Rank

#24-sep
Rank of a matrix Let be a matrix,
column space of is the subspace of spanned by the columns of
if are the columns of and is in then,
column space of

So, column space of is all such that

has a solution.

In terms of liner transformation defined by the linear matrix

Column rank of is the dimension of the column space of
Row rank of is the dimension of the row space of

Theorem.

For a matrix the row rank is same as the column space of and is called the rank of

Rank of product of matrices

Let denote the rank of
Let us try to find an upper bound for

Theorem.

Let and be matrices with same number of rows
Then, column space column space of iff for some matrix C

Let
Let columns space of
Let column space of
Conversely assume column space column space of
In particular every column of lies in the column space of .
Which implies every column of can be written as the linear combination of the columns of
So if are the column vectors of then there exist scalars

Let and

Theorem.

Let , be matrices with same number of columns then the row space of row space of A iff for a matrix
complete it

Definition (let

be a

matrix

is said to be of full column rank if ).

and is said to be full row rank is

Lemma.

if have full column rank, then has a left inverse, ie, a such that

has a full column rank, the columns of are linear independent

#26-sep

Theorem.

Let be a matrix. Then the following are equivalent
i) has a right inverse
ii)
iii)
iv) ie, is a full row rank
V) column space of A is

Proof.
ii
If has a right inverse such that
if

ii) iii)
clear
iii) => iv)

let the rows of be

by iii)
the rows are linear independent
iv)v)

column space of
But

V)i)

Col space of A
let be the unit vectors in

as columns space each is a linear combination of columns in :
so there exist scalars such that

so,

Theorem.

Let be a matrix. The following are equvalent
i) has a left inverse
ii)
iii)
iv) ie, A has full column rank
row space of

iii)iv)
let the columns be
let be a linear relation

iii) that is

iv) V)

v) i)
As row space each of the basis vectors of in a linear combination of the rows of A let the rows be . So, there exist scalars
Take and

Theorem.

Let a square matrix of order A TFAE
i) A has a right inverse
ii)
iii) A has a left inverse
iv) A has an inverse

Theorem.

Proof.
by ^8cc1ad Theorem 3.2
also by ^72e467 Theorem 3.3
thus
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Corollary.

if is of full column rank

Proof.
has a left inverse say

ii) if is a full row rank
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Exercise.

A is a matrix of rank 1 iff where and are column vectors

Proof.
If then rank as and as

Conversely let
That is, the dimension of column space

#1-oct

Let denote the column space
denote row space
denote the null space of A

Theorem.

Let be an matrix over and be a subspace of is a field let
Let be null space of then is a subspace of and
Then

Proof.

Rank Nullity for says that
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Theorem.

If is defined, then

Proof.

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Theorem (Sylvester's inequality).

Let be matrices of order and ,
then
the equality holds iff

Proof.

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