Question

If A and B are linear transformations such that AB =0 , A\(\neq 0\), B\(\neq 0\), then det A = det B =0
Please, help me prove it.

Answer 1

Linear transformations are basically matrices, right?

Answer 2

kind of, not exactly the same. But in this case, can consider as the same. Help me, TJ

Answer 3

Thinking... lol
I think I already told you before that matrixes and I do not get along ^_^

But let's see...

Answer 4

Maybe this can help...
\[\Large \det(AB) = \det(A)\det(B) = 0 \]

Answer 5

And then, we can be sure that at least one of either det(A) and det(B) being equal to zero.

Answer 6

hahaha... TJ is the best, just one line!!!

Answer 7

That doesn't quite do it yet, L66... I have only thus far shown that at least one of det(A) and det(B) is zero... not both... yet...

Answer 8

Let's try a different tack... a zero determinant is equivalent to saying that a matrix is singular, right? Maybe that works...

(I'm just tossing ideas here...)

Answer 9

Say something, L66... D:

Answer 10

We're basically done, L66... if you would only see it.... ;)

Answer 11

I have nothing to say, hihihi....

Answer 12

Okay, scratch what I said about determinants... highly unnecessary.

We just have to show that both A and B are singular. Ready? ^_^

Answer 13

ok, let me "say something"
case 1 , det A =0
this implied A x =0
the same with AB , so that A = AB or A - AB =0 --> A(1 - B) = 0
A \(\neq 0\) --> 1 - B =0 or B = I --> det B = 0

Answer 14

do the same if det B = 0

Answer 15

No... det A = 0 does not imply Ax = 0
^_^

Answer 16

The proof is far simpler than what you're thinking, I bet ^_^

Answer 17

can do it because they are linear transformations, not matrices

Answer 18

Ax means the linear transformation A apply on x. not A*x

Answer 19

Linear transformations and the matrix go hand-in-hand.

Even in the case of linear transformations, having a zero determinant doesn't mean A(x) = 0.

Trust me ^_^

Answer 20

So... are you ready to have your mind blown? XD
LOL JK

Let's proceed by contradiction.

Suppose det(A) is not zero.
then A is a non-singular matrix, yes?

Answer 21

so what do you think if
[A] is a matrix and A is a linear transformation
the theorem said that [A] similar to [B] iff there exists an invertible matrix T such that \([A]=[T^{-1}]*[B]*[T]\)
but if they are linear transformation, then A = TBT^-
so that, they work just...... slightly different, but that is important

Answer 22

Obviously, T^- BT cannot = T B T^-, right?

Answer 23

You're overcomplicating things.... I just asked for your affirmation...
Were you paying attention? D:

Answer 24

the problem is not as we think, hehehe... ok, give me your idea. I am waiting for you.

Answer 25

TJ, contradict works well, I need other method to get extra credit

Answer 26

Okay, once again, all we have to show is that both A and B are singular matrices... ok?
So suppose det(A) \(\ne\) 0, then that means A is a non-singular matrix, right?

Answer 27

Yes, Sir

Answer 28

Okay, then since A is a non-singular matrix, its inverse exists, right?

Answer 29

Yes, again.

Answer 30

Then...
\[\Large AB = 0\]
We could multiply both sides on the left by the inverse of A:
\[\Large A^{-1}AB = A^{-1}0 = 0\]

And then... do you see what happens?

Answer 31

contradict to the given information B \(\neq 0\) bingo. so that A cannot be invertible , right?

Answer 32

That is correct.
So we have proved that det(A) = 0

Now can you do the rest, by assuming det(B) \(\ne\) 0 and looking for a contradiction? ^_^

Answer 33

we do the same to get det B =0
and make conclusion, right?

Answer 34

Yes, only you multiply to the right instead of to the left.

And yes, you get the conclusion that both A and B must not be invertible, and therefore, both of their determinants must be zero.


Not so bad... apparently ^_^

Answer 35

Thank you TJ. There is not many students here can help me. That's the reason why I posted the question right after seeing you online. hihihihi

Answer 36

Well, I got lucky this time... don't ever rely on me for linear algebra... it simply isn't my cup of tea :D