The inverse of $(A+A^-1)$ when $A=A^-1$
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I have a matrix that is its own inverse, $A=A^-1$. I want to calculate the inverse of $(A+A^-1)$, for which I would like to use the following chain of equalities:
$$(A+A^-1)^-1=(2A)^-1=2A^-1=2A$$
It doesn't seem to work though, why is that? I can't find anything related among the arithmetic rules in our course literature.
linear-algebra matrices matrix-equations
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up vote
2
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I have a matrix that is its own inverse, $A=A^-1$. I want to calculate the inverse of $(A+A^-1)$, for which I would like to use the following chain of equalities:
$$(A+A^-1)^-1=(2A)^-1=2A^-1=2A$$
It doesn't seem to work though, why is that? I can't find anything related among the arithmetic rules in our course literature.
linear-algebra matrices matrix-equations
3
You need to invert the $2$ as well. $(2^-1A)(A+A^-1)=2^-1(AA+AA^-1)=2^-1(2I)=I$.
â user583896
Aug 14 at 11:44
Ah, of course, thanks!
â Chisq
Aug 14 at 11:46
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have a matrix that is its own inverse, $A=A^-1$. I want to calculate the inverse of $(A+A^-1)$, for which I would like to use the following chain of equalities:
$$(A+A^-1)^-1=(2A)^-1=2A^-1=2A$$
It doesn't seem to work though, why is that? I can't find anything related among the arithmetic rules in our course literature.
linear-algebra matrices matrix-equations
I have a matrix that is its own inverse, $A=A^-1$. I want to calculate the inverse of $(A+A^-1)$, for which I would like to use the following chain of equalities:
$$(A+A^-1)^-1=(2A)^-1=2A^-1=2A$$
It doesn't seem to work though, why is that? I can't find anything related among the arithmetic rules in our course literature.
linear-algebra matrices matrix-equations
asked Aug 14 at 11:43
Chisq
1467
1467
3
You need to invert the $2$ as well. $(2^-1A)(A+A^-1)=2^-1(AA+AA^-1)=2^-1(2I)=I$.
â user583896
Aug 14 at 11:44
Ah, of course, thanks!
â Chisq
Aug 14 at 11:46
add a comment |Â
3
You need to invert the $2$ as well. $(2^-1A)(A+A^-1)=2^-1(AA+AA^-1)=2^-1(2I)=I$.
â user583896
Aug 14 at 11:44
Ah, of course, thanks!
â Chisq
Aug 14 at 11:46
3
3
You need to invert the $2$ as well. $(2^-1A)(A+A^-1)=2^-1(AA+AA^-1)=2^-1(2I)=I$.
â user583896
Aug 14 at 11:44
You need to invert the $2$ as well. $(2^-1A)(A+A^-1)=2^-1(AA+AA^-1)=2^-1(2I)=I$.
â user583896
Aug 14 at 11:44
Ah, of course, thanks!
â Chisq
Aug 14 at 11:46
Ah, of course, thanks!
â Chisq
Aug 14 at 11:46
add a comment |Â
2 Answers
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up vote
2
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accepted
$A+A^-1 = 2A$
The inverse of $2A$ is $2^-1 A^-1$ i.e the inverse of $A+A^-1$ is $2^-1 A^-1$
Also known as $frac12A$
â stuart stevenson
Aug 14 at 11:55
You mean $frac12A^-1$?
â A. Pongrácz
Aug 14 at 12:06
add a comment |Â
up vote
0
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We just need to find a matrix $B$ so as $B(A+A^-1)=I$.After some trials we see that $1/2A$ is what we are looking for.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$A+A^-1 = 2A$
The inverse of $2A$ is $2^-1 A^-1$ i.e the inverse of $A+A^-1$ is $2^-1 A^-1$
Also known as $frac12A$
â stuart stevenson
Aug 14 at 11:55
You mean $frac12A^-1$?
â A. Pongrácz
Aug 14 at 12:06
add a comment |Â
up vote
2
down vote
accepted
$A+A^-1 = 2A$
The inverse of $2A$ is $2^-1 A^-1$ i.e the inverse of $A+A^-1$ is $2^-1 A^-1$
Also known as $frac12A$
â stuart stevenson
Aug 14 at 11:55
You mean $frac12A^-1$?
â A. Pongrácz
Aug 14 at 12:06
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$A+A^-1 = 2A$
The inverse of $2A$ is $2^-1 A^-1$ i.e the inverse of $A+A^-1$ is $2^-1 A^-1$
$A+A^-1 = 2A$
The inverse of $2A$ is $2^-1 A^-1$ i.e the inverse of $A+A^-1$ is $2^-1 A^-1$
edited Aug 14 at 12:02
A. Pongrácz
3,682624
3,682624
answered Aug 14 at 11:48
Sadil Khan
3297
3297
Also known as $frac12A$
â stuart stevenson
Aug 14 at 11:55
You mean $frac12A^-1$?
â A. Pongrácz
Aug 14 at 12:06
add a comment |Â
Also known as $frac12A$
â stuart stevenson
Aug 14 at 11:55
You mean $frac12A^-1$?
â A. Pongrácz
Aug 14 at 12:06
Also known as $frac12A$
â stuart stevenson
Aug 14 at 11:55
Also known as $frac12A$
â stuart stevenson
Aug 14 at 11:55
You mean $frac12A^-1$?
â A. Pongrácz
Aug 14 at 12:06
You mean $frac12A^-1$?
â A. Pongrácz
Aug 14 at 12:06
add a comment |Â
up vote
0
down vote
We just need to find a matrix $B$ so as $B(A+A^-1)=I$.After some trials we see that $1/2A$ is what we are looking for.
add a comment |Â
up vote
0
down vote
We just need to find a matrix $B$ so as $B(A+A^-1)=I$.After some trials we see that $1/2A$ is what we are looking for.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We just need to find a matrix $B$ so as $B(A+A^-1)=I$.After some trials we see that $1/2A$ is what we are looking for.
We just need to find a matrix $B$ so as $B(A+A^-1)=I$.After some trials we see that $1/2A$ is what we are looking for.
answered Aug 14 at 11:51
dmtri
571316
571316
add a comment |Â
add a comment |Â
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3
You need to invert the $2$ as well. $(2^-1A)(A+A^-1)=2^-1(AA+AA^-1)=2^-1(2I)=I$.
â user583896
Aug 14 at 11:44
Ah, of course, thanks!
â Chisq
Aug 14 at 11:46