Question

Show that \((A-I)^{-1}=I+A+A^{2}+A^{3}\) if \(A^{4}=0\)

Answer 1

Try pre-multiplying with (A-I) on both sides...

Answer 2

One second thought, this was not the question! I've done this one. Sorry!

Answer 3

On second thought*

Answer 4

I'm editing it!

Answer 5

Yeah sure!

Answer 6

So what's the new question?

Answer 7

I'm taking a picture. It's a big matrix and I'm supposed to use \(sgn(\sigma)\) to calculate it.

Answer 8

Are you familiar with how to derive the geometric series:

\[\frac{1}{1-x} = 1+x+x^2+x^3+x^4+\cdots\]

Answer 9

Kai does that work for matrices as well??

Answer 10

Yeah, I am.

Answer 11

@vishweshshrimali5 It does if you're careful :O Follow the steps you'd take to derive it for scalars but do it in a matrix sorta way. :D

Answer 12

:p Ohh well... basically like starting from beginning :P

Answer 13

I am supposed to use permutations and the signal function, as the Leibniz/Laplace formula

Answer 14

But a tree approach wouldn't work here? Like, all possible multiplications because it's a matrix with a lot of 0's.

Answer 15

Nevermind the Laplace method. Just the Leibniz one.

Answer 16

You have to show the equation for this matrix?

Answer 17

I'm forgetting something about the \(sgn(\sigma)\) function.

Answer 18

What are these words? Don't we just have to evaluate the determinant?

Answer 19

Just use alternative positive and negative coeff. that's what signum function is all about

Answer 20

I'm supposed to evalute using the Leibniz method! But might as well might find an easier way.

Answer 21

Wait! Let's go for Leibniz if it's specified

Answer 22

@vishweshshrimali5, the wiki article isn't much of help.

Answer 23

Convert it to a diagonal matrix. You'll just need two row operations.
Not familiar with the other one though.

Answer 24

\[\det(A)=\sum_{\sigma \in S_n}\text{sgn}(\sigma)\prod_{i=1}^{n} a_{\sigma(i),i}\]

Answer 25

Here we have 5 by 5 matrix

Answer 26

So there will be 5! combinations

Answer 27

Am I wrong or there will be a lot of operations?

Answer 28

For signum:

We have to see the oddness/evenness of the sequence...

Answer 29

There will be a big lot but because we have so many zeroes... many should become zero

Answer 30

Now, evenness or oddness can be defined as follows: the permutation is even
(or odd) if the new sequence can be obtained by an even number (odd, respectively) of switches of numbers. Right?

Answer 31

Hm... Sounds okay, but how about the number of switches? I'm assuming this is not arbitraty otherwise this method would fail.

Answer 32

Naah it's not arbitrary ... we find out the evenness/oddness w.r.t [1,2,3,4,5] set

Answer 33

Anyways so if I write the det using Leibniz formula:

\[det(A) = (-1)^0 a_{1,1} a_{2,2} a_{3,3} a_{4,4} a_{5,5} + (-1)^1 a_{1,1} a_{2,2} a_{3,3} a_{4,5} a_{5,4} + ...\]

Answer 34

Now only non zero terms are: \(a_{1,1}, a_{1,4}, a_{2,2}, a_{2,5}, a_{3,3}, a_{4,4}, a_{5,4}, a_{5,5}\)

Answer 35

So any combination having terms outside of the above will become zero....

Answer 36

This definitely makes it easier... doesn't it?

Answer 37

Affirmative. So \(sgn(\sigma)\) just grow by 1 for every switch?

Answer 38

The power of (-1) goes up by 1 for every switch

Answer 39

sgn will have only two possible values -1 or 1

Answer 40

But remember there will be 5! combinations = 120 ... so instead of trying to write the whole series... just use those 8 non zero terms to get the possible non zero combinations

Answer 41

I'm getting confused in the permutation. I'm not sure how to find if it's even or odd. For example, the set \({a,b,c,d,e}\) switched to \(a,c,b,e,d\) has which kind of parity?

Answer 42

Actually this is where I am getting confused too :P Never used Leibniz formula so have no idea... will have to look into it... @Kainui any ideas?

Answer 43

Just do \(R_5 \to R_5 - \frac{3}5 R_4\) followed by \(R_2 \to R_2 - \frac{1}2 R_5\). Now you're left with a diagonal matrix. And it's an obvious fact that for a diagonal matrix, the determinant is just the product of diagonal entries, which are \(2, 3, 4, 5, 2\).

Answer 44

I think they're expecting you to do this fast and use the Laplace expansion to expand along every row with zeroes in it except for one of the entries. The sign of the permutation doesn't matter for a 0 entry.

Answer 45

Yeah ... Leibniz is looking too messy to be used here...

Answer 46

Well, you can definitely use Leibniz, but don't spend time counting the 100 some odd terms that are all zero cause no matrix is worth that much time of your life haha

Answer 47

Actually the question asks to use the definition of determinant, but Laplace is shown shortly after it. That's why I took it back when I said I could use Laplace :P

Answer 48

And there comes the crux of the problem by our very own Kainui ;) Even if we go for Leibniz, we can directly use the 8 non negative terms to make our work easier... but it will be the sgn that will be a pain to find out..

