Elementary Linear Algebra Proof

  • Thread starter tylerc1991
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Homework Statement



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

Homework Equations



In this example, [itex]O[/itex] represents the zero vector.

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

The Attempt at a Solution



[itex] A \cdot B = A \cdot C \iff[/itex]

[itex] \displaystyle \sum_{k=1}^{n} a_kb_k = \sum_{k=1}^{n} a_kc_k \iff[/itex]

[itex] \displaystyle \sum_{k=1}^{n} a_kb_k - \sum_{k=1}^{n} a_kc_k = 0 \iff [/itex]

[itex] \displaystyle \sum_{k=1}^{n} a_kb_k - a_kc_k = 0 \iff[/itex]

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

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

But since [itex] A \neq O, \, a_k \neq 0.[/itex]

Hence, [itex] b_k - c_k = 0 \iff b_k = c_k \iff B = C.[/itex]

It's been a while since I wrote a proof and I felt a little shaky on line 4. Thank you for your time!
 

Answers and Replies

  • #2
I like Serena
Homework Helper
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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.
 
  • #3
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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 [itex]b_k - c_k \neq 0[/itex] 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?

[itex]A = (2,1)[/itex]
[itex]B = (b_1,b_2)[/itex]
[itex]C = (c_1,c_2)[/itex]

Clearly [itex]A \neq O.[/itex]

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

[itex]A \cdot B = A \cdot C[/itex]

and [itex]B \neq C.[/itex]

Whoops! Haha, lesson learned! Thank you for your time!
 
Last edited:
  • #4
I like Serena
Homework Helper
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Take for instance n=2 and try to find a counter example.
 
  • #5
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Here are three vectors in R2:
a = <1, 2>
b = <2, -1>
c = <-2, 1>

What is [itex]a \cdot b[/itex]?
What is [itex]a \cdot c[/itex]?
Is it reasonable to conclude that b = c?
 

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