Show that $A^-1 + B^-1$ is invertible when $A,B$ and $A+B$ are invertible Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Show that $A$ is invertible and that $AB=BA$How to show that B is not invertibleProof if $I+AB$ invertible then $I+BA$ invertible and $(I+BA)^-1=I-B(I+AB)^-1A$How to prove that If A is invertible then $A^-1$ and $A^2$ are invertible.When is the matrix $X^t A X$ invertible, with $A$ invertible.Solution to a matrix-valued ODE is invertible at all times assuming it is at a given time.Show that invertible matrices with an additional condition are diagonalizable.Showing Schur complement submatrices are invertible provided A has all its principal submatrices invertibleTwo invertible matricesProving matrices equation when all the matrices in it may not be invertible
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Show that $A^-1 + B^-1$ is invertible when $A,B$ and $A+B$ are invertible
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Show that $A$ is invertible and that $AB=BA$How to show that B is not invertibleProof if $I+AB$ invertible then $I+BA$ invertible and $(I+BA)^-1=I-B(I+AB)^-1A$How to prove that If A is invertible then $A^-1$ and $A^2$ are invertible.When is the matrix $X^t A X$ invertible, with $A$ invertible.Solution to a matrix-valued ODE is invertible at all times assuming it is at a given time.Show that invertible matrices with an additional condition are diagonalizable.Showing Schur complement submatrices are invertible provided A has all its principal submatrices invertibleTwo invertible matricesProving matrices equation when all the matrices in it may not be invertible
$begingroup$
I have the following issue: $A,Binmathbb C^ntimes n$ invertible, such that also $A + B$ is invertible. How is it shown that $A^-1 + B^-1$ is invertible?
matrices matrix-equations
edited Mar 28 at 2:15
FredH
3,6851024
asked Mar 27 at 15:44
Andreu Gooz BielAndreu Gooz Biel
162
$endgroup$
add a comment |
$begingroup$
I have the following issue: $A,Binmathbb C^ntimes n$ invertible, such that also $A + B$ is invertible. How is it shown that $A^-1 + B^-1$ is invertible?
matrices matrix-equations
edited Mar 28 at 2:15
FredH
3,6851024
asked Mar 27 at 15:44
Andreu Gooz BielAndreu Gooz Biel
162
$endgroup$
$begingroup$
See the introduction to posting mathematical notation.
$endgroup$
– hardmath
Mar 27 at 15:52
$begingroup$
Can we fix the title ?
$endgroup$
– T. Fo
Mar 27 at 19:57
$begingroup$
@Andreu Gooz Biel: If your question has been answered below, please, accept an answer. Otherwise your question remains open indefinitely. Thank you!
$endgroup$
– Moritz
Apr 1 at 20:59
add a comment |
$begingroup$
I have the following issue: $A,Binmathbb C^ntimes n$ invertible, such that also $A + B$ is invertible. How is it shown that $A^-1 + B^-1$ is invertible?
matrices matrix-equations
edited Mar 28 at 2:15
FredH
3,6851024
asked Mar 27 at 15:44
Andreu Gooz BielAndreu Gooz Biel
162
$endgroup$
I have the following issue: $A,Binmathbb C^ntimes n$ invertible, such that also $A + B$ is invertible. How is it shown that $A^-1 + B^-1$ is invertible?
matrices matrix-equations
matrices matrix-equations
edited Mar 28 at 2:15
FredH
3,6851024
asked Mar 27 at 15:44
Andreu Gooz BielAndreu Gooz Biel
162
edited Mar 28 at 2:15
FredH
3,6851024
asked Mar 27 at 15:44
Andreu Gooz BielAndreu Gooz Biel
162
edited Mar 28 at 2:15
FredH
3,6851024
edited Mar 28 at 2:15
FredH
3,6851024
edited Mar 28 at 2:15
FredH
3,6851024
3,6851024
asked Mar 27 at 15:44
Andreu Gooz BielAndreu Gooz Biel
162
asked Mar 27 at 15:44
Andreu Gooz BielAndreu Gooz Biel
162
asked Mar 27 at 15:44
Andreu Gooz BielAndreu Gooz Biel
162
162
$begingroup$
See the introduction to posting mathematical notation.
