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Why can the Cholesky decomposition be used to invert non-Hermitian matrices?
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The Wikipedia article and the paper it refers to have a blurb on it, but I haven't be able to find an explanation for this. More specifically, why does multiplying the transpose of the original matrix and the inverse of its square equal the inverse of the original matrix?
linear-algebra matrices inverse matrix-decomposition
edited Jul 26 at 9:09
Rodrigo de Azevedo
12.5k41751
asked Jul 26 at 9:03
Marko Grdinic
1369
add a comment |Â
up vote
0
down vote
favorite
The Wikipedia article and the paper it refers to have a blurb on it, but I haven't be able to find an explanation for this. More specifically, why does multiplying the transpose of the original matrix and the inverse of its square equal the inverse of the original matrix?
linear-algebra matrices inverse matrix-decomposition
edited Jul 26 at 9:09
Rodrigo de Azevedo
12.5k41751
asked Jul 26 at 9:03
Marko Grdinic
1369
Cholesky decomposition leads to upper and lower triangular matrices, which are easy to invert.
– nicomezi
Jul 26 at 9:09
1
If $B$ is invertible then so is $BB^*$, and $B^*(BB^*)^-1=B^*(B^*)^-1B^-1=B^-1$.
– Rahul
Jul 26 at 9:28
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The Wikipedia article and the paper it refers to have a blurb on it, but I haven't be able to find an explanation for this. More specifically, why does multiplying the transpose of the original matrix and the inverse of its square equal the inverse of the original matrix?
linear-algebra matrices inverse matrix-decomposition
edited Jul 26 at 9:09
Rodrigo de Azevedo
12.5k41751
asked Jul 26 at 9:03
Marko Grdinic
1369
The Wikipedia article and the paper it refers to have a blurb on it, but I haven't be able to find an explanation for this. More specifically, why does multiplying the transpose of the original matrix and the inverse of its square equal the inverse of the original matrix?
linear-algebra matrices inverse matrix-decomposition
edited Jul 26 at 9:09
Rodrigo de Azevedo
12.5k41751
asked Jul 26 at 9:03
Marko Grdinic
1369
edited Jul 26 at 9:09
Rodrigo de Azevedo
12.5k41751
edited Jul 26 at 9:09
Rodrigo de Azevedo
12.5k41751
edited Jul 26 at 9:09
Rodrigo de Azevedo
12.5k41751
12.5k41751
asked Jul 26 at 9:03
Marko Grdinic
1369
asked Jul 26 at 9:03
Marko Grdinic
1369
asked Jul 26 at 9:03
Marko Grdinic
1369
1369
Cholesky decomposition leads to upper and lower triangular matrices, which are easy to invert.
– nicomezi
Jul 26 at 9:09
1
If $B$ is invertible then so is $BB^*$, and $B^*(BB^*)^-1=B^*(B^*)^-1B^-1=B^-1$.
– Rahul
Jul 26 at 9:28
add a comment |Â
Cholesky decomposition leads to upper and lower triangular matrices, which are easy to invert.
– nicomezi
Jul 26 at 9:09
1
If $B$ is invertible then so is $BB^*$, and $B^*(BB^*)^-1=B^*(B^*)^-1B^-1=B^-1$.
– Rahul
Jul 26 at 9:28
Cholesky decomposition leads to upper and lower triangular matrices, which are easy to invert.
– nicomezi
Jul 26 at 9:09
Cholesky decomposition leads to upper and lower triangular matrices, which are easy to invert.
– nicomezi
Jul 26 at 9:09
1
1
If $B$ is invertible then so is $BB^*$, and $B^*(BB^*)^-1=B^*(B^*)^-1B^-1=B^-1$.
– Rahul
Jul 26 at 9:28
If $B$ is invertible then so is $BB^*$, and $B^*(BB^*)^-1=B^*(B^*)^-1B^-1=B^-1$.
– Rahul
Jul 26 at 9:28
add a comment |Â
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Cholesky decomposition leads to upper and lower triangular matrices, which are easy to invert.
– nicomezi
Jul 26 at 9:09
1
If $B$ is invertible then so is $BB^*$, and $B^*(BB^*)^-1=B^*(B^*)^-1B^-1=B^-1$.
– Rahul
Jul 26 at 9:28