My goals in these lectures are: ? To present enough linear algebra to understand the Singular Value Decomposition and related methods such as the Principal Components Analysis . ? To present some applications, especially concerning Latent Semantic Analysis and facial recognition . ? The linear algebra concepts that I’dliketoteachare: linear independence, span, basis, eigenvalue, eigenvector, singular value decomposition. ? Some of the material here maybe found in any textbook on the subject. I also included links on the coursewebsite, to papers on linear algebra and LSA and to a tutorial website on SVD that I’d like everyone to work through.Q520: Linear Algebra Vector Spaces In these lectures, Rnisthesetof vectors of lengthnofreal numbers. For example, R2 contains (1,1), (. 23234,0), (123,3). R10 contains (1,2,3,. 234,5, . 8643645 ,7,8,4,10). We could also write these as columns, but it’seasiertotype them as rows. We’lldefine dimension soon, but it will turnout that Rnhas dimensionn. Sometimes one wants to consider complex numbers instead of just real numbers. But we don’tneedthis.
Q520: Linear Algebra Vector Spaces and Subspaces We add vectors in the obvious way: (1,3,5) + (2,4,0) = (1,7,5) . And we also multiply by numbers: 6(1,3,4) = (6,18,24) . A subset VofRniscalleda vector space if two conditions hold: ? Wheneveruandv belong to V, so does the sumu+v. ? Whenever vbelongs to Vandcisareal number, thencv belongs to Vas well. The easiest example is when Vis all of Rn, but there are lots of others. Another way to say that Visa vector space would be to say that Va subspace ofRn.
Q520: Linear Algebra Example of a Subspace of R2 Consider the set of vectors (a, 2 a), where a 2 R. We write this as { (a, 2a) : a 2R} This contains, for example (3,6), (1,2), and (0,0). It does not contain (1,1) or (1,2). We can check that it is a subspace: First, (x, 2x) + (y, 2y) = (x+y, 2x 2y) = (x+y, 2(x+y)) . Secondc (x, 2 x) = (cx, c (2 x)).
Q520: Linear Algebra Linear Combinations Given a set v 1 , v 2 ,…, v k of vectors, a linear combination of these is any vector of the form X i c i v i =c 1 v 1 +···+c n v n where the c’sareanyreal numbers. For example, ifn=3 and v 1 = (2,1,0) v 2 = (1,3,1) v 3 = (1,1,1) , Then one vector in the span of these three would be 4v 1 +0v 2 3v 3 =4(2,1,0) +0(1,3,1) + (3)(1,1,1) = (8,4,0) + (0,0,0) + (3,3,3) = (5,1,3) There are many other vectors in the span, of course…

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