[discussion] taylor expansion of matrix inverse

8/19/2019 [Discussion] Taylor Expansion of Matrix Inverse

1/12

Aug 23, 2010#1

Aug 23, 2010#2

Aug 23, 2010#3

Matrix inverse equals power-series

Hi!

In an economics book about input-output analysis the following statement

is presented, but I cannot find the proof:

Can someone help me show why this is the case?

P.s. I think there is some assumptions made about such that all

elements is less than 1 and greater than 0. Btw, n goes to infty.

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So you mean . This is true if the right side

converges, which is true if and only if all of the eigenvalues of A have

absolute value smaller than 1.

To prove it, multiply both sides by .

Forums Mathematics Linear and Abstract Algebra

Mårten

( I − A )

− 1

= ( I + A + A

2

+ A

3

+ . . . + A

)

A

a

adriank

( I − A )

− 1

= ∑

∞

= 0

A

I − A

Mårten

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2/12

Aug 23, 2010#4

Aug 23, 2010#5

Aug 23, 2010#6

Aug 23, 2010#7

Thanks a lot! Easier than I thought.

Now I found another assumption regarding A, and that is that the row

sums are all less than 1, and the column sums are also all less than 1.

Could that be the same as saying that all eigenvalues have to be less than

|1|, and in that case why are these statements equivalent?

What exactly do you mean by row sums and column sums?

If the matrix A is

then the rowsums are and . The columnsums

are and .

Well then that's certainly not true. Maybe you meant absolute values, orsomething?

An equivalent condition to the eigenvalue condition is that for

all such that .

Last edited: Aug 23, 2010

adriank

Mårten

a

1 1

a

1 2

a

2 1

a

2 2

a

1 1

+ a

1 2

< 1 a

2 1

+ a

2 2

< 1

a

1 1

+ a

2 1

< 1 a

1 2

+ a

2 2

< 1

adriank

∣ A ∣ < 1

∣ ∣ = 1

Mårten

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3/12

Aug 24, 2010#8

But if, at the same time, all , it doesn't work then?

Anyhow, is there a way to set up criteria for A making all its eigenvalues

less than |1|?

adriank said: ↑

Mårten said: ↑

This isn't a valid proof. If you multiply both sides by I-A and simplify the

right-hand side, you're making assumptions about the infinite sum that you

can't make at this point. And if you intend to consider the equality in post

#1 with a finite number of terms in the sum, the two sides aren't actually

equal, so you'd be starting with an equality that's false.

(Maybe you understand this already, but it didn't look that way to me, and

you did say that you're an economy student. )

What you need to do is to calculate

a

> 0

Fredrik

Staff Emeritus Science Advisor Gold Member

So you mean . This is true if the right side converges, which is

true if and only if all of the eigenvalues of A have absolute value smaller than 1.

To prove it, multiply both sides by .

( I − A )

− 1

= ∑

∞

= 0

A

I − A

Thanks a lot! Easier than I thought.

Now I found another assumption regarding A, and that is that the row sums are all less

than 1, and the column sums are also all less than 1. Could that be the same as saying

that all eigenvalues have to be less than |1|, and in that case why are these statements

equivalent?

( l i m

→ ∞

( I + A + A

2

+ ⋯ + A

) ( I − A )

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4/12

Aug 24, 2010#9

Aug 24, 2010#10

Start by rewriting it as

and then prove that this is =I. This result implies that I-A is invertible, and

that the inverse is that series. The information you were given about the

components of A is needed to see that the unwanted term goes to zero in

that limit.


Thanks a lot! I really appreciate this help, I think I understand it pretty well

now.

Adriank, it seems that my conditions given about A satisfy your condition

that for all such that . I cannot prove it formally that

these two conditions are the same (or that my condition follows from

Adriank's), but some easy calculations and checks I've made, makes itreasonable. (If someone has the energy to prove it, I wouldn't be late to

look at it.)

My conditions about A once more (sorry to not have given them at the

same time before): Row sums and column sums are all, one by one, less

than 1, and at the same time .

P.s. Actually, I didn't say I'm an economics student, I just said I read aneconomics book.

Btw, it occurred to me that I don't understand really why all eigenvalues of

A have to have absolute value less than 1 in order to make the series

above converge. Why does that eigenvalue condition affect the

= l i m

→ ∞

( ( I + A + A

2

+ ⋯ + A

) ( I − A ) )

Mårten

∣ A ∣ < 1 ∣ ∣ = 1

0 ≤ a

< 1

Mårten

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5/12

Aug 24, 2010#11

convergence property of the series?

Another follow-up question: Now when we have this nice power expansion

for the inverse matrix, is that actually the way (some) matrix inverses is

calculated in computers?

Following on what Fredrik said, note that

Now we want to take the limit as to show that

But for that to be true, you need that

This is true if and only if the operator norm of A is less than 1. And it turns

out that the operator norm of A is the largest absolute value of the

eigenvalues of A. If you have some other condition that implies this, then

that works too.

As for computation, usually that's a very inefficient way to calculate it

directly, unless A is nilpotent (so that the series only has finitely manynonzero terms). However, if A is n by n, then you can express An in terms

of lower powers of A by the Cayley-Hamilton theorem. You could also

apply various numerical techniques to directly find the inverse of I - A; a

lot of times, all you care about is (I - A)-1x for some vector x, which can

often be done even more efficiently than calculating the full matrix

inverse.

adriank

(

∑

= 0

A

( I − A ) = I − A

+ 1

.

→ ∞

(

∞

∑

= 0

A

( I − A ) = I .

l i m

→ ∞

A

+ 1

= 0 .

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6/12

Aug 24, 2010#12

Let's try the Hilbert-Schmidt norm, defined by

This definition and the information given together imply that if A is a m×m

matrix,

The norm satisfies

so

Fredrik

Staff Emeritus Science Advisor Gold Member

⟨ A , B ⟩ = T r A

†

B

∥ A ∥

2

= ⟨ A , A ⟩ = ∑

,

a

∗

a

= ∑

,

| a

|

2

∥ A ∥ ≥ 0

∥ A ∥

2


7/12

Aug 24, 2010#13

Aug 24, 2010#14

Aug 25, 2010#15

but this doesn't converge as n→∞. It looks like the assumption

isn't strong enough to guarantee convergence. It looks like we would need

something like (or another norm to define convergence).

Hmm...


Fredrik said: ↑

Did you take into account that my assumption wasn't just that

(actually ), but also that the row sums and column sums are,

one by one, less than 1? In a way, the latter thing seems to imply that

, "on the average" at least.

I'll take a look at the operator room in the meantime.

Well, the Hilbert-Schmidt norm is always bigger than the operator

norm, so if A has Hilbert-Schmidt norm less than 1, then its operator norm

is also less than 1.

And the Hilbert-Schmidt norm happens to satisfy

Mårten said: ↑

| a

| < 1

| a

| < 1 /

Mårten

It looks like the assumption isn't strong enough to guarantee convergence. It

looks like we would need something like (or another norm to define

convergence). Hmm...

| a

| < 1

| a

| < 1 /

| a

| < 1

0 ≤ a

< 1

| a

| < 1 /

adriank

∥ ⋅ ∥

H S

∥ A ∥

2

H S

= ∑

,

∣ a

∣

2

.

Fredrik

Did you take into account that my assumption wasn't just that (actually| a

| < 1

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8/12

Aug 26, 2010#16

No, I didn't take that into account. When I do, I get

which is better, but not good enough. We want . The condition on

the "row sums" and "column sums" allows the possibility of a diagonal

matrix with entries close to 1 on the diagonal. Such a matrix doesn't have

a Hilbert-Schmidt norm


9/12

Aug 26, 2010#17

equation above imply that all the components of An+1 go to zero when n

goes to infty? Where exactly in the equation above does one see that? It

was a couple of years ago I took my classes in linear algebra...

(Edit: I do understand that it's enough to show that the elements of A2 is

less than the elements of A, because then the elements in A3 should be

even less, and so on, but I cannot yet see why the elements of A2 is less

than the elements of A.)


Mårten said: ↑

It doesn't, and you don't. For this method of proof to work, we needed the

norm of A to be


10/12

Sep 12, 2010#18

Since the left-hand side is the largest component of An+1, all components

of An+1 go to zero.


Thanks a lot for this derivation, it was really a nifty one! Just a minor thing,

I think the rightmost inequality in the uppermost equation above should

say "= ||B||a" not "


11/12

Feb 22, 2011#19

Correction

I looked through this again, and realized that I was wrong about the

condition I gave that the rowsums are less than 1. Fortunately, this doesn't

have any implications for the proof above, since it just makes use of that

the column sums are less than 1. So the proof still holds!

For anyone surfing by this thread, the following background information

could therefore be useful: What you start out with in input-output analysis,

is the input-output table which could be written as

,

where f ij represents input of products from industry i to industry j , i is a

unit column vector (with just ones in it), C is a column vector representing

final demand (non-industry use from households, government, et.c.), and x

is a column vector representing total ouput (as well as total input). Not

included in the above equation is a row of value-added (VA) below the F-

matrix - this row could be interpreted as the labour force input (i.e.,

salaries) required for the industries to produce their products. The different

inputs of products and labor for a certain industry j is shown in the rows of

column j in the F-matrix and the VA-row; the sum of this column equals

the total inputs x j . This value is the same as the total output x j from that

industry, which is the rowsum of F and C for row j . That is, total costs equal

total revenues.

Now, if the matrix A is generated via a ij = f ij /x j , the cells in a certain

column j in A, represent the shares of total input x j . That implies that each

column sum of A are less than 1. And that's not the same as saying that

the row sums of A are less than 1, since the cells in a row of A doesn't

represent the shares of that row's rowsum, according to a ij = f ij /x j . Finally,

the above equation can now be written as

Mårten

F + C =

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(where x becomes a function of the final demand C ), and that explains why

the topic of this thread was about matrix inverses.

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A + C = ⇔ = ( I − A )

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[discussion] taylor expansion of matrix inverse

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