Are Mutually Perpendicular Vectors Always Linearly Independent?

[A.Magnus]

In a problem I am working on, it is given that $V_1, V_2, ... , V_n$ are mutually perpendicular vectors in a space defined with a certain scalar product. I need to prove or disprove that $V_i$ are linearly independence regardless of any definition of scalar product.

I think the solution should go like these: Assume that the vectors are linearly independent such that there exist number $c_i$, not all of them being trivial, so that $c_1V_1 + c_2V_2 + ... c_iV_i + ... + c_nV_n = 0.$ And then
$$\begin{align}
V_i (c_1V_1 + c_2V_2 + ... c_iV_i + ... + c_nV_n) &= 0\\
c_iV_iV_i &=0\\
\end{align}$$

Am I on the right track? How do I tie this up to proving or disproving the claim? I am lost on writing it mathematically. Many thanks in advance for your gracious help. ~MA

 

[I like Serena]

In a problem I am working on, it is given that $V_1, V_2, ... , V_n$ are mutually perpendicular vectors in a space defined with a certain scalar product. I need to prove or disprove that $V_i$ are linearly independence regardless of any definition of scalar product.

I think the solution should go like these: Assume that the vectors are linearly independent such that there exist number $c_i$, not all of them being trivial, so that $c_1V_1 + c_2V_2 + ... c_iV_i + ... + c_nV_n = 0.$ And then
$$\begin{align}
V_i (c_1V_1 + c_2V_2 + ... c_iV_i + ... + c_nV_n) &= 0\\
c_iV_iV_i &=0\\
\end{align}$$

Am I on the right track? How do I tie this up to proving or disproving the claim? I am lost on writing it mathematically. Many thanks in advance for your gracious help. ~MA

Hi again MaryAnn! ;)

You're already there.
If not all numbers $c_i$ are zero, there must be at least one that is non-zero, let's say $c_i \ne 0$.
Furthermore, one of the requirements for an inner product is that $\langle \mathbf v, \mathbf v \rangle > 0$ iff $\mathbf v \ne \mathbf 0$.
Since we have $c_i\mathbf V_i\mathbf V_i =0$ we have a contradiction.
Therefore the $\mathbf V_i$ are linearly independent.

 

[A.Magnus]

Hi again MaryAnn! ;)

You're already there.
If not all numbers $c_i$ are zero, there must be at least one that is non-zero, let's say $c_i \ne 0$.
Furthermore, one of the requirements for an inner product is that $\langle \mathbf v, \mathbf v \rangle > 0$ if $\mathbf v \ne \mathbf 0$.
Since we have $c_i\mathbf V_i\mathbf V_i =0$ we have a contradiction.
Therefore the $\mathbf V_i$ are linearly independent.

Thank you for your gracious help! But the problem is not asking for linear independence, the question is asking "Is it true that for any scalar product these vectors are linearly independent?" The solution I wrote you above may mis-lead you to think that the problem is asking for linear independence but it is not. Sorry for any misunderstanding! It is my bad! Thank you again. ~MA

 

[I like Serena]

Thank you for your gracious help! But the problem is not asking for linear independence, the question is asking "Is it true that for any scalar product these vectors are linearly independent?" The solution I wrote you above may mis-lead you to think that the problem is asking for linear independence but it is not. Sorry for any misunderstanding! It is my bad! Thank you again. ~MA

The reasoning above is valid for any scalar product.
So it is true that for any scalar product these vectors are linearly independent.
What am I missing?

 

[A.Magnus]

The reasoning above is valid for any scalar product.
So it is true that for any scalar product these vectors are linearly independent.
What am I missing?

Now you got it right! Is there mathematical detail that you may want to suggest, instead of just writing "... the $V_i$ are linear independent for any scalar product..."? My prof is notoriously very picky about detail. Sorry to take off your time answering my question again. ~MA

 

[I like Serena]

Now you got it right! Is there mathematical detail that you may want to suggest, instead of just writing "... the $V_i$ are linear independent for any scalar product..."? My prof is notoriously very picky about detail. Sorry to take off your time answering my question again. ~MA

Sorry, no more detail to add.
(Oh, do make sure you write "Assume... dependent" instead of "Assume... independent" in your opening post if we're talking about being nitpicky.)
I believe it is full-proof, so I'd be interested in your prof finding something (else) to nitpick about. (Wink)

 

[A.Magnus]

Sorry, no more detail to add.
(Oh, do make sure you write "Assume... dependent" instead of "Assume... independent" in your opening post if we're talking about being nitpicky.)
I believe it is full-proof, so I'd be interested in your prof finding something (else) to nitpick about. (Wink)

Thank you again, sorry for my dumb question. I hope it does not dumbfound you! ~MA