# Find two 2x2 matrices that multiply to give 0

• Physics Slayer
In summary, the conversation discusses two ways of multiplying matrices, one by assuming specific values for the elements and the other using trial and error. The concept of rank and its importance in matrix multiplication is also mentioned. The conversation ends with a suggestion to experiment with simple matrices to better understand the relationship between the matrices and their resulting product.
Physics Slayer
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.

I would think of it in terms of images and kernels.

To start, what do the rank of A and B need to be?

topsquark
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?

topsquark
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?

Maarten Havinga, WWGD and topsquark
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:
vela
Just in case, related to what Mathwonk said, look up the Fundamental Theorem of Linear Algebra.

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