## Vector Spaces !!!!!!

Finding a suitable text is proving difficult. This is (nearly) teaching to the exam, a state of which I disapprove. This sheet will change if I become more clear what it is we need to know and what it makes sense to know.

First, solving questions:

To find the rank of a matrix, we must manipulate the rows (or columns) by adding, subtracting and multiplying by a factor to try to make one row (or column) be filled with zeros. If we can do that, then the rank of the matrix is the same as the number of rows or columns that are non-zero. The rank of a matrix is very much the same as the dimension of the vector space.
So we could possibly find the size of the row space of a rectangular matrix. However, most of our questions are about square matrices.

When doing this manipulation, it helps the later parts of the question to try to reduce the matrix to echelon form (as many 1 and 0 entries as possible and zeros below the leading diagonal).

A vector space is a commutative group of vectors under addition; its scalar elements are reals (technically, a field) with the operations of multiplication mixing scalar k and vector a to give ka. The group must be closed (do the operation and the result lies in the vector space); there must be an identity element (typically 0 for addition, 1 for multiplication), the operators must be distributive and associative. If also commutative, then the group is commutative. The mix of number and vector may be ‘sided’. So the vector space may not be associative or commutative under multiplication.

A basis for a space looks very like a set of eigenvectors and will be the same as the column vectors that make up your echelon form. The definition of a basis is the smallest set of vectors which can be used in linear combination to produce any vector in the space (called spanning the space). If a basis is finite then every basis has that same number of elements; this defines the dimension of the space.

Exercise: find the dimension of each of these matrices, change it to row echelon form (manipulate rows not columns), find a basis for the space and find the kernel (where its dimension is non-zero).

⎛ 2   1  -3  ⎞      ⎛ 3  -1  -1 ⎞     ⎛ 2  -5  2  ⎞     ⎛  1  -1  2  ⎞
⎢ 1  -2   1  ⎥       ⎢ 1   1  1  ⎥     ⎢ 9   3 -4  ⎥     ⎢  4   1   1  ⎥
⎝ 2  1   -1  ⎠       ⎝ 4  -1  1  ⎠     ⎝ 7  3  -2  ⎠     ⎝  3   4   1  ⎠

⎛  4  2  -1  ⎞     ⎛ 4  -7  6  ⎞    ⎛ 4  -1  5  ⎞
⎢ 2  3    2  ⎥    ⎢ 5  1  -4  ⎥    ⎢ 4   1   1  ⎥
⎝  5  -1   8  ⎠   ⎝  3  -2  3  ⎠    ⎝ 3   4  1  ⎠

We can use these matrices in a similar way to solve sets of equations by elimination. Do this by extending the matrix by one column, made of the values of the equations. Put a vertical dotted line where the = signs went. Maximise the number of zeros. Write the answer.

So for example and basing on Q1 above, if we want to solve the system of equations
2x+y-3z = -20 ; x-2y+z= 20 ; 2x+y-z= -6  ; then you can reduce the extended matrix in the same way the original is reduced to echelon form,by manipulating rows as follows:

⎛2  1  -3  -20⎞   leave it alone ⎛2  1  -3  -20 ⎞ add 3R3    ⎛2 1 0    1 ⎞
⎢1 -2   1   20 ⎥  do (R₁-2R₂)/5⎢0   1   -1  -12⎥ add last R
3⎢0 1 0   -5⎥
⎝2  1  -1   -6 ⎠   do (R-R₁)/2   ⎝0  0    1     7 ⎠   done        ⎝ 0 0 1   7 ⎠
and finish this off, so (x,y,z) = (3,-5,7).  It is surprising how poor one’s
arithmetic turn’s out to be; go slowly and carefully. recording what you did, as shown, is invaluable both in marking and finding errors. Usually examiners pick integer values (but you should test that statement).

The null space, or kernel, of the transformations is the set of vectors that is transformed to the zero vector [M.x=0]. The dimension of the null space is the same as the difference between the apparent rank of the matrix (the number of rows, for our work), typically 4, and the dimension of the range space, typically 2 or 3. Finding a basis for the kernel may involve algebra [M.λe=0] to find e. Clearly you may (unlikely, but possible) find a different basis from another student. Where different bases (plural of basis) produce the same matrix, then the bases are similar. Any transformation that can be represented by a matrix is a linear transformation, hence (from here) we refer to ‘our’ vector spaces as linear spaces, a subset of the bigger subject to be studied at university.

So how do you find a kernel?? You use the reduced row-echelon form say E, of the matrix of your originating transformation:  then E.x = 0 and generally x is easy to see from this. It is not easy to do this from the original form.

In general, the solution to Ax=b is the sum of any particular solution Ax=b and the general solution to Ae=0, just as with differential equations. A particular set of exam questions works on the assumption that any vector x is such that x=x0+ λe
i.e. made of some fundamental vector and a multiple of the basis (a basis) of the kernel.

Stick to row operations because that will not alter the row space or null space of the matrix.

I suspect that doing the parallel work on the identity matrix while manipulating into echelon form will give the kernel automatically. I will investigate.

http://www.maths.lse.ac.uk/personal/martin/fme2a.pdf  University text, so it covers what is needed but at great length.
http://filestore.aqa.org.uk/subjects/AQA-MFP4-TEXTBOOK.PDF FP4, stops before vector spaces. Good material for the other A-levels. Try jumping to Chapter 6, around page 70. Ch 7 talks about eigenvectors.PDP-1 appears after page 100.  The Exercises have answers!!
https://www.scribd.com/document/90740117/2011-Vector-Spaces-Matrices Zen Harper’s version of what CIE requires one to know. Heavy going, buty only 11 pages.

A field (you don’t need to know, but some of you would ask) is a set with two binary operations + and x, Abelian over + and x (a little problem there with zero) and x is distributive over +.   See the previous page, groups rings and fields.