Linear Algebra for year 1 university students by sir Zevia - AcadEx COGNITIVE

This course is aimed at imparting knowledge on year 1 students in all state universities

Course Content

Matrices and matrix operations

Matrices
Matrices are a powerful tool for calculations involving linear transformations. It is important to understand how to find the matrix of a linear transforma- tion and the properties of matrices. 7.1 Linear Transformations and Matrices Ordered, finite-dimensional, bases for vector spaces allows us to express linear operators as matrices. 7.1.1 Basis Notation A basis allows us to efficiently label arbitrary vectors in terms of column vectors. Here is an example. Example 74 Let V = a b c d a, b, c, d ∈ R be the vector space of 2 × 2 real matrices, with addition and scalar multiplication defined componentwise. One choice of basis is the ordered set (or list) of matrices B = 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 =: (e 1 1 , e1 2 , e2 1 , e2 2 ).Given a particular vector and a basis, your job is to write that vector as a sum of multiples of basis elements. Here an arbitrary vector v ∈ V is just a matrix, so we write v = a b c d = a 0 0 0 + 0 b 0 0 + 0 0 c 0 + 0 0 0 d = a 1 0 0 0 + b 0 1 0 0 + c 0 0 1 0 + d 0 0 0 1 = a e1 1 + b e1 2 + c e2 1 + d e2 2 . The coefficients (a, b, c, d) of the basis vectors (e 1 1 , e1 2 , e2 1 , e2 2 ) encode the information of which matrix the vector v is. We store them in column vector by writing v = a e1 1 + b e1 2 + c e2 1 + d e2 2 =: (e 1 1 , e1 2 , e2 1 , e2 2 )   a b c d   =:   a b c d   B . The 4-vector   a b c d   ∈ R 4 encodes the vector a b c d ∈ V but is NOT equal to it! (After all, v is a matrix so could not equal a column vector.) Both notations on the right hand side of the above equation really stand for the vector obtained by multiplying the coefficients stored in the column vector by the corresponding basis element and then summing over them. Next, lets consider a tautological example showing how to label column vectors in terms of column vectors: Example 75 (Standard Basis of R 2 ) The vectors e1 = 1 0 , e2 = 0 1 are called the standard basis vectors of R 2 = R {1,2} . Their description as functions of {1, 2} are e1(k) = 1 if k = 1 0 if k = 2 , e2(k) = 0 if k = 1 1 if k = 2 .It is natural to assign these the order: e1 is first and e2 is second. An arbitrary vector v of R 2 can be written as v = x y = xe1 + ye2. To emphasize that we are using the standard basis we define the list (or ordered set) E = (e1, e2), and write x y E := (e1, e2) x y := xe1 + ye2 = v. You should read this equation by saying: “The column vector of the vector v in the basis E is x y .” Again, the first notation of a column vector with a subscript E refers to the vector obtained by multiplying each basis vector by the corresponding scalar listed in the column and then summing these, i.e. xe1 +ye2. The second notation denotes exactly the same thing but we first list the basis elements and then the column vector; a useful trick because this can be read in the same way as matrix multiplication of a row vector times a column vector–except that the entries of the row vector are themselves vectors! You should already try to write down the standard basis vectors for R n for other values of n and express an arbitrary vector in R n in terms of them. The last example probably seems pedantic because column vectors are al- ready just ordered lists of numbers and the basis notation has simply allowed us to “re-express” these as lists of numbers. Of course, this objection does not apply to more complicated vector spaces like our first matrix example. Moreover, as we saw earlier, there are infinitely many other pairs of vectors in R 2 that form a basis. Example 76 (A Non-Standard Basis of R 2 = R {1,2} ) b = 1 1 , β = 1 −1 . As functions of {1, 2} they read b(k) = 1 if k = 1 1 if k = 2 , β(k) = 1 if k = 1 −1 if k = 2 .

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KUMA ZEVIA NKWI

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Price:Free
Instructor:KUMA ZEVIA NKWI
Duration:72 hours45 minutes
Lessons:2
Students:0
Level:Intermediate