Help with full rank factorization

In summary, the conversation discusses the existence of a full rank factorization for an arbitrary m x n matrix. It is stated that if \textit{A} \in \textbf{R}^{m x n} has a rank of \textit{r}, then there exist matrices \textit{B} \in \textbf{R}^{m x r} and \textit{C} \in \textbf{R}^{r x n} such that \textit{A = BC}. It is also mentioned that \textit{rank(A) = rank(B) = r}. The conversation then proceeds to discuss proving the second property by assuming the first and using the properties of rank.
  • #1
learningstill
3
0
I've been tasked with proving the existence of a full rank factorization for an arbitrary m x n matrix, namely:

Let [itex]\textit{A}[/itex] [itex]\in[/itex] [itex]\textbf{R}^{m x n}[/itex] with [itex]\textit{rank(A) = r}[/itex] then there exist matrices [itex]\textit{B}[/itex] [itex]\in[/itex] [itex]\textbf{R}^{m x r}[/itex] and [itex]\textit{C}[/itex] [itex]\in[/itex] [itex]\textbf{R}^{r x n}[/itex] such that [itex]\textit{A = BC}[/itex]. Furthermore [itex]\textit{rank(A) = rank(B) = r}[/itex].

I think I can prove the second property if I assume the first using [itex]\it{rank(AB)}[/itex] [itex]\leq[/itex] [itex]\it{rank(A)}[/itex] and [itex]\it{rank(AB)}[/itex] [itex]\leq[/itex] [itex]\it{rank(B)}[/itex].

I'd appreciate a push in the right direction. Thanks.
 
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  • #2
We have the following situation:
$$
V_n \stackrel{C}{\twoheadrightarrow} V_n/\operatorname{ker}A \cong V_r \cong \operatorname{im}A \stackrel{B}{\rightarrowtail} V_m
$$
which together ##BC## transform as ##A##. We have split the domain of ##A\, : \,V_n \longrightarrow V_m ## into ##V_n \cong \operatorname{ker}A \oplus V_n/\operatorname{ker}A## and the codomain in ##\operatorname{im}A \oplus V_{m-r}\,.##
 

1. What is full rank factorization?

Full rank factorization is a mathematical method used to decompose a matrix into a product of two or more matrices. It is often used in linear algebra and can be useful in solving systems of equations and finding eigenvalues.

2. Why is full rank factorization important?

Full rank factorization allows us to break down a complex matrix into simpler parts, making it easier to understand and manipulate. It is also useful in various applications, such as data analysis, image processing, and computer graphics.

3. How is full rank factorization different from other matrix decomposition methods?

Full rank factorization is unique in that it requires the matrix to have full rank, meaning that all of its columns are linearly independent. Other decomposition methods, such as LU decomposition or QR decomposition, do not have this requirement.

4. What are the benefits of using full rank factorization?

One of the main benefits of full rank factorization is that it can simplify complex calculations and make them more efficient. It can also provide insight into the structure of a matrix, which can be useful in understanding the underlying data or system being represented.

5. Can full rank factorization be performed on any type of matrix?

No, full rank factorization can only be performed on matrices that have full rank. This means that the matrix must have at least as many linearly independent rows as columns. If a matrix does not have full rank, other decomposition methods may be used instead.

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