Fast calculation of the Hadamard transform of a 4X4 matrix
$begingroup$
I'm reading a book envolves Hadamard transformations, and in one of the examples they are trying to reconstruct Y from RX+Z where all are 4X4 matrixes.
R is given by:
R= 1/4 * begin{array}{l}5&3&3&-3\3&5&-3&3\3&-3&5&3\-3&3&3&5end{array}
And one step further (without any explanation they write that
H(R) = HRH =
begin{array}{l}2&0&0&0\0&2&0&0\0&0&2&0\0&0&0&-1end{array}
Did they calculate HRH "behind the scenes" or there's a fast way of calculating the transform gives certain characteristics to the matrix?
hadamard-matrices
asked Jan 6 at 9:03
DsCppDsCpp
1065
$endgroup$
add a comment |
$begingroup$
I'm reading a book envolves Hadamard transformations, and in one of the examples they are trying to reconstruct Y from RX+Z where all are 4X4 matrixes.
R is given by:
R= 1/4 * begin{array}{l}5&3&3&-3\3&5&-3&3\3&-3&5&3\-3&3&3&5end{array}
And one step further (without any explanation they write that
H(R) = HRH =
begin{array}{l}2&0&0&0\0&2&0&0\0&0&2&0\0&0&0&-1end{array}
Did they calculate HRH "behind the scenes" or there's a fast way of calculating the transform gives certain characteristics to the matrix?
hadamard-matrices
asked Jan 6 at 9:03
DsCppDsCpp
1065
$endgroup$
add a comment |
$begingroup$
I'm reading a book envolves Hadamard transformations, and in one of the examples they are trying to reconstruct Y from RX+Z where all are 4X4 matrixes.
R is given by:
R= 1/4 * begin{array}{l}5&3&3&-3\3&5&-3&3\3&-3&5&3\-3&3&3&5end{array}
And one step further (without any explanation they write that
H(R) = HRH =
begin{array}{l}2&0&0&0\0&2&0&0\0&0&2&0\0&0&0&-1end{array}
Did they calculate HRH "behind the scenes" or there's a fast way of calculating the transform gives certain characteristics to the matrix?
hadamard-matrices
asked Jan 6 at 9:03
DsCppDsCpp
1065
$endgroup$
I'm reading a book envolves Hadamard transformations, and in one of the examples they are trying to reconstruct Y from RX+Z where all are 4X4 matrixes.
R is given by:
R= 1/4 * begin{array}{l}5&3&3&-3\3&5&-3&3\3&-3&5&3\-3&3&3&5end{array}
And one step further (without any explanation they write that
H(R) = HRH =
begin{array}{l}2&0&0&0\0&2&0&0\0&0&2&0\0&0&0&-1end{array}
Did they calculate HRH "behind the scenes" or there's a fast way of calculating the transform gives certain characteristics to the matrix?
hadamard-matrices
hadamard-matrices
asked Jan 6 at 9:03
DsCppDsCpp
1065
asked Jan 6 at 9:03
DsCppDsCpp
1065
edited Jan 6 at 9:15
DsCpp
asked Jan 6 at 9:03
DsCppDsCpp
1065
asked Jan 6 at 9:03
DsCppDsCpp
1065
asked Jan 6 at 9:03
DsCppDsCpp
1065
1065
add a comment |
add a comment |
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