Dimension and Orthogonality in Vector Spaces: A Proof of the Inequality m ≤ n

[roam]

Homework Statement

If {u₁, u₂,...,u_m} are nonzero pairwise orthogonal vectors of a subspace W of dimension n, prove that m \leq n.

The Attempt at a Solution

I look at all my notes but I still can't understand what this qurstion asks or what definitions I need to be using for this... I'm stuck and appreciate some guidance so I can get started. Thank you.

 

[matt grime]

Well, what is n? It's the dimension. So we're asking if a set of mutually orthogonal vectors can have cardinality greater then n. Can it? Hmm, what is dimension? It is the cardinality (size) of a maximally linearly independent set. Does that help? (I.e. you should now think: if I can show that the set is ________ then by what's gone before it must have at most n elements.)

 

[roam]

Well, what is n? It's the dimension. So we're asking if a set of mutually orthogonal vectors can have cardinality greater then n. Can it? Hmm, what is dimension? It is the cardinality (size) of a maximally linearly independent set. Does that help? (I.e. you should now think: if I can show that the set is ________ then by what's gone before it must have at most n elements.)

Do you mean I have to prove that the set is linearly independent? A set of mutually orthogonal non-zero vectors is always linearly independent.

OK so the vectors {u₁, u₂,...,u_m} are in R^m and they are orthogonal (i.e u_j · u_i=0 if i =/= j).

Let's take a linear combination of the vectors in this set that gives the zero vector:

k₁u₁+k₂u₂+...+k_mu_m = 0

I just need to show that there's only one value for all the constants, k₁ = k₂ = … = k_m = 0.

If I take the dot product of both sides of the equation with u₁:

u₁ · (k₁u₁+k₂u₂+...+k_mu_m) = 0 · u₁

Factoring out the constant.

k₁(u₁ · u₁) + k₂(u₁ · u₂) +...+k_m(u₁ · u_m) = 0

k₁(u₁)² + k₂(0) +...+k_m(0) = 0

k₁=0 & all scalar coefficients are zeros and therefore vectors are independent.

I showed that the vectors in the set are linearly independent but I can't see exactly how this is useful in showing that m \leq n...
Could you please provide me with more explanation of what I need to do next?

 

[matt grime]

I stated the definition of dimension in my first post: it is the maximum size of *any* set of linearly independent vectors. You have a set of m linearly independent vectors. You know that *by definition* the maximum size a set of linearly independent vectors can have is n.

 

[roam]

It proves that m \leq n, end of proof? Is what I've done sufficient to write as a proof or do I need to also further prove the definition of dimension that you stated?

 

[matt grime]

Yes it obviously proves m<=n. Was that really a question? Do I need to explain it more?

What I used above is the definition of dimension. What other one do you have?

 

[roam]

OK, thank you very much for your help Matt.

 

[matt grime]

Seriously, are you using a different definition? You could be using the minimal size of a spanning set as the definition, so you probably ought to show that this agrees with maximal size of a linearly independent set. Or whatever definition you're using (I can't think of another elementary one).