Linear Algebra - Finding Rank

[TickleTackleTock]

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.

 

[fresh_42]

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?

 

[TickleTackleTock]

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?

 

[fresh_42]

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

 

[TickleTackleTock]

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!

 

[StoneTemplePython]

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?