My Math Forum Projection of a vector

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 November 28th, 2010, 08:21 AM #1 Newbie   Joined: Nov 2010 Posts: 1 Thanks: 0 Projection of a vector I like to spend my free-time reading about the mathematical formalism of Quantum mechanics. I'm at the moment studying linear vector spaces, but I encountered a problem whilst trying to understand the Gram-Schmidt proces. I understand, that given 2 linearly independent vectors la1> and la2> you can orthonormalize them with this proces. My teacher explained it this way. 1) Choose any of the 2 vectors and divide by its norm. This will in all cases provide you with the first unit vector, le1>. 2) Dot this unit vector with the other vector, say I chose la1> as my first vector, which gives you this number: . This is the portion of la2> which is along le1>, he told me. However, I'm not sure i understand why that is. It makes sense for me if le1>= (0,1) or le1> = (1,0), because then it seems quite elementary. However I'm not sure I understand why it is so in other cases, i.e. when it's just a generel unit vector le1>=(cosv,sinv). How do you proof this matemathically. I know it's a simple question, but you must understand, that I'm a high-school student whose only teacher is one on youtube lectures about QM. That was my first question. My second question is why, you always get the unit vector when u divide by the norm of a vector, say we have vector la1> then the norm is ll la1>ll = ^0,5. Because i can see this work for 2-dimensional euclidean space due to pythagoras, because then the norm is just the length and then any vector is just an expansion of the length of the unit vector by it's own length. But how do you proof this for n-dimensional spaces, where the norm is not really the length as we understand it.
 November 29th, 2010, 09:29 PM #2 Senior Member   Joined: Nov 2010 Posts: 502 Thanks: 0 Re: Projection of a vector Why isn't the norm equal to the length as we understand it? It seems reasonable to me. The norm behaves just like the length in 2-space.

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