Multiply block matrices with uneven blocks The Next CEO of Stack OverflowLinear Algebra: Block MultiplyHow to multiply a matrix of block matrices?Proofs of Determinants of Block matricesdiagonalize block matrices with special type of unitary matricesVery basic computations in determinants and matricesForming unitary matrices from identity relationJordan blocks, similarity classDeterminants of block tridiagonal matrices when off-diagonal blocks are not m*m matricesDo matrix multiplication rules apply when multiplying matrices made up of smaller matrices?Question about elementary row operations with block matrices
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Multiply block matrices with uneven blocks
The Next CEO of Stack OverflowLinear Algebra: Block MultiplyHow to multiply a matrix of block matrices?Proofs of Determinants of Block matricesdiagonalize block matrices with special type of unitary matricesVery basic computations in determinants and matricesForming unitary matrices from identity relationJordan blocks, similarity classDeterminants of block tridiagonal matrices when off-diagonal blocks are not m*m matricesDo matrix multiplication rules apply when multiplying matrices made up of smaller matrices?Question about elementary row operations with block matrices
$begingroup$
Is there a way to multiply matrices which are divided into blocks of the form
beginbmatrix
A&B
endbmatrix
For some $ntimes n $ matrix A and $n times 1$ matrix B with its transpose from the left? Without having to multiply every entry.
linear-algebra
asked yesterday
TwistTwist
636
$endgroup$
add a comment |
$begingroup$
Is there a way to multiply matrices which are divided into blocks of the form
beginbmatrix
A&B
endbmatrix
For some $ntimes n $ matrix A and $n times 1$ matrix B with its transpose from the left? Without having to multiply every entry.
linear-algebra
asked yesterday
TwistTwist
636
$endgroup$
add a comment |
$begingroup$
Is there a way to multiply matrices which are divided into blocks of the form
beginbmatrix
A&B
endbmatrix
For some $ntimes n $ matrix A and $n times 1$ matrix B with its transpose from the left? Without having to multiply every entry.
linear-algebra
asked yesterday
TwistTwist
636
$endgroup$
Is there a way to multiply matrices which are divided into blocks of the form
beginbmatrix
A&B
endbmatrix
For some $ntimes n $ matrix A and $n times 1$ matrix B with its transpose from the left? Without having to multiply every entry.
linear-algebra
linear-algebra
asked yesterday
TwistTwist
636
asked yesterday
TwistTwist
636
asked yesterday
TwistTwist
636
asked yesterday
TwistTwist
636
asked yesterday
TwistTwist
636
636
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I assume you mean $beginbmatrix A & B endbmatrix^Tbeginbmatrix A & B endbmatrix$.
Then multiplication can be done using block matrices
$beginbmatrix A & B endbmatrix^Tbeginbmatrix A & B endbmatrix=beginbmatrix A^T \ B^T endbmatrix beginbmatrix A & B endbmatrix =beginbmatrix A^TA & A^TB \ B^TA & B^TBendbmatrix $
Operation is possible if dimensions of block matrices are appropriate for multiplication operation. In this case they are appropriate:
$ A^TA: $ $n times n , n times n$ result $n times n $
$ A^TB: $ $n times n , n times 1$ result $n times 1 $
$ B^TA: $ $1 times n , n times n$ result $1 times n $
$ B^TB: $ $1 times n , n times 1$ result $1 times 1 $.
answered yesterday
WidawensenWidawensen
4,72831446
$endgroup$
add a comment |
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$begingroup$
I assume you mean $beginbmatrix A & B endbmatrix^Tbeginbmatrix A & B endbmatrix$.
Then multiplication can be done using block matrices
$beginbmatrix A & B endbmatrix^Tbeginbmatrix A & B endbmatrix=beginbmatrix A^T \ B^T endbmatrix beginbmatrix A & B endbmatrix =beginbmatrix A^TA & A^TB \ B^TA & B^TBendbmatrix $
Operation is possible if dimensions of block matrices are appropriate for multiplication operation. In this case they are appropriate:
$ A^TA: $ $n times n , n times n$ result $n times n $
$ A^TB: $ $n times n , n times 1$ result $n times 1 $
$ B^TA: $ $1 times n , n times n$ result $1 times n $
$ B^TB: $ $1 times n , n times 1$ result $1 times 1 $.
answered yesterday
WidawensenWidawensen
4,72831446
$endgroup$
add a comment |
$begingroup$
I assume you mean $beginbmatrix A & B endbmatrix^Tbeginbmatrix A & B endbmatrix$.
Then multiplication can be done using block matrices
$beginbmatrix A & B endbmatrix^Tbeginbmatrix A & B endbmatrix=beginbmatrix A^T \ B^T endbmatrix beginbmatrix A & B endbmatrix =beginbmatrix A^TA & A^TB \ B^TA & B^TBendbmatrix $
Operation is possible if dimensions of block matrices are appropriate for multiplication operation. In this case they are appropriate:
$ A^TA: $ $n times n , n times n$ result $n times n $
$ A^TB: $ $n times n , n times 1$ result $n times 1 $
$ B^TA: $ $1 times n , n times n$ result $1 times n $
$ B^TB: $ $1 times n , n times 1$ result $1 times 1 $.
answered yesterday
WidawensenWidawensen
4,72831446
$endgroup$
add a comment |
$begingroup$
I assume you mean $beginbmatrix A & B endbmatrix^Tbeginbmatrix A & B endbmatrix$.
Then multiplication can be done using block matrices
$beginbmatrix A & B endbmatrix^Tbeginbmatrix A & B endbmatrix=beginbmatrix A^T \ B^T endbmatrix beginbmatrix A & B endbmatrix =beginbmatrix A^TA & A^TB \ B^TA & B^TBendbmatrix $
Operation is possible if dimensions of block matrices are appropriate for multiplication operation. In this case they are appropriate:
$ A^TA: $ $n times n , n times n$ result $n times n $
$ A^TB: $ $n times n , n times 1$ result $n times 1 $
$ B^TA: $ $1 times n , n times n$ result $1 times n $
$ B^TB: $ $1 times n , n times 1$ result $1 times 1 $.
answered yesterday
WidawensenWidawensen
4,72831446
$endgroup$
I assume you mean $beginbmatrix A & B endbmatrix^Tbeginbmatrix A & B endbmatrix$.
Then multiplication can be done using block matrices
$beginbmatrix A & B endbmatrix^Tbeginbmatrix A & B endbmatrix=beginbmatrix A^T \ B^T endbmatrix beginbmatrix A & B endbmatrix =beginbmatrix A^TA & A^TB \ B^TA & B^TBendbmatrix $
Operation is possible if dimensions of block matrices are appropriate for multiplication operation. In this case they are appropriate:
$ A^TA: $ $n times n , n times n$ result $n times n $
$ A^TB: $ $n times n , n times 1$ result $n times 1 $
$ B^TA: $ $1 times n , n times n$ result $1 times n $
$ B^TB: $ $1 times n , n times 1$ result $1 times 1 $.
answered yesterday
WidawensenWidawensen
4,72831446
answered yesterday
WidawensenWidawensen
4,72831446
answered yesterday
WidawensenWidawensen
4,72831446
answered yesterday
WidawensenWidawensen
4,72831446
4,72831446
add a comment |
add a comment |
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