Inverse of an “anti-Jordan” matrix

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Call a matrix anti-Jordan if it has $-1$s on the subdiagonal, values on the diagonal, and zeroes elsewhere; that is, it is written
$$beginbmatrix
lambda_1 & 0 & 0 & dots \
-1 & lambda_2 & 0 & dots\
0 & -1 & lambda_3 & dots\
vdots &&& ddots
endbmatrix$$
Is there a general formula for the inverse of such a matrix?

If it helps, I'm trying to compute the spectral radius of an $N times N$ matrix $FV^-1$ where
$$F = beginbmatrixbeta_1 & beta_2 & beta_3 & dots\0 & 0 & 0 &dots\0 &0 & 0 & dots\vdots &&& ddotsendbmatrix$$
$N$ is "very large" and $$V = beginbmatrix
1 + mu_1 & 0 & 0 & dots \
-1 & 1 + mu_2 & 0 & dots\
0 & -1 & 1 + mu_3 & dots\
vdots &&& ddots
endbmatrix.$$

linear-algebra matrices

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asked Jul 18 at 21:52

Aidan Backus

6716

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up vote
3
down vote

favorite

Call a matrix anti-Jordan if it has $-1$s on the subdiagonal, values on the diagonal, and zeroes elsewhere; that is, it is written
$$beginbmatrix
lambda_1 & 0 & 0 & dots \
-1 & lambda_2 & 0 & dots\
0 & -1 & lambda_3 & dots\
vdots &&& ddots
endbmatrix$$
Is there a general formula for the inverse of such a matrix?

If it helps, I'm trying to compute the spectral radius of an $N times N$ matrix $FV^-1$ where
$$F = beginbmatrixbeta_1 & beta_2 & beta_3 & dots\0 & 0 & 0 &dots\0 &0 & 0 & dots\vdots &&& ddotsendbmatrix$$
$N$ is "very large" and $$V = beginbmatrix
1 + mu_1 & 0 & 0 & dots \
-1 & 1 + mu_2 & 0 & dots\
0 & -1 & 1 + mu_3 & dots\
vdots &&& ddots
endbmatrix.$$

linear-algebra matrices

share|cite|improve this question

asked Jul 18 at 21:52

Aidan Backus

6716

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Call a matrix anti-Jordan if it has $-1$s on the subdiagonal, values on the diagonal, and zeroes elsewhere; that is, it is written
$$beginbmatrix
lambda_1 & 0 & 0 & dots \
-1 & lambda_2 & 0 & dots\
0 & -1 & lambda_3 & dots\
vdots &&& ddots
endbmatrix$$
Is there a general formula for the inverse of such a matrix?

If it helps, I'm trying to compute the spectral radius of an $N times N$ matrix $FV^-1$ where
$$F = beginbmatrixbeta_1 & beta_2 & beta_3 & dots\0 & 0 & 0 &dots\0 &0 & 0 & dots\vdots &&& ddotsendbmatrix$$
$N$ is "very large" and $$V = beginbmatrix
1 + mu_1 & 0 & 0 & dots \
-1 & 1 + mu_2 & 0 & dots\
0 & -1 & 1 + mu_3 & dots\
vdots &&& ddots
endbmatrix.$$

linear-algebra matrices

share|cite|improve this question

asked Jul 18 at 21:52

Aidan Backus

6716

Call a matrix anti-Jordan if it has $-1$s on the subdiagonal, values on the diagonal, and zeroes elsewhere; that is, it is written
$$beginbmatrix
lambda_1 & 0 & 0 & dots \
-1 & lambda_2 & 0 & dots\
0 & -1 & lambda_3 & dots\
vdots &&& ddots
endbmatrix$$
Is there a general formula for the inverse of such a matrix?

If it helps, I'm trying to compute the spectral radius of an $N times N$ matrix $FV^-1$ where
$$F = beginbmatrixbeta_1 & beta_2 & beta_3 & dots\0 & 0 & 0 &dots\0 &0 & 0 & dots\vdots &&& ddotsendbmatrix$$
$N$ is "very large" and $$V = beginbmatrix
1 + mu_1 & 0 & 0 & dots \
-1 & 1 + mu_2 & 0 & dots\
0 & -1 & 1 + mu_3 & dots\
vdots &&& ddots
endbmatrix.$$

linear-algebra matrices

share|cite|improve this question

asked Jul 18 at 21:52

Aidan Backus

6716

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 21:52

Aidan Backus

6716

asked Jul 18 at 21:52

Aidan Backus

6716

asked Jul 18 at 21:52

Aidan Backus

6716

6716

add a comment | 

add a comment | 

2 Answers
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6
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The inverse is the matrix $B = A^-1$ with $b_ij=0$ for $i < j$ and $b_ij = lambda_j^-1dotslambda_i^-1$ for $ige j$.

