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Given an <math>m \times n</math> [[matrix (mathematics)|matrix]] <math>A</math> of [[rank (linear algebra)|rank]] <math>r</math>, a '''rank decomposition''' or '''rank factorization''' of <math>A</math> is a product <math>A=CF</math>, where <math>C</math> is an <math>m \times r</math> matrix and <math>F</math> is an <math>r \times n</math> matrix.
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'''Every finite dimensional matrix has a rank decomposition:''' Let <math>A</math> be an <math>m\times n</math> matrix whose [[column rank]] is <math>r</math>. Therefore, there are <math>r</math> [[linearly independent]] columns in <math>A</math>; equivalently, the [[dimension]] of the [[column space]] of <math>A</math> is <math>r</math>. Let <math>c_1,c_2,\ldots,c_r</math> be any [[basis (linear algebra)|basis]] for the column space of <math>A</math> and place them as column vectors to form the <math>m\times r</math> matrix <math>C = [c_1:c_2:\ldots:c_r]</math>. Therefore, every column vector of <math>A</math> is a [[linear combination]] of the columns of <math>C</math>. To be precise, if <math>A = [a_1:a_2:\ldots:a_n]</math> is an <math>m\times n</math> matrix with <math>a_j</math> as the <math>j</math>-th column, then
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:<math>a_j = f_{1j}c_1 + f_{2j}c_2 + \cdots + f_{rj}c_r,</math> 
where <math>f_{ij}</math>'s are the scalar coefficients of <math>a_j</math> in terms of the basis <math>c_1,c_2,\ldots,c_r</math>. This implies that <math>A = CF</math>, where <math>f_{ij}</math> is the <math>(i,j)</math>-th element of <math>F</math>.  
 
== rank(<math>A</math>) = rank(<math>A^\text{T}</math>) ==
An immediate consequence of rank factorization is that the rank of <math>A</math> is equal to the rank of its transpose <math>A^\text{T}</math>. Since the columns of <math>A</math> are the rows of <math>A^\text{T}</math>, the [[column rank]] of <math>A</math> equals its [[row rank]].  
 
'''Proof:''' To see why this is true, let us first define rank to mean column rank. Since <math>A = CF</math>, it follows that <math>A^\text{T} = F^\text{T}C^\text{T}</math>. From the definition of [[matrix multiplication]], this means that each column of <math>A^\text{T}</math> is a [[linear combination]] of the columns of <math>F^\text{T}</math>. Therefore, the column space of <math>A^\text{T}</math> is contained within the column space of <math>F^\text{T}</math> and, hence, rank(<math>A^\text{T}</math>) ≤ rank(<math>F^\text{T}</math>). Now, <math>F^\text{T}</math> is <math>n</math>×<math>r</math>, so there are <math>r</math> columns in <math>F^\text{T}</math> and, hence, rank(<math>A^\text{T}</math>) ≤ <math>r</math> = rank(<math>A</math>). This proves that rank(<math>A^\text{T})</math> ≤ rank(<math>A</math>). Now apply the result to <math>A^\text{T}</math> to obtain the reverse inequality: since <math>(A^\text{T})^\text{T}</math> = <math>A</math>, we can write rank(<math>A</math>) = rank(<math>(A^\text{T})^\text{T})</math> ≤ rank(<math>A^\text{T}</math>). This proves rank(<math>A)</math> ≤ rank(<math>A^\text{T}</math>). We have, therefore, proved rank(<math>A^\text{T})</math> ≤ rank(<math>A</math>) and rank(<math>A</math>) ≤ rank(<math>A^\text{T}</math>), so rank(<math>A</math>) = rank(<math>A^\text{T}</math>). (Also see the first proof of column rank = row rank under [[rank (linear algebra)|rank]]).
 
