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Inverse of $aI - frac{a}{b}J$

0

$begingroup$

How would I find the inverse of this matrix?

I know we can pull the a out so that:

begin{equation}
aI - frac{a}{b}J = aleft[I - frac{1}{b}Jright]
end{equation}

and then we can focus on the inverse of the matrix in the brackets, but this seems to be outside the scope of how I know how to invert a matrix.

linear-algebra inverse

share|cite|improve this question

edited Mar 26 at 1:42

Brian

1,495416

asked Mar 26 at 1:40

user1992460user1992460

505

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $J$ the all-ones matrix?
  $endgroup$
  – M. Vinay
  Mar 26 at 2:20
  

  
  
  
  

add a comment | 

0

$begingroup$

How would I find the inverse of this matrix?

I know we can pull the a out so that:

begin{equation}
aI - frac{a}{b}J = aleft[I - frac{1}{b}Jright]
end{equation}

and then we can focus on the inverse of the matrix in the brackets, but this seems to be outside the scope of how I know how to invert a matrix.

linear-algebra inverse

share|cite|improve this question

edited Mar 26 at 1:42

Brian

1,495416

asked Mar 26 at 1:40

user1992460user1992460

505

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $J$ the all-ones matrix?
  $endgroup$
  – M. Vinay
  Mar 26 at 2:20
  

  
  
  
  

add a comment | 

$begingroup$

How would I find the inverse of this matrix?

I know we can pull the a out so that:

begin{equation}
aI - frac{a}{b}J = aleft[I - frac{1}{b}Jright]
end{equation}

and then we can focus on the inverse of the matrix in the brackets, but this seems to be outside the scope of how I know how to invert a matrix.

linear-algebra inverse

share|cite|improve this question

edited Mar 26 at 1:42

Brian

1,495416

asked Mar 26 at 1:40

user1992460user1992460

505

$endgroup$

share|cite|improve this question

edited Mar 26 at 1:42

Brian

1,495416

asked Mar 26 at 1:40

user1992460user1992460

505

share|cite|improve this question

edited Mar 26 at 1:42

Brian

1,495416

asked Mar 26 at 1:40

user1992460user1992460

505

edited Mar 26 at 1:42

Brian

1,495416

edited Mar 26 at 1:42

Brian

1,495416

asked Mar 26 at 1:40

user1992460user1992460

505

asked Mar 26 at 1:40

user1992460user1992460

505

1 Answer
1

active

oldest

votes

0

$begingroup$

Observe that $J^2 = nJ$, when $J$ is of order $n times n$.

Then $(I - rJ)(I - sJ) = I - (r + s)J + rs J^2 = I + (nrs - r - s)J$.

Now, for any given $r$, we can find $s$ such that the above becomes $I$ by solving $nrs - r - s = 0$. That is, $s = frac{r}{nr - 1}$. This does not work if $r = frac 1 n$. But when $r = frac 1 n$, we have $I - rJ = I - frac 1 n J$, which is non-invertible — there are multiple ways to see this — observe that it is idempotent (but is not the identity matrix, and therefore must be singular), or observe that $0$ is an eigenvalue, since $n$ is an eigenvalue of $J$.

Thus, for any $r ne frac 1 n$, $(I - rJ)^{-1} = I - frac{r}{nr - 1}J$.

Therefore, $$left(a - dfrac a b Jright)^{-1} = dfrac 1 a I - dfrac{1}{a(n - b)} J$$ for any $a ne 0, b ne n$.

share|cite|improve this answer

answered Mar 26 at 2:28

M. VinayM. Vinay

7,35822136

$endgroup$

add a comment | 

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0

$begingroup$

Observe that $J^2 = nJ$, when $J$ is of order $n times n$.

Then $(I - rJ)(I - sJ) = I - (r + s)J + rs J^2 = I + (nrs - r - s)J$.

Now, for any given $r$, we can find $s$ such that the above becomes $I$ by solving $nrs - r - s = 0$. That is, $s = frac{r}{nr - 1}$. This does not work if $r = frac 1 n$. But when $r = frac 1 n$, we have $I - rJ = I - frac 1 n J$, which is non-invertible — there are multiple ways to see this — observe that it is idempotent (but is not the identity matrix, and therefore must be singular), or observe that $0$ is an eigenvalue, since $n$ is an eigenvalue of $J$.

Thus, for any $r ne frac 1 n$, $(I - rJ)^{-1} = I - frac{r}{nr - 1}J$.

Therefore, $$left(a - dfrac a b Jright)^{-1} = dfrac 1 a I - dfrac{1}{a(n - b)} J$$ for any $a ne 0, b ne n$.

share|cite|improve this answer

answered Mar 26 at 2:28

M. VinayM. Vinay

7,35822136

$endgroup$

add a comment | 

0

$begingroup$

Observe that $J^2 = nJ$, when $J$ is of order $n times n$.

Then $(I - rJ)(I - sJ) = I - (r + s)J + rs J^2 = I + (nrs - r - s)J$.

Now, for any given $r$, we can find $s$ such that the above becomes $I$ by solving $nrs - r - s = 0$. That is, $s = frac{r}{nr - 1}$. This does not work if $r = frac 1 n$. But when $r = frac 1 n$, we have $I - rJ = I - frac 1 n J$, which is non-invertible — there are multiple ways to see this — observe that it is idempotent (but is not the identity matrix, and therefore must be singular), or observe that $0$ is an eigenvalue, since $n$ is an eigenvalue of $J$.

Thus, for any $r ne frac 1 n$, $(I - rJ)^{-1} = I - frac{r}{nr - 1}J$.

Therefore, $$left(a - dfrac a b Jright)^{-1} = dfrac 1 a I - dfrac{1}{a(n - b)} J$$ for any $a ne 0, b ne n$.

share|cite|improve this answer

answered Mar 26 at 2:28

M. VinayM. Vinay

7,35822136

$endgroup$

add a comment | 

$begingroup$

Observe that $J^2 = nJ$, when $J$ is of order $n times n$.

Then $(I - rJ)(I - sJ) = I - (r + s)J + rs J^2 = I + (nrs - r - s)J$.

Now, for any given $r$, we can find $s$ such that the above becomes $I$ by solving $nrs - r - s = 0$. That is, $s = frac{r}{nr - 1}$. This does not work if $r = frac 1 n$. But when $r = frac 1 n$, we have $I - rJ = I - frac 1 n J$, which is non-invertible — there are multiple ways to see this — observe that it is idempotent (but is not the identity matrix, and therefore must be singular), or observe that $0$ is an eigenvalue, since $n$ is an eigenvalue of $J$.

Thus, for any $r ne frac 1 n$, $(I - rJ)^{-1} = I - frac{r}{nr - 1}J$.

Therefore, $$left(a - dfrac a b Jright)^{-1} = dfrac 1 a I - dfrac{1}{a(n - b)} J$$ for any $a ne 0, b ne n$.

share|cite|improve this answer

answered Mar 26 at 2:28

M. VinayM. Vinay

7,35822136

$endgroup$

share|cite|improve this answer

answered Mar 26 at 2:28

M. VinayM. Vinay

7,35822136

answered Mar 26 at 2:28

M. VinayM. Vinay

7,35822136

answered Mar 26 at 2:28

M. VinayM. Vinay

7,35822136