Your spectral radius will be the absolute value of
$$
sum_i=1^Nfracbeta_i(1+mu_1)dots(1+mu_i).
$$

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edited Jul 18 at 22:27

answered Jul 18 at 22:12

amsmath

1,613114

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up vote
3
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$$beginbmatrix
lambda_1 & 0 & 0 & dots \
-1 & lambda_2 & 0 & dots\
0 & -1 & lambda_3 & dots\
vdots &vdots &vdots & ddots
endbmatrix^-1=beginbmatrix
lambda_1^-1 & 0 & 0 & dots\
lambda_1^-1lambda_2^-1 & lambda_2^-1 & 0 & dots \
lambda_1^-1lambda_2^-1lambda^-1_3 & lambda_2^-1lambda_3^-1 & lambda_3^-1 & dots\
vdots & vdots & vdots & ddots
endbmatrix$$
I found this by applying row operations the augmented matrix $[A|I]$, where $A$ is your anti-Jordon matrix and $I$ is the $ntimes n$ identity, until it was in the form $[I|B]$.

Edit: My answer was originally wrong, and has been fixed to match amsmath's correct answer.

share|cite|improve this answer

edited Jul 18 at 22:20

community wiki

2 revs
Mike Earnest

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This is not correct. When I multiply your $A^-1$ to $A$, in the entry $(3,1)$ I get $-lambda_2^-1lambda_3^-1$ insetad of $0$.
  – amsmath
  Jul 18 at 22:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @amsmath Yep, I was too hasty.
  – Mike Earnest
  Jul 18 at 22:20
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
6
down vote



accepted

The inverse is the matrix $B = A^-1$ with $b_ij=0$ for $i < j$ and $b_ij = lambda_j^-1dotslambda_i^-1$ for $ige j$.

Your spectral radius will be the absolute value of
$$
sum_i=1^Nfracbeta_i(1+mu_1)dots(1+mu_i).
$$

share|cite|improve this answer

edited Jul 18 at 22:27

answered Jul 18 at 22:12

amsmath

1,613114

add a comment | 

up vote
6
down vote



accepted

The inverse is the matrix $B = A^-1$ with $b_ij=0$ for $i < j$ and $b_ij = lambda_j^-1dotslambda_i^-1$ for $ige j$.

Your spectral radius will be the absolute value of
$$
sum_i=1^Nfracbeta_i(1+mu_1)dots(1+mu_i).
$$

share|cite|improve this answer

edited Jul 18 at 22:27

answered Jul 18 at 22:12

amsmath

1,613114

add a comment | 

up vote
6
down vote



accepted

up vote
6
down vote



accepted

The inverse is the matrix $B = A^-1$ with $b_ij=0$ for $i < j$ and $b_ij = lambda_j^-1dotslambda_i^-1$ for $ige j$.

Your spectral radius will be the absolute value of
$$
sum_i=1^Nfracbeta_i(1+mu_1)dots(1+mu_i).
$$

share|cite|improve this answer

edited Jul 18 at 22:27

answered Jul 18 at 22:12

amsmath

1,613114

The inverse is the matrix $B = A^-1$ with $b_ij=0$ for $i < j$ and $b_ij = lambda_j^-1dotslambda_i^-1$ for $ige j$.

Your spectral radius will be the absolute value of
$$
sum_i=1^Nfracbeta_i(1+mu_1)dots(1+mu_i).
$$

share|cite|improve this answer

edited Jul 18 at 22:27

answered Jul 18 at 22:12

amsmath

1,613114

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 22:27

edited Jul 18 at 22:27

edited Jul 18 at 22:27

answered Jul 18 at 22:12

amsmath

1,613114

answered Jul 18 at 22:12

amsmath

1,613114

answered Jul 18 at 22:12

amsmath

1,613114

1,613114

add a comment | 

add a comment | 

up vote
3
down vote

$$beginbmatrix
lambda_1 & 0 & 0 & dots \
-1 & lambda_2 & 0 & dots\
0 & -1 & lambda_3 & dots\
vdots &vdots &vdots & ddots
endbmatrix^-1=beginbmatrix
lambda_1^-1 & 0 & 0 & dots\
lambda_1^-1lambda_2^-1 & lambda_2^-1 & 0 & dots \
lambda_1^-1lambda_2^-1lambda^-1_3 & lambda_2^-1lambda_3^-1 & lambda_3^-1 & dots\
vdots & vdots & vdots & ddots
endbmatrix$$
I found this by applying row operations the augmented matrix $[A|I]$, where $A$ is your anti-Jordon matrix and $I$ is the $ntimes n$ identity, until it was in the form $[I|B]$.