== Rank Factorization from Row Echelon Forms ==
In practice, we can construct one specific rank factorization as follows: we can compute <math>B</math>, the [[row echelon form|reduced row echelon form]] of <math>A</math>. Then <math>C</math> is obtained by removing from <math>A</math> all non-[[gaussian elimination|pivot columns]], and <math>F</math> by eliminating all zero rows of <math>B</math>.
 
== Example ==
 
Consider the matrix
:<math>A = \begin{bmatrix} 1 & 3 & 1 & 4 \\ 2 & 7 & 3 & 9 \\ 1 & 5 & 3 & 1 \\ 1 & 2 & 0 & 8 \end{bmatrix}\sim
\begin{bmatrix} 1 & 0 & -2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}=B\text{.}</math>
<math>B</math> is in reduced echelon form.
Then <math>C</math> is obtained by removing the third column of <math>A</math>, the only one which is not a pivot column, and <math>F</math> by getting rid of the last row of zeroes, so
:<math>C = \begin{bmatrix} 1 & 3 & 4 \\ 2 & 7 & 9 \\ 1 & 5 & 1 \\ 1 & 2 & 8 \end{bmatrix}\text{,}\qquad
F = \begin{bmatrix} 1 & 0 & -2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\text{.}</math>
It is straightforward to check that
:<math>A = \begin{bmatrix} 1 & 3 & 1 & 4 \\ 2 & 7 & 3 & 9 \\ 1 & 5 & 3 & 1 \\ 1 & 2 & 0 & 8 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 4 \\ 2 & 7 & 9 \\ 1 & 5 & 1 \\ 1 & 2 & 8 \end{bmatrix}\begin{bmatrix} 1 & 0 & -2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} = CF\text{.}</math>
 
== Proof ==
Let <math>P</math> be an <math>n\times n</math> permutation matrix such that <math>AP=(C,D)</math> in [[block matrix|block partitioned]] form, where the columns of <math>C</math> are the <math>r</math> pivot columns of <math>A</math>. Every column of <math>D</math> is a linear combination of the columns of <math>C</math>, so there is a matrix <math>G</math> such that <math>D=CG</math>, where the columns of <math>G</math> contain the coefficients of each of those linear combinations. So <math>AP=(C,CG)=C(I_r,G)</math>, <math>I_r</math> being the <math>r\times r</math> identity matrix. We will show now that <math>(I_r,G)=FP</math>.
 
Transforming <math>AP</math> into its reduced row echelon form amounts to left-multiplying by a matrix <math>E</math> which is a product of [[elementary matrix|elementary matrices]], so <math>EAP=BP=EC(I_r,G)</math>, where <math>EC=\begin{pmatrix} I_r \\ 0 \end{pmatrix}</math>. We then can write <math>BP=\begin{pmatrix} I_r & G \\ 0 & 0 \end{pmatrix}</math>, which allows us to identify <math>(I_r,G)=FP</math>, i.e. the nonzero <math>r</math> rows of the reduced echelon form, with the same permutation on the columns as we did for <math>A</math>. We thus have <math>AP=CFP</math>, and since <math>P</math> is invertible this implies <math>A=CF</math>, and the proof is complete.
 
==References==
{{refbegin}}
* {{Citation | last = Lay | first = David C. | date = 2005 | title = Linear Algebra and its Applications | publisher = Addison Wesley | edition = 3rd | isbn = 978-0-201-70970-4}}
* {{Citation | last = Golub | first = Gene H. | last2 = Van Loan | first2 = Charles F. | date = 1996 | title = Matrix Computations | series = Johns Hopkins Studies in Mathematical Sciences | publisher = The Johns Hopkins University Press | edition = 3rd | isbn = 978-0-8018-5414-9}}
* {{Citation | last = Stewart | first = Gilbert W. | date = 1998 | title = Matrix Algorithms. I. Basic Decompositions  | publisher = SIAM | isbn = 978-0-89871-414-2}}
{{refend}}
 
[[Category:Matrix decompositions]]
[[Category:Linear algebra]]

Latest revision as of 05:52, 11 January 2015

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