Answer 49

If you have read about elementary row operations then as @ParthKohli showed just use them... they are much easier here....

Answer 50

I've figured as much. They wouldn't give a matrix with lots of numbers \(\neq\)0

Answer 51

Yeah...

Answer 52

But the definition of determinant is so vague I.M.O.

Answer 53

I already had this typed up and didn't want to interrupt, but I am getting antsy so I'm going to drop this now. Ignore it and come back to it later please cause I want to talk about the determinant stuff now.

----

First let's start with this infinite sum. We can do the finite case and get the same answer, but since \(A^4=0\) it really isn't worth it imo

\[S=I+A+A^2+\cdots\]
Now we multiply both sides by \(A\)
\[AS=A+A^2+A^3+\cdots\]
Add \(I\) to both sides:
\[AS+I = I+A+A^2 +\cdots = S\]
Now we add and subtract to get S on one side
\[S-AS=I\]
Factor S out
\[(I-A)S=I\]
We were specifically told to prove something about \((I-A)^{-1}\) so it definitely exists, thus we can left multiply both sides of this equation by it to get what we always wanted deep in our hearts:
\[S=(I-A)^{-1}\]
Since S is that and it's also that sum, which is what we wanted to prove,
\[(I-A)^{-1}=I+A+A^2+A^3\]
We're done! If you don't believe this infinite thing you can walk through the exact same stuff with the finite sum and when you see \(A^4\) kill it. Also if you want you can add \(A^4\) when you need it if necessary, cause it's 0 and you can add 0 to anything and not hurt it.

Answer 54

Well I can't say that :P My whole work revolves around dealing with matrices, determinants for data storage and computation :p

Answer 55

Okay @Kainui _/\_ Wow!!

Answer 56

woah. Actually there's some problem in the set for \(A^{n+1}\). You've answered it too :P

Answer 57

so what definitions of determinants are you given and expected to work this problem with haha

Answer 58

Well guys I gtg... have an early class to attend tomorrow :(

Answer 59

Till then.. keep solving :P Good night :)

Answer 60

Nice seeing you around vishwesh, it's been a while :P good night

Answer 61

I was given Leibniz formula. Laplace method is given shortly afterwards :(

Answer 62

Thanks @vishweshshrimali5 for your effort!

Answer 63

Alright so does your book use this exact notation or something else, I want to be consistent with your book and class instead of some wikipedia article:

\[\det(A)=\sum_{\sigma \in S_n}\text{sgn}(\sigma)\prod_{i=1}^{n} a_{\sigma(i),i}\]

Answer 64

Yeah that's the notation.

Answer 65

I'm having trouble with the permutations

Answer 66

For example, let \(S=\{1,2,3\}\). It says that (3,2,1) has 3 switches!

Answer 67

So \(\sigma\) represents the numbers 1 through 5 in a specific order, like (3,5,2,1,4).

Since \(\det(A)=\det(A^\top)\) we don't have to worry about talking about rows or columns of a determinant, I'll just talk about rows to explain what that example above means.

It means we pick the 3rd entry from the first row, then the 5th entry from the second row, then the 2nd entry on the 3rd row, the 1st entry off of the 4th row and the 4th entry from the last row. These are what that little indexed sigma means: \(\sigma(i)\) so in my example, \(\sigma(2)=5\).

Kinda a lot going on, but the main takeaway here is that this severely limits the nonzero products you can pick from the matrix, so we have two choices for the first row we can pick from the first or 4th entry on that row. I'll go ahead and pick the 4th one in the row, it's a 2. Next I look at the second row, I have a choice between a 3 and 1, I'll pick the 3 why not. Then I get to the 3rd row and I have only one option. So I pick the 4 and keep going to the next row where I have a 5. It's in the 4th entry of the 4th row and it's the ONLY entry. But my very first decision was something from the 4th entry of the 1st row, so I'm forced to pick a 0 entry if I start with it. So here's what my permutation would have looked like: \((4,2,3, 4, \ )\)

I don'tknow if that makes it better or worse

Answer 68

For example, let \(S=\{1,2,3\}\). It says that (3,2,1) has 3 switches!

Yes, so every permutation has a "standard order" and this one's is (1,2,3). Apparently they're limiting you to only doing switches between adjacent pairs, although technically all that really matters is the parity of the switch which is the same if you just immediately flip the 1 and 3, that's an odd number of flips. The way they're doing it is one of two possible ways:

(3,2,1) -> (3,1,2) -> (1,3,2) -> (1,2,3)

idk why they're making you do flips between adjacent ones only, but it's more concrete I suppose since this is all pretty whacked out and they're trying not to confuse you... So if I've confused you, well, happens I guess lol.

Answer 69

This method is so confusing. I might leave it aside because... Just because. I'm grateful for your answers and your help but I am unlikely to revisit this method ever again, lol. Back to elementary rows and Laplace... :(

Answer 70

Yeah, this method is terrible I agree haha. The only upside of the Leibniz formula is that it's a formula that you can use in equations... Which is still pretty disgusting, but at least it can be used in higher level physics to "plug and chug" determinants with stuff if needed ahaha