Elementary Linear Algebra Proof

tylerc1991
Messages
158
Reaction score
0

Homework Statement



Prove or disprove the following about vectors in \mathbb{R}^n: If A \cdot B = A \cdot C and A \neq O, then B = C.

Homework Equations



In this example, O represents the zero vector.

Let the vectors be represented as:
A = (a_1,a_2,\dots,a_n)
B = (b_1,b_2,\dots,b_n)
C = (c_1,c_2,\dots,c_n)

The Attempt at a Solution



A \cdot B = A \cdot C \iff

\displaystyle \sum_{k=1}^{n} a_kb_k = \sum_{k=1}^{n} a_kc_k \iff

\displaystyle \sum_{k=1}^{n} a_kb_k - \sum_{k=1}^{n} a_kc_k = 0 \iff

\displaystyle \sum_{k=1}^{n} a_kb_k - a_kc_k = 0 \iff

\displaystyle \sum_{k=1}^{n} a_k(b_k - c_k) = 0 \iff

So either a_k = 0, or b_k-c_k = 0 for k = 1,2,\dots,n.

But since A \neq O, \, a_k \neq 0.

Hence, b_k - c_k = 0 \iff b_k = c_k \iff B = C.

It's been a while since I wrote a proof and I felt a little shaky on line 4. Thank you for your time!
 
Physics news on Phys.org
Hi tylerc1991! :smile:

It's only the 6th and 7th line of your proof that are shaky.
If a sum of terms is zero, that does not mean that each individual term has to be zero.
Also, if at least one component of A is not zero, then A is not the zero-vector.

Let me know if you want a hint.
 
I like Serena said:
It's only the 6th and 7th line of your proof that are shaky.
If a sum of terms is zero, that does not mean that each individual term has to be zero.
Also, if at least one component of A is not zero, then A is not the zero-vector.

This is true. I have been dallying around with a few ideas (assuming b_k - c_k \neq 0 and trying to arive at a contradiction), but it seems like there is always a counterexample. I am thinking I should try to break it up into cases, but I don't want to overcomplicate it. Small hint please! :shy:

EDIT: How about a counterexample to the original claim?

A = (2,1)
B = (b_1,b_2)
C = (c_1,c_2)

Clearly A \neq O.

If we try b_1 = 1, \, c_1 = \frac{1}{2}, \, b_2 = 3, \, c_2 = 4, then

A \cdot B = A \cdot C

and B \neq C.

Whoops! Haha, lesson learned! Thank you for your time!
 
Last edited:
Take for instance n=2 and try to find a counter example.
 
Here are three vectors in R2:
a = <1, 2>
b = <2, -1>
c = <-2, 1>

What is a \cdot b?
What is a \cdot c?
Is it reasonable to conclude that b = c?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top