_n + b_n) is another vector, which is distinct from the vector b, which is a solution to

x_1 * v_1 + ... + x_n * v_n = u.

But this is not possible since we know that that equation has a UNIQUE solution.

Therefore, our assumption has lead to an impossibility and hence we conclude that it is an incorrect assumption.

Please let me know if this was helpful.

RichardEmt is offline  

  My Math Forum > College Math Forum > Linear Algebra

answer, dimension, feedback, independence, involving, linear, linear dependence, span

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
Should span be a finite linear combination only? zzzhhh Linear Algebra 3 September 26th, 2017 10:00 PM
Linear Independence and Linear Dependence. Luiz Linear Algebra 1 August 26th, 2015 09:22 AM
Linear Independence and Linear Dependence. Luiz Linear Algebra 5 August 24th, 2015 02:10 PM
Find dimension and span kap Linear Algebra 1 January 18th, 2012 09:57 AM
Linear Combination and Span WeMo123 Linear Algebra 1 January 11th, 2011 04:04 PM

Copyright © 2019 My Math Forum. All rights reserved.