Matrices review

# Matrices review - Matrices We have already defined what we...

• Notes
• crupani10
• 24

This preview shows pages 1–5. Sign up to view the full content.

Matrices We have already defined what we mean by a matrix. In this section, we introduce algebraic operations into the set of matrices. Definition. If A and B are two matrices of the same size, say k by n, we define A t B to be the k by n matrix obtained by adding the corresponding entries of A and B, md we define cA to be the matrix obtained from A by multiplying each entry of A by c. That is, if aij and bij are the entries of A and B, respectively, in row i and column j, then the entries of A + B and of cA in row i and column j are aij bij and ca ij ' respectively. PJot? that for fixed 1: ar:d n , the set of all k by n matrices satisfies all the properties of a linear space. This fact is hardly surprising, for a k by n matrix is very much like a Ic-n tuple; that only difference is that the components are written in a rectangular array instead of a linear array. Unlike tuples, however, matrices have a further operation, a product operation. It is defined as follows: Cefinition. If A is a k by n mtrix, and B is an n by p matrix, we define the product D = A * B of A and B to be the matrix of size k by p whose entry dij in row i and column j is given by the formula n Here i = k and j = I,...,p.

This preview has intentionally blurred sections. Sign up to view the full version.

The entry di j is computed, roughly speaking, by tak- ing the "dot product" of the 1- 'th row of A with the j- th column of B. Schematically, This definition seems rather strange, but it is in fact e,xtremely useful. Motivation will come later! One important justification for this definition is the fact that this product operation satisfies some of the familar "laws of algebra" : Theorem '1. Matrix - multiplication has the following properties: k t -- A, B, C ,-D be matrices. (1) (Distributivihy) If A-(B + C) is defined, then - - - Similarly; .- if (B + C). D is - defined, then .- (El + C)*D = B*D + C.D. (2) (Homogeneity) If - A * B - is defined, then (3) (Associativity) - If A * B - and B6C are - defined,then .-
(4) (Existenceof identities) Fcr -- each m, there is an m by m matrix s~chthat for - I m . -- - matrices A and B, we have - -- I .A = A and B *Im = B li'i whenever these products are defined. --. - Proof. We verify the first distributivity formula. In order for B + C to be defined, B and C must have the same size, say n by p. men in order for A-(B + C) to be defined, A must have n columns. Suppose A has size 'x by n. .Then A - B and A - C aredefinedand have size k by p; thus their sum is also defined. The distributivity formula now follows from the equation n n n 1 a ( b . + c ) = I a b + I a c s=l is SJ sj s=l is sj s=l is sj' The other distributivity formula and the homogeneity formula are proved similarly. We leave them as exercises. Now let us verify associativity. - . If A is k by n and B is n by p, then A B is k by p. The product (AeB) C is thus defined provided C has size p by q. The product A (B*C) is defined in precisely the same circumstances. Proof of equality I ( is an exercise in summation symbols: The entry in row i and column j of (A*B) C is and the corresponding entry of A ( B = C ) is

This preview has intentionally blurred sections. Sign up to view the full version.

These t w o expressions are equal.
This is the end of the preview. Sign up to access the rest of the document.
• Spring '14
• FlorianEnescu

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern