I am struggling to understand what it actually means to imbue a vector space with an inner product. Earlier you define the "canonical unit vectors" as (1,0,0) etc. This is clearly basis dependent so I'm guessing a "canonical" unit vector is a unit vector represented in a canonical basis. And a general unit vector is one with =1. This length should be coordinate independent, correct? Or is the numerical value always a coordinate value? If it is coordinate independent, is then the definition of the inner product equivalent with choosing the unit vectors? I ask because I often see the metric used to translate the "length" of a vector between basis, but this always implies a choice of basis to start in which should have the "correct" value. This seems arbitrary. Is that "choice of correct value" the definition of the inner product? Choosing a family of bases(the canonical ones?) where the metric is the identity?
The thing is that R^n comes with a canonical basis even if you don't care about lengths and angles. I can still call these vectors canonical "unit" vectors despite that we don't measure lengths, simply because there is only one 1 involved.
@@brightsideofmaths I learned a bit more and I think I understand better now. The way we define the inner product defines which vectors get length 1, (which tells us which vectors map to the canonical basis in R^n with the coordinate isomorphosm?). For example on the space of Polynomials the standard way is the integral from 0 to 1 f(x)g(x). That choice defines the unit vectors. We can make a different choice and get different unit vectors. All length are then(obviously) relative to the basis which is picked out by the function that defines the length. We can then find an orthogonal set of these (Grahm-Schmidt) that spans the space and for this as a basis set the metric tensor is the identity since it just contains the inner products of the basis vectors, which are by definition of orthogonality the Kronecker delta. I stumbled pretty hard by just using the monomials as a basis and expecting them to be orthogonal...
For me it's helpful not to think in bases and representations. Just take the inner product and the vectors as abstract definitions. I do this approach in the Abstract Linear Algebra Series :) @@narfwhals7843
@@brightsideofmaths I agree with this approach, which is why I stumbled so hard on this topic. Because any explicit definition of the inner product explicitly picks out a "preferred" kind of basis. But the only thing special about them is the unit vectors and at some point we have to plug in numbers to do calculations either way. I just originally assumed _which_ numbers we plug in is completely arbitrary, but it depends on the definition of the inner product.
![Linear Algebra 9 | Inner Product and Norm [dark version]](http://i.ytimg.com/vi/fXWdBineqWI/mqdefault.jpg)
Excellent pace and clarity for mere mortals like myself. Thank you very much!!
Glad it was helpful! :)
inner products are the best!
A weekend with another video from you, is a good weekend :)
Yeah, I also think it's great weekend :)
Very helpful Thank you so much 🙏🏻
Great vid, bouta watch the whole playlist
Yeah, please do it :)
From the definition of norm, can we say that the norm is induced by the inner product?
Yes!
nice explanation. Thanks
Glad you liked it
is there a video that deals with cyclic vector subspaces?
Not yet, but it will come :)
Thank you...and one more thing I'm not clear on what r^n Is....
Is it a Cartesian product with same nos or different nos
Yes, it's the cartesian product with n factors.
so there are more norm and inner products other than standarts?
Yes, there are a lot :)
@@brightsideofmaths thank you for the answer. Damn u mathematics. :-)
I am struggling to understand what it actually means to imbue a vector space with an inner product.
Earlier you define the "canonical unit vectors" as (1,0,0) etc. This is clearly basis dependent so I'm guessing a "canonical" unit vector is a unit vector represented in a canonical basis. And a general unit vector is one with =1.
This length should be coordinate independent, correct?
Or is the numerical value always a coordinate value?
If it is coordinate independent, is then the definition of the inner product equivalent with choosing the unit vectors?
I ask because I often see the metric used to translate the "length" of a vector between basis, but this always implies a choice of basis to start in which should have the "correct" value.
This seems arbitrary.
Is that "choice of correct value" the definition of the inner product? Choosing a family of bases(the canonical ones?) where the metric is the identity?
The thing is that R^n comes with a canonical basis even if you don't care about lengths and angles. I can still call these vectors canonical "unit" vectors despite that we don't measure lengths, simply because there is only one 1 involved.
@@brightsideofmaths I learned a bit more and I think I understand better now. The way we define the inner product defines which vectors get length 1, (which tells us which vectors map to the canonical basis in R^n with the coordinate isomorphosm?). For example on the space of Polynomials the standard way is the integral from 0 to 1 f(x)g(x). That choice defines the unit vectors. We can make a different choice and get different unit vectors.
All length are then(obviously) relative to the basis which is picked out by the function that defines the length.
We can then find an orthogonal set of these (Grahm-Schmidt) that spans the space and for this as a basis set the metric tensor is the identity since it just contains the inner products of the basis vectors, which are by definition of orthogonality the Kronecker delta.
I stumbled pretty hard by just using the monomials as a basis and expecting them to be orthogonal...
For me it's helpful not to think in bases and representations. Just take the inner product and the vectors as abstract definitions. I do this approach in the Abstract Linear Algebra Series :) @@narfwhals7843
@@brightsideofmaths I agree with this approach, which is why I stumbled so hard on this topic. Because any explicit definition of the inner product explicitly picks out a "preferred" kind of basis.
But the only thing special about them is the unit vectors and at some point we have to plug in numbers to do calculations either way.
I just originally assumed _which_ numbers we plug in is completely arbitrary, but it depends on the definition of the inner product.
The explicit abstract definition of an inner product does not pick any basis.@@narfwhals7843
Hi prof, may i have your prefered reference for linear algebra
My book :D
Prove that if V is an inner product space and u \in V then ||u|| = sqrt\langleu, u
angle is a norm on V😢
You can do it!
👍
What's norm
A map to measure lengths of vectors :)
Thank you...and one more thing I'm not clear on what r^n Is....
Is it a Cartesian product with same nos or different nos
In field@@brightsideofmaths
Cute accent ... "Summawuie" ,
Haha, thanks :D
If you have vectors (-1, 0) and (1, 1) the inner product becomes -1
Yes!