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?

- anonymous

- schrodinger

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- myininaya

What is An?

- anonymous

the matrix

- anonymous

commonly seen as Ax=b.
this is An=0.

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## More answers

- myininaya

Oh I thought there might have been a difference
because you called the matrix A then you called it An

- myininaya

Are you saying n is an eigenvector ?

- anonymous

honesly i dont know wat to do. this is what the teacher asked us

- myininaya

Or are you say An is the matrix?

- anonymous

idk..lol

- myininaya

Ok I think
|dw:1340595284758:dw|
and
|dw:1340595304816:dw|

- myininaya

Is that what you think?

- myininaya

But that doesn't make since that A_n would be the matrix with nothing but zero entries

- myininaya

do you mean the determinant is 0?

- myininaya

|A_n|=0?

- myininaya

that would make since

- myininaya

sense*

- myininaya

I think that is what you mean

- myininaya

so do you know how to find the determinant of a matrix?

- anonymous

ya ad-bc

- myininaya

Try finding the determinant of A_2 ?
What do you get?

- anonymous

zero

- myininaya

Ok what about A_3

- anonymous

same

- myininaya

Ok so we have convinced ourselves that |A_n|=0
But we must prove it

- myininaya

|dw:1340595692432:dw|

- myininaya

I would just show a little work for this
show a pattern you know

- myininaya

like how you did for A_3 and then do the nth term
you know what I mean?

- anonymous

how do u find determinant for 3x3 or bigger matrix?

- myininaya

oh ok for an A_3
|dw:1340595913293:dw|

- myininaya

|dw:1340595950507:dw|

- myininaya

The signs alternate

- myininaya

Like you take top entries

- anonymous

only the first row?

- myininaya

And take everything that isn't below that entry

- anonymous

like if its 10x10 u still only do the first row /?

- myininaya

For |A_4|
|dw:1340596049002:dw|

- myininaya

|dw:1340596102188:dw|

- myininaya

and you already know how to find the determinant for a 3 by 3

- myininaya

Same thing just take the top entries and do the signs alternating

- anonymous

\[
A=\left(
\begin{array}{cccc}
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 4 \\
0 & 0 & 0 & 3 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)\\
A^2=\left(
\begin{array}{cccc}
0 & 0 & 1 & 10 \\
0 & 0 & 0 & 3 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)\\
A^3=\left(
\begin{array}{cccc}
0 & 0 & 0 & 3 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)\\
A^4=\left(
\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)
\]

- anonymous

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

- myininaya

yes just like i did above for the 3 by 3

- myininaya

But not all the top entries are 0

- anonymous

@eliassaab i dont really understnad wat u wrote

- anonymous

right...

- anonymous

See my example below and examine what is going on?

- anonymous

you have zeros. but what are you doing to the matrix?

- myininaya

Do you think he means to raise A to a power @eliassaab ?

- myininaya

Instead of finding the determinant ?

- anonymous

You raise it to the power 2, then 3, then 4.

- myininaya

Ok I'm sorry @liliy I don't know what your question is asking anymore.

- anonymous

You do not need to deal with determina

- anonymous

determinant

- anonymous

so can you start over with me?

- anonymous

what does a^n=0 even mean?

- anonymous

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.

- anonymous

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.

- myininaya

So you are just giving another way right @eliassaab Do you think I interpreted is question correctly?

- anonymous

@myininaya, you do not need determinant to do that,

- myininaya

Yeah I know, but I'm asking you if I interpreted it correctly?

- anonymous

@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..

- anonymous

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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