Rank Factorization of Matrix

Definition (Rank Factorization of Matrix).

Let be a matrix of rank . Rank factorization of is a pair of matrices where is of order and is of order , such that

Theorem.

Let is and be then also, TFAE
i)k=r(A), , ( ,Q) is a rank factorization
ii) ie, is a full column matrix and is a full row matrix
iii) columns of form a basis
iv) rows of form a basis of

Proof.

not to prove the equivalence of condition i to iv

ii
let

ii iii
is a full column rank matrix ie, columns of are linearly independent
is a full row rank matrix ie, has right inverse say, .

we need to show
Its easy to see that
if

as columns of is linearly independent columns of form basis of

iii i) easy

Complete this by showing ii=>iv and iv<=ii
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Corollary.

If is a rank factorization of A then
i)
ii)
iii)

Proof.

If
if
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Theorem.

Let A be of the same order then

equality holds iff and

Proof.

So the column space of is a subspace of the sum of the column space of A and column space of

suppose equality holds then

As equality holds

Conversely, let and
Let and
if we are done
so lets assume
let be a rank factorization of and that of B

then
if and
then

Case I if
also columns of and are linearly independent
columns of are linearly independent from ()
the columns of are linearly independent columns of .
They form a basis of
so is a rank factorization of and

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Theorem.

Let A be of the same order then

equality holds iff and is a direct sum

Proof.

If
then

Suppose equality holds then the sum is a direct sum
Conversely assume that the sum is a direct sum

Let if its less than 1 ie, 0 ignore it
and let be a rank fact of

let

As each of the 's are full column rank
the set of all columns of is a linearly independent set as the sum is a direct sum
similarly all rows of are linear independent
Hence is a full column matrix and is a full row matrix hence is a the Rank Factorization Rank Factorization of

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