Facebook Page
Twitter
RSS
+ Reply to Thread
Results 1 to 9 of 9
  1. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    2,588
    Thanks
    2,073 times
    Thanked
    663 times
    Trophies
    1 Highscore
    Awards
    MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #1
    Hey!!

    Let $\mathbb{K}$ be a fiels and $A\in \mathbb{K}^{p\times q}$ and $B\in \mathbb{K}^{q\times r}$.
    I want to show that $\text{Rank}(AB)\leq \text{Rank}(A)$ and $\text{Rank}(AB)\leq \text{Rank}(B)$.

    We have that every column of $AB$ is a linear combination of the columns of $A$, or not?

    What can we do to show the inequality?

  2. MHB Master
    MHB Global Moderator
    MHB Math Scholar
    MHB POTW Director
    Euge's Avatar
    Status
    Offline
    Join Date
    Jun 2014
    Posts
    1,387
    Thanks
    1,367 time
    Thanked
    3,484 times
    Thank/Post
    2.512
    Awards
    MHB Advanced Algebra Award (2016)  

MHB Topology and Advanced Geometry Award (2015)  

MHB Analysis Award (2015)  

Graduate POTW Award (2014)  

MHB Advanced Algebra Award (2014)
    #2
    It follows from the fact that the column space of $AB$ is contained in the column space of $A$, and the row space of $AB$ is contained in the row space of $B$. I'll prove the latter. Given $y$ in the row space of $AB$, there is a row vector $x$ such that $y=xAB$. The vector $z := xA$ is a row vector such that $y=zB$. Thus, $y$ belongs to the row space of $B$.

  3. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    2,588
    Thanks
    2,073 times
    Thanked
    663 times
    Trophies
    1 Highscore
    Awards
    MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #3 Thread Author
    Quote Originally Posted by Euge View Post
    Given $y$ in the row space of $AB$, there is a row vector $x$ such that $y=xAB$.
    The row space of $AB$ contains the rows of $AB$ that spans the rows, right?
    Then $y\in R(AB)$ means that we can write $y$ a linear combination of the vectors of $R(AB)$, right?
    We can write this as $y=xAB$ because:
    Let $x^T=(x_1, x_2, \ldots , x_{n-1}, x_n)$, then we have that $$xAB=\left (x_1 \cdot (1. \text{ row of }AB))+(x_2 \cdot (2. \text{ row of }AB))+\ldots +(x_{n-1} \cdot ((n-1). \text{ row of }AB))+(x_n \cdot (n. \text{ row of }AB))\right )$$ where some of the $x_i$'s might be $0$.
    Is this correct?

  4. MHB Master
    MHB Global Moderator
    MHB Math Scholar
    MHB POTW Director
    Euge's Avatar
    Status
    Offline
    Join Date
    Jun 2014
    Posts
    1,387
    Thanks
    1,367 time
    Thanked
    3,484 times
    Thank/Post
    2.512
    Awards
    MHB Advanced Algebra Award (2016)  

MHB Topology and Advanced Geometry Award (2015)  

MHB Analysis Award (2015)  

Graduate POTW Award (2014)  

MHB Advanced Algebra Award (2014)
    #4
    It looks good, although $x^T$ should just be written as $x$, since $x$ is already a row vector.

  5. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    2,588
    Thanks
    2,073 times
    Thanked
    663 times
    Trophies
    1 Highscore
    Awards
    MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #5 Thread Author
    Quote Originally Posted by Euge View Post
    It looks good, although $x^T$ should just be written as $x$, since $x$ is already a row vector.
    Ah ok!!

    Let $y\in C(AB)$ then there is a column vector $x$ such that $y=ABx$. The vector $z:Bx$ is a column vector such that $y=Az$. Therefore, $y$ belongs to the column space of $A$.

    Is this correct?


    Having that $C(AB)\subseteq C(A)$ and $R(AB)\subseteq R(B)$, how exactly does it imply that $\text{Rank}(AB)\leq \text{Rank}(A)$ and $\text{Rank}(AB)\leq \text{Rank}(B)$ ?

  6. MHB Master
    MHB Global Moderator
    MHB Math Scholar
    MHB POTW Director
    Euge's Avatar
    Status
    Offline
    Join Date
    Jun 2014
    Posts
    1,387
    Thanks
    1,367 time
    Thanked
    3,484 times
    Thank/Post
    2.512
    Awards
    MHB Advanced Algebra Award (2016)  

MHB Topology and Advanced Geometry Award (2015)  

MHB Analysis Award (2015)  

Graduate POTW Award (2014)  

MHB Advanced Algebra Award (2014)
    #6
    Your work is correct. To finish the argument, use the fact that the row rank of a matrix equals its column rank.

  7. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    2,588
    Thanks
    2,073 times
    Thanked
    663 times
    Trophies
    1 Highscore
    Awards
    MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #7 Thread Author
    Quote Originally Posted by Euge View Post
    Your work is correct. To finish the argument, use the fact that the row rank of a matrix equals its column rank.
    So, we have that $\text{Rank}(AB)=|C(AB)|=|R(AB)|$ and $\text{Rank}(A)=|C(A)|$ and $\text{Rank}(B)=|R(B)|$, right?

    Since $C(AB)\subseteq C(A)$, it follows that $|C(AB)|\leq |C(A)|$, or not?

    And since $R(AB)\subseteq R(B)$, it follows that $|R(AB)|\leq |R(B)|$.

    Therefore, we have that $\text{Rank}(AB)=|C(AB)|\leq |C(A)|=\text{Rank}(A)$ and $\text{Rank}(AB)=|R(AB)|\leq |R(B)|=\text{Rank}(B)$.

    Is this correct?

  8. MHB Master
    MHB Global Moderator
    MHB Math Scholar
    MHB POTW Director
    Euge's Avatar
    Status
    Offline
    Join Date
    Jun 2014
    Posts
    1,387
    Thanks
    1,367 time
    Thanked
    3,484 times
    Thank/Post
    2.512
    Awards
    MHB Advanced Algebra Award (2016)  

MHB Topology and Advanced Geometry Award (2015)  

MHB Analysis Award (2015)  

Graduate POTW Award (2014)  

MHB Advanced Algebra Award (2014)
    #8
    Looks great!

  9. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    2,588
    Thanks
    2,073 times
    Thanked
    663 times
    Trophies
    1 Highscore
    Awards
    MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #9 Thread Author
    Thank you very much!!

Similar Threads

  1. Replies: 10
    Last Post: February 19th, 2015, 09:00
  2. Rotations, Complex Matrices and Real Matrices
    By Peter in forum Linear and Abstract Algebra
    Replies: 7
    Last Post: February 18th, 2015, 08:38
  3. Replies: 0
    Last Post: November 12th, 2014, 16:45
  4. Rank of adj(A)
    By Yankel in forum Linear and Abstract Algebra
    Replies: 1
    Last Post: December 4th, 2013, 11:09
  5. Rank of the product of two matrices
    By aukie in forum Linear and Abstract Algebra
    Replies: 1
    Last Post: July 20th, 2012, 17:13

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards