How do I solve linear algebra questions involving matrices?

[vg19]

Hi,
I am having trouble with the following questions
1) Let U be such that AU=0 implies that A=0. If AU=BU, show that A=B.
So far, I did this, but it doesn't seem right to me.
To Show:
A=B
0=A
=AU
=BU
=0
Therefore A = B.
2)If A=
[a b]
[c d] (2x2 matrix, sorry not to sure on how to place them here)

where a is not equal to 0, show that A factors in the form A =
[1 0][y z]
[x 1][0 w] (those are two 2x2 matricies multiplied together)

Im not too sure on how to start on this question at all.

Thanks in advance

 

[LeonhardEuler]

Your mistake in the first one is in assuming A=0. This is not necessarily the case. You want to show that $AU=BU \rightarrow A=B$, or equivalently $AU-BU=0 \rightarrow A=B$. You never said what kind of objects A, B and U are, but presumably they have some sort of linearity property, correct? If this is the case, then it is easy to rearange the equation to get it to say something times U = 0. You are told in the problem that this implies that something is zero. Take it from there.

 

[vg19]

Sorry, A, U, and B are matricies. I am not too sure I understand the part when you said A may not be 0. What does it mean when it says "implies that A = 0"?

Thanks!

 

[LeonhardEuler]

Sorry, A, U, and B are matricies. I am not too sure I understand the part when you said A may not be 0. What does it mean when it says "implies that A = 0"?
Thanks!

When they say "AU=0 implies that A=0" they mean that if AU=0, then A=0 where A is any matrix that can be multiplied by U. Don't get confused by the fact that they later use the same letter when they say "If AU=BU, show that A=B". When they say A, they mean any matrix.

 

[HallsofIvy]

If AU= BU then AU-BU= 0. That's what you need to use "If AU= 0,...".

 

[vg19]

Great! I am pretty sure I understand it now. Thanks a lot