Elementary Linear Algebra Proof

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Homework Help Overview

The discussion revolves around a proof regarding vectors in \(\mathbb{R}^n\), specifically addressing the claim that if \(A \cdot B = A \cdot C\) and \(A \neq O\), then \(B = C\). Participants are exploring the validity of this statement and examining the implications of the dot product.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to prove the statement using algebraic manipulation of the dot product. Some participants question the validity of the steps taken, particularly regarding the implications of a sum equating to zero. Others suggest considering counterexamples to challenge the original claim.

Discussion Status

Participants are actively engaging with the problem, with some providing hints and suggestions for exploring counterexamples. There is recognition of the need to clarify assumptions and the conditions under which the original claim holds true.

Contextual Notes

Some participants note the importance of considering specific cases and counterexamples, particularly in lower dimensions like \(n=2\), to better understand the claim's limitations.

tylerc1991
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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!
 
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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?
 

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