$endgroup$
– hardmath
Mar 27 at 15:52
$begingroup$
Can we fix the title ?
$endgroup$
– T. Fo
Mar 27 at 19:57
$begingroup$
@Andreu Gooz Biel: If your question has been answered below, please, accept an answer. Otherwise your question remains open indefinitely. Thank you!
$endgroup$
– Moritz
Apr 1 at 20:59
add a comment |
$begingroup$
See the introduction to posting mathematical notation.
$endgroup$
– hardmath
Mar 27 at 15:52
$begingroup$
Can we fix the title ?
$endgroup$
– T. Fo
Mar 27 at 19:57
$begingroup$
@Andreu Gooz Biel: If your question has been answered below, please, accept an answer. Otherwise your question remains open indefinitely. Thank you!
$endgroup$
– Moritz
Apr 1 at 20:59
$begingroup$
See the introduction to posting mathematical notation.
$endgroup$
– hardmath
Mar 27 at 15:52
$begingroup$
See the introduction to posting mathematical notation.
$endgroup$
– hardmath
Mar 27 at 15:52
$begingroup$
Can we fix the title ?
$endgroup$
– T. Fo
Mar 27 at 19:57
$begingroup$
Can we fix the title ?
$endgroup$
– T. Fo
Mar 27 at 19:57
$begingroup$
@Andreu Gooz Biel: If your question has been answered below, please, accept an answer. Otherwise your question remains open indefinitely. Thank you!
$endgroup$
– Moritz
Apr 1 at 20:59
$begingroup$
@Andreu Gooz Biel: If your question has been answered below, please, accept an answer. Otherwise your question remains open indefinitely. Thank you!
$endgroup$
– Moritz
Apr 1 at 20:59
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$
A^-1+B^-1 = A^-1(A+B)B^-1
$$
By the way: (Spanish) demonstración --> (English) proof. ;-)
answered Mar 27 at 15:48
amsmathamsmath
3,387421
$endgroup$
add a comment |
$begingroup$
Here is a more pedantic approach to @amsmath's slick approach:
Suppose we want to solve $(A^-1 + B^-1) x = A^-1x + B^-1 x = y$.
Then $Ay=x + AB^-1 x $, and letting $x'=B^-1 x$ we get
$Ay = B x' + A x' = (A+B)x'$ and so $x'= (A+B)^-1 Ay$ and finally
$x=Bx' = B(A+B)^-1 Ay$.
Hence $(A^-1 + B^-1)^-1 = B(A+B)^-1 A$.
answered Mar 27 at 16:05
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
$begingroup$
Of course you know that a square matrix has a two sided inverse if you have found an inverse from one side.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$
A^-1+B^-1 = A^-1(A+B)B^-1
$$
By the way: (Spanish) demonstración --> (English) proof. ;-)
answered Mar 27 at 15:48
amsmathamsmath
3,387421
$endgroup$
add a comment |
$begingroup$
$$
A^-1+B^-1 = A^-1(A+B)B^-1
$$
By the way: (Spanish) demonstración --> (English) proof. ;-)
answered Mar 27 at 15:48
amsmathamsmath
3,387421
$endgroup$
add a comment |
$begingroup$
$$
A^-1+B^-1 = A^-1(A+B)B^-1
$$
By the way: (Spanish) demonstración --> (English) proof. ;-)
answered Mar 27 at 15:48
amsmathamsmath
3,387421
$endgroup$
$$
A^-1+B^-1 = A^-1(A+B)B^-1
$$
By the way: (Spanish) demonstración --> (English) proof. ;-)
answered Mar 27 at 15:48
amsmathamsmath
3,387421
answered Mar 27 at 15:48
amsmathamsmath
3,387421
answered Mar 27 at 15:48
amsmathamsmath
3,387421
answered Mar 27 at 15:48
amsmathamsmath
3,387421
3,387421
add a comment |
add a comment |
$begingroup$
Here is a more pedantic approach to @amsmath's slick approach:
Suppose we want to solve $(A^-1 + B^-1) x = A^-1x + B^-1 x = y$.
