Normalized basis when taking inner product

[kent davidge]

Consider that a vector can be represented in two different basis. My question is do we need to normalize both basis before taking the inner product?

What motivates this question is because I found out that the inner product of a vector having components $a,b$ in the normalized polar basis of $\mathbb{R}^2$ will be $a^2 + b^2$ whereas it will be $r^2(a^2 + b^2)$ if the basis is not normalized.

 

[Mark44]

Consider that a vector can be represented in two different basis. My question is do we need to normalize both basis before taking the inner product?

No. The vectors in a basis don't have to be normalized (i.e., of unit length).

 

[kent davidge]

No. The vectors in a basis don't have to be normalized (i.e., of unit length).

Hi Mark. Please consider re reading my post after I added context to my question.

 

[Mark44]

What motivates this question is because I found out that the inner product of a vector having components $a,b$ in the normalized polar basis of $\mathbb{R}^2$ will be $a^2 + b^2$ whereas it will be $r^2(a^2 + b^2)$ if the basis is not normalized.

Here's a link to a page I found that might be helpful -- http://www.math.tamu.edu/~fulling/coalweb/polar.htm

BTW, saying "inner product of a vector" with no mention of another vector is meaningless. In this case, I think you meant the inner product of a vector with itself.

 

[kent davidge]

BTW, saying "inner product of a vector" with no mention of another vector is meaningless. In this case, I think you meant the inner product of a vector with itself

That's what I meant

Here's a link to a page I found that might be helpful -- http://www.math.tamu.edu/~fulling/coalweb/polar.htm

They don't specify the reasons for using unit polar vectors, but the conclusion I draw from their example is that by using the unit vectors we are not affecting our numerical quantities, we merely use the vectors in order to get a direction, in this case.