Matrix Problem: Find A and B such that A = O, B =O, AB= O and BA =O

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The discussion centers on finding two non-null matrices A and B such that both products AB and BA yield the zero matrix, while maintaining that neither A nor B is a null matrix. Participants suggest that both matrices should be singular, with at least one being nilpotent. The use of triangular matrices is recommended for simplicity, along with experimenting with symmetric rank one matrices and full rank diagonal matrices. The relevant equations include A! = O, B! = O, AB = O, and BA! = O.

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Crystal037
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Homework Statement
Find a matrix such that A and B such that A not equal to null matrix B not equal to null matrix AB= null matrix and BA not equal to null matrix
Relevant Equations
A! = O, B! =O, AB= O and BA! =O
Let
A=
[ a b]
[ c d ]
B =
[ w x]
[ y z]
Then aw +by=0 bx+dz=0
cw+dy=0 cx+dz=0
aw+cx! =0 bw+x! =0
ya+cz!=0 by+dz! =0
But I don't get the answer after this
 
Last edited:
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Crystal037 said:
Problem Statement: Find a matrix such that A and B such that A not equal to null matrix B not equal to null matrix AB= null matrix and BA not equal to null matrix
I'm not sure I follow your problem statement. Do ##A## and ##B## both need to be square? The fact that ##BA \neq \mathbf 0## implies that ##A## and ##B## are both non-zero.

The way to think about this is both matrices are singular. You'll want at least one to be nilpotent... in either case they both have 'a lot' of zero eigenvalues. What's the simplest way to come up with an example fitting that? Make both of them triangular matrices

edit:
another approach would be to select a mix of other 'easy' matrices -- perhaps a symmetric rank one matrix, and full rank diagonal matrix... playing around with triangular matrices is probably easier though.
 
Last edited:
Crystal037 said:
Problem Statement: Find a matrix such that A and B such that A not equal to null matrix B not equal to null matrix AB= null matrix and BA not equal to null matrix
Relevant Equations: A! = O, B! =O, AB= O and BA! =O

Let
A=
[ a b]
[ c d ]
B =
[ w x]
[ y z]
Then aw +by=0 bx+dz=0
cw+dy=0 cx+dz=0
aw+cx! =0 bw+x! =0
ya+cz!=0 by+dz! =0
But I don't get the answer after this
Pick ##a,\ b,\, w,\, y## such that ##aw = -by## and have all other entries be zero.
 
The matrix multiplies seem wrong. I'm thinking:

$$AB = \begin{bmatrix} aw+by & ax+bz \\ cw+dy & cx+dz \end{bmatrix} = 0$$
$$BA = \begin{bmatrix} aw+cx & bw+dx \\ ay+cz & by+dz \end{bmatrix} \neq 0$$
 
Last edited:
It certainly is possible.

Here's a screen shot from a Ti-84 graphics calculator.

244085


Of course, that does not give the entries for the matrices, but I pretty much followed the advice in Post #3 in constructing them..
 
Last edited:

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