Show that if A is nxn and has all 0’s on and below the diagonal then An = 0. Hint: do not at first be too ambitious. First find A2 and observe something useful about it. What about A3?

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Show that if A is nxn and has all 0’s on and below the diagonal then An = 0. Hint: do not at first be too ambitious. First find A2 and observe something useful about it. What about A3?

Mathematics
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What is An?
the matrix
commonly seen as Ax=b. this is An=0.

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Oh I thought there might have been a difference because you called the matrix A then you called it An
Are you saying n is an eigenvector ?
honesly i dont know wat to do. this is what the teacher asked us
Or are you say An is the matrix?
idk..lol
Ok I think |dw:1340595284758:dw| and |dw:1340595304816:dw|
Is that what you think?
But that doesn't make since that A_n would be the matrix with nothing but zero entries
do you mean the determinant is 0?
|A_n|=0?
that would make since
sense*
I think that is what you mean
so do you know how to find the determinant of a matrix?
ya ad-bc
Try finding the determinant of A_2 ? What do you get?
zero
Ok what about A_3
same
Ok so we have convinced ourselves that |A_n|=0 But we must prove it
|dw:1340595692432:dw|
I would just show a little work for this show a pattern you know
like how you did for A_3 and then do the nth term you know what I mean?
how do u find determinant for 3x3 or bigger matrix?
oh ok for an A_3 |dw:1340595913293:dw|
|dw:1340595950507:dw|
The signs alternate
Like you take top entries
only the first row?
And take everything that isn't below that entry
like if its 10x10 u still only do the first row /?
For |A_4| |dw:1340596049002:dw|
|dw:1340596102188:dw|
and you already know how to find the determinant for a 3 by 3
Same thing just take the top entries and do the signs alternating
\[ A=\left( \begin{array}{cccc} 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)\\ A^2=\left( \begin{array}{cccc} 0 & 0 & 1 & 10 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)\\ A^3=\left( \begin{array}{cccc} 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)\\ A^4=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \]
but the determinant of the new 3x3 is gonna also be broken down right? .. im ur case its zero bec the coefficient is zero so it odsnt really mater
yes just like i did above for the 3 by 3
But not all the top entries are 0
@eliassaab i dont really understnad wat u wrote
right...
See my example below and examine what is going on?
you have zeros. but what are you doing to the matrix?
Do you think he means to raise A to a power @eliassaab ?
Instead of finding the determinant ?
You raise it to the power 2, then 3, then 4.
Ok I'm sorry @liliy I don't know what your question is asking anymore.
You do not need to deal with determina
determinant
so can you start over with me?
what does a^n=0 even mean?
Any matrix like yours, when you raise it to the power 2, you get what is first above the diagonal is zero When you raise it to the power 2, you get the first and the second above the diagonal to be zero. When you raise it to the power 3, you get the first and the second and third above the diagonal to be zero. When you raise it to the power 4, you get everything zero.
Look at A^4 in my example above to see that A^4=0, this means all the entries of the matrix A^4 are zeros.
So you are just giving another way right @eliassaab Do you think I interpreted is question correctly?
@myininaya, you do not need determinant to do that,
Yeah I know, but I'm asking you if I interpreted it correctly?
@eliassaab i dont undesrtnad how you started to do the problem. my teacher said start with a^2 .. and move to bigger ones... so wat is a= to a 4x4 and then writing a^2... a^3..
Here is a quick proof using the characteristic polynomial f(x) of the matrix A that says the f(A)=0. Our matrix has\( f(x)=x^n\), hence \(f(A)=A^n=0\)

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