Linear Algebra - Finding Rank

  • #1

Homework Statement



Suppose that AB = 0, where A is a 3 x 7 full rank matrix and B is 7 x 53. What is the highest possible rank of matrix B.

Homework Equations




The Attempt at a Solution


Since each column of B is in the null space of A, the rank of B is at most 4.

I don't understand why it is 4.

What operation do I need to perform here to understand this? I am not a student, I am just trying to remember Linear Algebra.
 

Answers and Replies

  • #2
fresh_42
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Instead of these unwieldy matrices, let us consider the linear transformations they represent. For the sake of simplicity, let's denote the dimension of a vector space by an index. Then we have ##A\, : \,V_7 \longrightarrow V_3## and ##B\, : \, V_{53} \longrightarrow V_7\,.## So ##B## sends at least ##53-7=46## dimensions to zero anyway. Thus its rank must be between ##7## and ##0##. ##A## on the other hand sends exactly ##4## dimensions to zero, as it is of full rank. But both applied in a row: $$V_{53} \stackrel{B}{\longrightarrow} V_7 \stackrel{A}{\longrightarrow} V_3$$ sends all ##53## dimensions to zero. Now how many can ##B## leave from those ##7##, which will be left for ##A## to be sent to zero?
 
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  • #3
So, if Matrix A were instead a 8 x 3 matrix, then B would have a rank of five? What I am seeing is that it is the subtraction of columns from rows in order to find the rank of B that maps A to 0?
 
  • #4
fresh_42
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So, if Matrix A were instead a 8 x 3 matrix, then B would have a rank of five? What I am seeing is that it is the subtraction of columns from rows in order to find the rank of B that maps A to 0?
In this case you couldn't multiply the matrices in the first place. This means for the transformations, that we need to explain, what happens between the 8-dimensional image of ##B## and the 7-dimensional space, on which ##A## is defined.

You can translate it into row and column actions, too. But as we don't have a certain example, we can assume that the matrices are already in a form which is nice:
$$
A=\begin{bmatrix}I_3 & | & 0_4 \end{bmatrix}\, , \,B=\begin{bmatrix} B_7^{\,'}&| &0_{46}\end{bmatrix}
$$
with the ##(3\times 3)## identity matrix ##I_3## and any ##(7 \times 7)## matrix ##B_7^{\,'}##. This wouldn't change the result but is easier to handle. Now calculate ##AB =0## and see what does this mean for ##B_7^{\,'}## and its maximal rank.

Here are a couple of formulas which also might help occasionally: https://en.wikipedia.org/wiki/Rank_(linear_algebra)#Properties
In your example above I basically used https://en.wikipedia.org/wiki/Rank–nullity_theorem
 
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  • #5
Sorry about that. I mean a a 2 x 7 matrix. That way the numbers would change a bit.

So, in this case(Your new example), the Rank(A) = 3, Rank(B) = unknown and the Rank(AB) = 0. Which is exactly the same as the previous example. I appreciate your help but I think that I am going to read through an introductory book again. It is apparent that I don't remember enough. Thank you!
 
  • #6
StoneTemplePython
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A couple other approaches:

1.) Work through the Sylvester rank Inequality proof. Then apply it here.
2.) Less general, using gramm schmidt in reals, you can reason that all columns both B are orthogonal to the rows of A, and hence this implies what?
 

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