Then $Ay=x + AB^-1 x $, and letting $x'=B^-1 x$ we get
$Ay = B x' + A x' = (A+B)x'$ and so $x'= (A+B)^-1 Ay$ and finally
$x=Bx' = B(A+B)^-1 Ay$.
Hence $(A^-1 + B^-1)^-1 = B(A+B)^-1 A$.
answered Mar 27 at 16:05
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
$begingroup$
Here is a more pedantic approach to @amsmath's slick approach:
Suppose we want to solve $(A^-1 + B^-1) x = A^-1x + B^-1 x = y$.
Then $Ay=x + AB^-1 x $, and letting $x'=B^-1 x$ we get
$Ay = B x' + A x' = (A+B)x'$ and so $x'= (A+B)^-1 Ay$ and finally
$x=Bx' = B(A+B)^-1 Ay$.
Hence $(A^-1 + B^-1)^-1 = B(A+B)^-1 A$.
answered Mar 27 at 16:05
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
$begingroup$
Here is a more pedantic approach to @amsmath's slick approach:
Suppose we want to solve $(A^-1 + B^-1) x = A^-1x + B^-1 x = y$.
Then $Ay=x + AB^-1 x $, and letting $x'=B^-1 x$ we get
$Ay = B x' + A x' = (A+B)x'$ and so $x'= (A+B)^-1 Ay$ and finally
$x=Bx' = B(A+B)^-1 Ay$.
Hence $(A^-1 + B^-1)^-1 = B(A+B)^-1 A$.
answered Mar 27 at 16:05
copper.hatcopper.hat
128k561161
$endgroup$
Here is a more pedantic approach to @amsmath's slick approach:
Suppose we want to solve $(A^-1 + B^-1) x = A^-1x + B^-1 x = y$.
Then $Ay=x + AB^-1 x $, and letting $x'=B^-1 x$ we get
$Ay = B x' + A x' = (A+B)x'$ and so $x'= (A+B)^-1 Ay$ and finally
$x=Bx' = B(A+B)^-1 Ay$.
Hence $(A^-1 + B^-1)^-1 = B(A+B)^-1 A$.
answered Mar 27 at 16:05
copper.hatcopper.hat
128k561161
answered Mar 27 at 16:05
copper.hatcopper.hat
128k561161
answered Mar 27 at 16:05
copper.hatcopper.hat
128k561161
answered Mar 27 at 16:05
copper.hatcopper.hat
128k561161
128k561161
add a comment |
add a comment |
$begingroup$
Of course you know that a square matrix has a two sided inverse if you have found an inverse from one side.
$endgroup$
add a comment |
$begingroup$
Of course you know that a square matrix has a two sided inverse if you have found an inverse from one side.
$endgroup$
add a comment |
$begingroup$
Of course you know that a square matrix has a two sided inverse if you have found an inverse from one side.
$endgroup$
Of course you know that a square matrix has a two sided inverse if you have found an inverse from one side.
answered Mar 27 at 16:05
community wiki
Moritz
add a comment |
add a comment |
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$begingroup$
See the introduction to posting mathematical notation.
$endgroup$
– hardmath
Mar 27 at 15:52
$begingroup$
Can we fix the title ?
$endgroup$
– T. Fo
Mar 27 at 19:57
$begingroup$
@Andreu Gooz Biel: If your question has been answered below, please, accept an answer. Otherwise your question remains open indefinitely. Thank you!
$endgroup$
– Moritz
Apr 1 at 20:59