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When is AXB equal to BXA in the case of square matrices?
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My question is motivated by the inverse of $A^-1+B^-1$.
$$A^-1+B^-1=A^-1(A+B)B^-1implies(A^-1+B^-1)^-1=B(A+B)^-1A$$
$$A^-1+B^-1=B^-1+A^-1=B^-1(A+B)A^-1implies(A^-1+B^-1)^-1=A(A+B)^-1B$$
$$B(A+B)^-1A=A(A+B)^-1B$$
Is this a special result due to the fact that $(A+B)^-1$ is sandwiched between $A$ and $B$, or does it hold for other cases as well, i.e. $AXB=BXA$ where $X$ has some special properties?
If it's the former, then some intuition for why it holds beyond the math mechanics that I have shown above would be appreciated.
matrices inverse inverse-function
asked Jul 16 at 8:55
Amrit Prasad
828
add a comment |Â
up vote
0
down vote
favorite
My question is motivated by the inverse of $A^-1+B^-1$.
$$A^-1+B^-1=A^-1(A+B)B^-1implies(A^-1+B^-1)^-1=B(A+B)^-1A$$
$$A^-1+B^-1=B^-1+A^-1=B^-1(A+B)A^-1implies(A^-1+B^-1)^-1=A(A+B)^-1B$$
$$B(A+B)^-1A=A(A+B)^-1B$$
Is this a special result due to the fact that $(A+B)^-1$ is sandwiched between $A$ and $B$, or does it hold for other cases as well, i.e. $AXB=BXA$ where $X$ has some special properties?
If it's the former, then some intuition for why it holds beyond the math mechanics that I have shown above would be appreciated.
matrices inverse inverse-function
asked Jul 16 at 8:55
Amrit Prasad
828
I'm not sure if I'd call it a duplicate, but related: math.stackexchange.com/questions/2852002/…
– Theo Bendit
Jul 16 at 9:23
Thanks for pointing that particular thread out. Searching in stackexchange isn't particularly great, else I would have probably left the above as a comment on the top answer on that thread. The second part of my question is to figure out all $X$ values for which this is true. This was not answered on that thread. I'm happy to merge this with the other question if needed though.
– Amrit Prasad
Jul 16 at 9:45
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My question is motivated by the inverse of $A^-1+B^-1$.
$$A^-1+B^-1=A^-1(A+B)B^-1implies(A^-1+B^-1)^-1=B(A+B)^-1A$$
$$A^-1+B^-1=B^-1+A^-1=B^-1(A+B)A^-1implies(A^-1+B^-1)^-1=A(A+B)^-1B$$
$$B(A+B)^-1A=A(A+B)^-1B$$
Is this a special result due to the fact that $(A+B)^-1$ is sandwiched between $A$ and $B$, or does it hold for other cases as well, i.e. $AXB=BXA$ where $X$ has some special properties?
If it's the former, then some intuition for why it holds beyond the math mechanics that I have shown above would be appreciated.
matrices inverse inverse-function
asked Jul 16 at 8:55
Amrit Prasad
828
My question is motivated by the inverse of $A^-1+B^-1$.
$$A^-1+B^-1=A^-1(A+B)B^-1implies(A^-1+B^-1)^-1=B(A+B)^-1A$$
$$A^-1+B^-1=B^-1+A^-1=B^-1(A+B)A^-1implies(A^-1+B^-1)^-1=A(A+B)^-1B$$
$$B(A+B)^-1A=A(A+B)^-1B$$
Is this a special result due to the fact that $(A+B)^-1$ is sandwiched between $A$ and $B$, or does it hold for other cases as well, i.e. $AXB=BXA$ where $X$ has some special properties?
If it's the former, then some intuition for why it holds beyond the math mechanics that I have shown above would be appreciated.
matrices inverse inverse-function
asked Jul 16 at 8:55
Amrit Prasad
828
asked Jul 16 at 8:55
Amrit Prasad
828
asked Jul 16 at 8:55
Amrit Prasad
828
asked Jul 16 at 8:55
Amrit Prasad
828
828
I'm not sure if I'd call it a duplicate, but related: math.stackexchange.com/questions/2852002/…
– Theo Bendit
Jul 16 at 9:23
Thanks for pointing that particular thread out. Searching in stackexchange isn't particularly great, else I would have probably left the above as a comment on the top answer on that thread. The second part of my question is to figure out all $X$ values for which this is true. This was not answered on that thread. I'm happy to merge this with the other question if needed though.
– Amrit Prasad
Jul 16 at 9:45
add a comment |Â
I'm not sure if I'd call it a duplicate, but related: math.stackexchange.com/questions/2852002/…
– Theo Bendit
Jul 16 at 9:23
Thanks for pointing that particular thread out. Searching in stackexchange isn't particularly great, else I would have probably left the above as a comment on the top answer on that thread. The second part of my question is to figure out all $X$ values for which this is true. This was not answered on that thread. I'm happy to merge this with the other question if needed though.
– Amrit Prasad
Jul 16 at 9:45
I'm not sure if I'd call it a duplicate, but related: math.stackexchange.com/questions/2852002/…
– Theo Bendit
Jul 16 at 9:23
I'm not sure if I'd call it a duplicate, but related: math.stackexchange.com/questions/2852002/…
– Theo Bendit
Jul 16 at 9:23
Thanks for pointing that particular thread out. Searching in stackexchange isn't particularly great, else I would have probably left the above as a comment on the top answer on that thread. The second part of my question is to figure out all $X$ values for which this is true. This was not answered on that thread. I'm happy to merge this with the other question if needed though.
– Amrit Prasad
Jul 16 at 9:45
Thanks for pointing that particular thread out. Searching in stackexchange isn't particularly great, else I would have probably left the above as a comment on the top answer on that thread. The second part of my question is to figure out all $X$ values for which this is true. This was not answered on that thread. I'm happy to merge this with the other question if needed though.
– Amrit Prasad
Jul 16 at 9:45
add a comment |Â
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I'm not sure if I'd call it a duplicate, but related: math.stackexchange.com/questions/2852002/…
– Theo Bendit
Jul 16 at 9:23
Thanks for pointing that particular thread out. Searching in stackexchange isn't particularly great, else I would have probably left the above as a comment on the top answer on that thread. The second part of my question is to figure out all $X$ values for which this is true. This was not answered on that thread. I'm happy to merge this with the other question if needed though.
– Amrit Prasad
Jul 16 at 9:45