Question

For a Diagonal matrix A=[aij]m*n, how do you prove that A^n=[aij^n]m*n?

Answer 1

sorry but you have to write the problem more clearly...

Answer 2

what is aij
?
and is\[m\times n\]just the row and column notation?

Answer 3

yes ij refers to row and column

Answer 4

m x n is the row column notation and aij is the general element located at row i and column j

Answer 5

prove it?
Hm...
maybe by induction, but I'm hoping to circumvent that

Answer 6

it's visually obvious...

Answer 7

that is a pretty difficult job.

Answer 8

I wouldn't know how to write it in LaTeX anyway I don't think :/

Answer 9

well what is a LaTeX anyway

Answer 10

the language to type equations:

\frac{-b\pm\sqrt{b^2-4ac}}{2a}

This is the quadratic formula, just enclose it in \[ with another set of brackets on the right and...

Answer 11

\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]tadah!

Answer 12

here's a cheat sheet
http://omega.albany.edu:8008/Symbols.html

Answer 13

OK even i saw that subscript was not available in the equation editor

Answer 14

the underline symbol:

a_i

\[a_i\]

Answer 15

the fraction thing I did is not in the editor either

Answer 16

Well thank you, but that still does not solve the problem beforehand

Answer 17

but I don't know how to write a matrix larger than 3X3

Answer 18

no I don't think I know an easy proof...
I bet Zarkon Does

Answer 19

We're gonna have to prove that a diagonal matrix multiplied by itself is going to multiply each element in it by itself.
Like I said this is obvious, but how to show it...

Answer 20

us the definition of matrix multiplication

Answer 21

Can we do it using the summation notation?

Answer 22

Row by column stuff you mean?

Answer 23

\[\sum_{1}^{n}a_ir*b_jr\]

Answer 24

close :)

Answer 25

i=j
for all elements in the diagonal matrix
will that fix it?

Answer 26

sorry got it wrong \[\sum_{r=1}^{n}a_ir*a_jr\]

Answer 27

you need to write a_{ir}

\[a_{ir}\]

Answer 28

also your subscripts are not completly correct

Answer 29

and then \[\forall i \neq j , a_{ir}*a_{jr} = 0\]

Answer 30

Well subscripts in summation are still my nightmares

Answer 31

the inside subscripts need to match

Answer 32

I don't believe so .. because row has to be multiplied with the column i.e i*j

Answer 33

I would say listen to Zarkon
He know his stuff

Answer 34

so
i=j=n or whatever
is the idea?

Answer 35

Aceept my apologies Zarkon

Answer 36

not n we already used that
but i and j are the same, and we are using that?

Answer 37

I am saying all products where \[i \neq j\] are zero and where \[i = j\] are a_(ii)^2

Answer 38

i mean {a_{ii}^2}

Answer 39

\[a_{ii}^2\]

Answer 40

exactly, would that help?

Answer 41

since, then for a diagonal matrix of order n it would be \[a_{ii}^n\]

Answer 42

i am using order n because a diagonal matrix is square

Answer 43

I'm not sure how to demonstrate it beyond that either

Answer 44

Is that proof sufficient?
Have we proven it o-0

Answer 45

since !st element of \[A^2\] would be \[a_{ii}^2 +0+0+0+0...\]

Answer 46

\[A^2=\{a_{i,j}\}^2=\left\{\sum_{r}a_{ir}a_{rj}\right\}\]

\[\sum_{r}a_{ir}a_{rj}\] is zero if \(i\ne r\) or \(j\ne r\)
so the only possible non zero values occur when i=r=j

Answer 47

when i=j we have

\[\sum_{r}a_{ir}a_{ri}=0+0+\cdots +a_{ii}^2+0+\cdots +0=a_{ii}^2\]

Answer 48

thank you Zarkon. I believe that finishes it doesn't it?

Answer 49

All hail

Answer 50

I would use induction to be complete

Answer 51

I knew it...

Answer 52

basis step is clearly true now

Answer 53

Well I'm afraid of that. Still a sophomore

Answer 54

So is the assumption step.

Answer 55

just assume \[A^n=\{a_{ij}^n\}\]

prove that \[A^{n+1}=\{a_{ij}^{n+1}\}\]

Answer 56

\[A^{n+1}=A^{n}A\]

Answer 57

then use our assumption

Answer 58

and the definition of matrix multiplaction

Answer 59

\[A^n = A^{n-1}*A\]

Answer 60

It wolud then be \[\sum_{r}^{}a_{ir}^na_{jr}\]

Answer 61

your subscripts are still not correct

Answer 62

I believe substituting value of \[a_{ij}^n\] from our assumption

Answer 63

darn it

Answer 64

but keep it with\[a_{ii}\]yes?

Answer 65

\[\sum_{r}a_{ir}^na_{rj}\]

Answer 66

well how old are the two of you ?

Answer 67

Zarkon is a big boy
I'm just a student like you

Answer 68

how old...old enough :)

Answer 69

Now I realize it was rj in , not jr. Thank you

Answer 70

and again the terms will be zero unless i=r=j

Answer 71

then \[\sum_{r}a_{ir}^na_{ri}=0+\cdots 0+a^n_{ii}a_{ii}+0+\cdots 0=a_{ii}^{n+1}\]

Answer 72

and zero otherwise
so \[A^{n+1}=\{a^{n+1}_{ij}\}\]

Answer 73

Therefore by the principle of mathematical induction...

Answer 74

...we are done
QED

Answer 75

i wish

Answer 76

...we are done :)

Answer 77

Praise to Zarkon!

Answer 78

Bow to Zarkon

Answer 79

if you really wanted to know....I'm 39

Answer 80

and im 17

Answer 81

24
Do you mind saying what you do for a living Zarkon?
I'm just curious to know what kind of stuff you apply your math too.

Answer 82

So am I

Answer 83

I'm a mathematics professor

Answer 84

|gasp|

Answer 85

Well that makes sense :D

Answer 86

Anyone know about the Tranjugate of a matrix?

Answer 87

Never even heard of it.

Answer 88

the conjugate transpose .

Answer 89

oh, then I have, but I can't do anything with it I don't think.

Answer 90

Don't you need that to prove Cramer's rule?

Answer 91

no

Answer 92

damn, taking linear algebra for me next semester

Answer 93

How do you write a conjugate in equation?

Answer 94

have fun...it is a good class

Answer 95

not sure what you are asking

Answer 96

Oh I was thinking of the cofactor matrix...

Answer 97

the bar over a complex number, which denotes it conjugate... z bar for a complex no. z

Answer 98

yes...that you would use