Edit: My answer was originally wrong, and has been fixed to match amsmath's correct answer.

share|cite|improve this answer

edited Jul 18 at 22:20

community wiki

2 revs
Mike Earnest

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This is not correct. When I multiply your $A^-1$ to $A$, in the entry $(3,1)$ I get $-lambda_2^-1lambda_3^-1$ insetad of $0$.
  – amsmath
  Jul 18 at 22:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @amsmath Yep, I was too hasty.
  – Mike Earnest
  Jul 18 at 22:20
  

  
  
  
  

add a comment | 

up vote
3
down vote

$$beginbmatrix
lambda_1 & 0 & 0 & dots \
-1 & lambda_2 & 0 & dots\
0 & -1 & lambda_3 & dots\
vdots &vdots &vdots & ddots
endbmatrix^-1=beginbmatrix
lambda_1^-1 & 0 & 0 & dots\
lambda_1^-1lambda_2^-1 & lambda_2^-1 & 0 & dots \
lambda_1^-1lambda_2^-1lambda^-1_3 & lambda_2^-1lambda_3^-1 & lambda_3^-1 & dots\
vdots & vdots & vdots & ddots
endbmatrix$$
I found this by applying row operations the augmented matrix $[A|I]$, where $A$ is your anti-Jordon matrix and $I$ is the $ntimes n$ identity, until it was in the form $[I|B]$.

Edit: My answer was originally wrong, and has been fixed to match amsmath's correct answer.

share|cite|improve this answer

edited Jul 18 at 22:20

community wiki

2 revs
Mike Earnest

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This is not correct. When I multiply your $A^-1$ to $A$, in the entry $(3,1)$ I get $-lambda_2^-1lambda_3^-1$ insetad of $0$.
  – amsmath
  Jul 18 at 22:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @amsmath Yep, I was too hasty.
  – Mike Earnest
  Jul 18 at 22:20
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

$$beginbmatrix
lambda_1 & 0 & 0 & dots \
-1 & lambda_2 & 0 & dots\
0 & -1 & lambda_3 & dots\
vdots &vdots &vdots & ddots
endbmatrix^-1=beginbmatrix
lambda_1^-1 & 0 & 0 & dots\
lambda_1^-1lambda_2^-1 & lambda_2^-1 & 0 & dots \
lambda_1^-1lambda_2^-1lambda^-1_3 & lambda_2^-1lambda_3^-1 & lambda_3^-1 & dots\
vdots & vdots & vdots & ddots
endbmatrix$$
I found this by applying row operations the augmented matrix $[A|I]$, where $A$ is your anti-Jordon matrix and $I$ is the $ntimes n$ identity, until it was in the form $[I|B]$.

Edit: My answer was originally wrong, and has been fixed to match amsmath's correct answer.

share|cite|improve this answer

edited Jul 18 at 22:20

community wiki

2 revs
Mike Earnest

$$beginbmatrix
lambda_1 & 0 & 0 & dots \
-1 & lambda_2 & 0 & dots\
0 & -1 & lambda_3 & dots\
vdots &vdots &vdots & ddots
endbmatrix^-1=beginbmatrix
lambda_1^-1 & 0 & 0 & dots\
lambda_1^-1lambda_2^-1 & lambda_2^-1 & 0 & dots \
lambda_1^-1lambda_2^-1lambda^-1_3 & lambda_2^-1lambda_3^-1 & lambda_3^-1 & dots\
vdots & vdots & vdots & ddots
endbmatrix$$
I found this by applying row operations the augmented matrix $[A|I]$, where $A$ is your anti-Jordon matrix and $I$ is the $ntimes n$ identity, until it was in the form $[I|B]$.

Edit: My answer was originally wrong, and has been fixed to match amsmath's correct answer.

share|cite|improve this answer

edited Jul 18 at 22:20

community wiki

2 revs
Mike Earnest

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 22:20

edited Jul 18 at 22:20

edited Jul 18 at 22:20

community wiki

2 revs
Mike Earnest

community wiki

2 revs
Mike Earnest

community wiki

2 revs
Mike Earnest

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This is not correct. When I multiply your $A^-1$ to $A$, in the entry $(3,1)$ I get $-lambda_2^-1lambda_3^-1$ insetad of $0$.
  – amsmath
  Jul 18 at 22:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @amsmath Yep, I was too hasty.
  – Mike Earnest
  Jul 18 at 22:20
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This is not correct. When I multiply your $A^-1$ to $A$, in the entry $(3,1)$ I get $-lambda_2^-1lambda_3^-1$ insetad of $0$.
  – amsmath
  Jul 18 at 22:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @amsmath Yep, I was too hasty.
  – Mike Earnest
  Jul 18 at 22:20
  

  
  
  
  

2

2

This is not correct. When I multiply your $A^-1$ to $A$, in the entry $(3,1)$ I get $-lambda_2^-1lambda_3^-1$ insetad of $0$.
– amsmath
Jul 18 at 22:14

This is not correct. When I multiply your $A^-1$ to $A$, in the entry $(3,1)$ I get $-lambda_2^-1lambda_3^-1$ insetad of $0$.
– amsmath
Jul 18 at 22:14

@amsmath Yep, I was too hasty.
– Mike Earnest
Jul 18 at 22:20

@amsmath Yep, I was too hasty.
– Mike Earnest
Jul 18 at 22:20

add a comment | 

 

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