# Vector Spaces, Sub-spaces and Singularity- The Universe of Linear Algebra

In mathematics we have concept of abstract objects i.e. it is an object which in real sense does not exist but to make things intuitive to humans we use these analogies to describe concepts.For e.g. Hyperplane is a subspace in an N- dimensional vector space which segregates the data points in that space.Here hyperplane is an abstract object.

The most beautiful part of Physics and Mathematics is that they both are body and soul. One has no meaning without the presence of the other and when they get together, they answer all the questions that we have in the universe.So, let’s dive into the universe of Linear Algebra and try to understand these abstract concepts and objects.

# VECTOR SPACE

Within Vector space, we have concept of span and sub-spaces. A span is the linear combination of subset of vectors (W) in a vector space (V) whereas sub-space (S) is a linear sub-space of vector space (V) if and only if :

1. the zero vector is in S, 0∈S
2. If u,v ∈ S then u+v ∈ S
3. If u ∈ S, then ku ∈ S for all k∈R

It is important to note that subset W is not a sub-space but span(W) is a sub-space of vector space V.

Thus, if W = span(W) then W is a subspace of vector space V.

To sum up , Vector spaces have sub-spaces which have subset of vectors which satisfy the above mentioned conditions to make up a vector-space of their own. Span of a subset of vectors leads to a vector space but the subset of vectors in that span individually do not make a vector-space.

Keep these concepts in mind because we are moving towards the bigger picture of solving the linear equations using the above concepts.

SOLVING LINEAR EQUATIONS

The most fundamental use of linear algebra is to solve linear equations.

A X = B ,studying linear algebra we have come across this equation but most of us really don’t understand this equation from vector spaces perspective.

Here we have a matrix A of dimension 5x4 , X of dimension 4X1 and B of dimension 5X1.

In vector space language , we can interpret as : We apply a linear transformation (A) on a 4 dimensional vector (X) to have a 5 dimensional vector as output.

Essentially , we solve this equation for X ,provided we have A and B.

To solve this equation, we have to take inverse of A , and to find inverse of A ,we have to check the whether inverse of A exists or not and the result boils down to understanding whether B is in the span of A.

Condition for A inverse to exist :

1. For inverse to exist, the column space of A transpose should be equal to the number of rows of B. Therefore, A transpose should have atleast m number of columns , where m is the number of rows of B.

for eg. if A is 3x2 matrix then X is 2 dimensional and B is 3D. It is possible to value of B out of the plane where X is present and in that case we cannot solve for X.

But 1 is not sufficient condition

2. The column space of A transpose should be linearly independent.

means , there should not be a column in the A transpose which can be linear combination of any other columns.If we have linearly dependent columns then column space of A shrinks down to one less dimension or in the worst case it shrinks down to single dimension which leads to singularity.

Overall,

The column space of A to encompass all possible values of B , A transpose must contain atleast m linearily independent columns where m is the number of rows in the B.

“MATRIX A SHOULD HAVE A SUBSPACE WHERE ALL B CAN BE FOUND FOR X TO EXIST”

It is like , A has to be United States of America , if you are looking to shift from New York to D.C. . If A is India , you cannot shift from New York to D.C. by calling it “intra country address transformation”

Homo Bayesian