Find two 2x2 matrices that multiply to give 0

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SUMMARY

To find two 2x2 matrices A and B such that their product AB equals the zero matrix, one must consider the ranks of both matrices. The rank of A and B must be less than 2, allowing for multiple solutions due to the freedom in selecting variable combinations. A practical approach involves experimenting with simple matrices containing only 0s and 1s, and understanding the relationship between the rows of A and the columns of B. The discussion highlights the importance of the Fundamental Theorem of Linear Algebra in this context.

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Homework Statement
Find two ##2x2## matrices ##A## and ##B## such that ##AB = 0## but ##BA\neq0##
Relevant Equations
AB=0
One way would be to assume
$$A= \begin{bmatrix}a_1 & a_2\\a_3 & a_4 \end{bmatrix}$$ and $$B=\begin{bmatrix}b_1 & b_2\\b_3 & b_4\end{bmatrix}$$ and then multiply but then you end up with 4 equations and 8 variables, how would that work?

the other way would be to use trial and error, any help would be appreciated.
 
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I would think of it in terms of images and kernels.

To start, what do the rank of A and B need to be?
 
Office_Shredder said:
I would think of it in terms of images and kernels.

To start, what do the rank of A and B need to be?
I am unfamiliar with terms like images and kernels.

both A and B are 2x2 matrices
 
Do you know what the rank of a matrix is?
 
Office_Shredder said:
Do you know what the rank of a matrix is?
I thought its 2x2 its given in the question
 
do you know that AB = 0 means the rows of A are perpendicular to the columns of B?
 
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Physics Slayer said:
you end up with 4 equations and 8 variables, how would that work?
That just means that there are a lot of solutions because you have a lot of freedom to pick variable value combinations that work.
Physics Slayer said:
the other way would be to use trial and error, any help would be appreciated.
Experiment with simple matrices with only 0s and 1s as elements. Find how to make a matrix that will zero a row. Find out how to make a matrix that will move a row. Suppose ##B## zeros a row and ##A## moves that row. What happens if ##B## zeros the row before ##A## moves it versus ##A## moving the row before ##B## zeros it? You can use that to get the two cases ##AB = 0## and ##BA \ne 0##.
 
Last edited:
Just in case, related to what Mathwonk said, look up the Fundamental Theorem of Linear Algebra.
 

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