Perform Singular Value Decomposition of (2 0 1;0 1 0)
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I've got the questions to perform a singular value decomposition for the matrix A = (2 0 1;0 1 0). I know the method and calculated it to be,
A = PDQ, where
P = (5 0;0 1)
Q = (1/sqrt(5)).(2 0 -1;0 1 0;1 0 2)
D = (sqrt(5) 0 0; 0 1 0)
my only issue is when I put it back into the equation where A = PDQ, I dont get the original matrix back, can someone figure out which one of my 3 variables are wrong? Or if im just being silly
Thanks
matrices
asked Jan 12 at 23:46
Mitul SuchakMitul Suchak
65
$endgroup$
add a comment |
$begingroup$
I've got the questions to perform a singular value decomposition for the matrix A = (2 0 1;0 1 0). I know the method and calculated it to be,
A = PDQ, where
P = (5 0;0 1)
Q = (1/sqrt(5)).(2 0 -1;0 1 0;1 0 2)
D = (sqrt(5) 0 0; 0 1 0)
my only issue is when I put it back into the equation where A = PDQ, I dont get the original matrix back, can someone figure out which one of my 3 variables are wrong? Or if im just being silly
Thanks
matrices
asked Jan 12 at 23:46
Mitul SuchakMitul Suchak
65
$endgroup$
$begingroup$
You need to calculate $PDQ^T$. Also, $P=I_2$.
$endgroup$
– obareey
Jan 13 at 9:57
$begingroup$
See the way they teach it to me at uni is PDQ, but enter Q as row vectors instead of column, this is so when calculating the pseudoinverse we can do A+ = P(T)D+Q(T)
$endgroup$
– Mitul Suchak
Jan 13 at 16:01
$begingroup$
IM SO SILLY, my P was wrong, no idea how I got 5 0 for the first row, dont think i divided by anything
$endgroup$
– Mitul Suchak
Jan 13 at 16:26
add a comment |
$begingroup$
I've got the questions to perform a singular value decomposition for the matrix A = (2 0 1;0 1 0). I know the method and calculated it to be,
A = PDQ, where
P = (5 0;0 1)
Q = (1/sqrt(5)).(2 0 -1;0 1 0;1 0 2)
D = (sqrt(5) 0 0; 0 1 0)
my only issue is when I put it back into the equation where A = PDQ, I dont get the original matrix back, can someone figure out which one of my 3 variables are wrong? Or if im just being silly
Thanks
matrices
asked Jan 12 at 23:46
Mitul SuchakMitul Suchak
65
$endgroup$
I've got the questions to perform a singular value decomposition for the matrix A = (2 0 1;0 1 0). I know the method and calculated it to be,
A = PDQ, where
P = (5 0;0 1)
Q = (1/sqrt(5)).(2 0 -1;0 1 0;1 0 2)
D = (sqrt(5) 0 0; 0 1 0)
my only issue is when I put it back into the equation where A = PDQ, I dont get the original matrix back, can someone figure out which one of my 3 variables are wrong? Or if im just being silly
Thanks
matrices
matrices
asked Jan 12 at 23:46
Mitul SuchakMitul Suchak
65
asked Jan 12 at 23:46
Mitul SuchakMitul Suchak
65
asked Jan 12 at 23:46
Mitul SuchakMitul Suchak
65
asked Jan 12 at 23:46
Mitul SuchakMitul Suchak
65
asked Jan 12 at 23:46
Mitul SuchakMitul Suchak
65
65
$begingroup$
You need to calculate $PDQ^T$. Also, $P=I_2$.
$endgroup$
– obareey
Jan 13 at 9:57
$begingroup$
See the way they teach it to me at uni is PDQ, but enter Q as row vectors instead of column, this is so when calculating the pseudoinverse we can do A+ = P(T)D+Q(T)
$endgroup$
– Mitul Suchak
Jan 13 at 16:01
$begingroup$
IM SO SILLY, my P was wrong, no idea how I got 5 0 for the first row, dont think i divided by anything
$endgroup$
– Mitul Suchak
Jan 13 at 16:26
add a comment |
$begingroup$
You need to calculate $PDQ^T$. Also, $P=I_2$.
$endgroup$
– obareey
Jan 13 at 9:57
$begingroup$
See the way they teach it to me at uni is PDQ, but enter Q as row vectors instead of column, this is so when calculating the pseudoinverse we can do A+ = P(T)D+Q(T)
$endgroup$
– Mitul Suchak
Jan 13 at 16:01
$begingroup$
IM SO SILLY, my P was wrong, no idea how I got 5 0 for the first row, dont think i divided by anything
$endgroup$
– Mitul Suchak
Jan 13 at 16:26
$begingroup$
You need to calculate $PDQ^T$. Also, $P=I_2$.
$endgroup$
– obareey
Jan 13 at 9:57
$begingroup$
You need to calculate $PDQ^T$. Also, $P=I_2$.
$endgroup$
– obareey
Jan 13 at 9:57
$begingroup$
See the way they teach it to me at uni is PDQ, but enter Q as row vectors instead of column, this is so when calculating the pseudoinverse we can do A+ = P(T)D+Q(T)
$endgroup$
– Mitul Suchak
Jan 13 at 16:01
$begingroup$
See the way they teach it to me at uni is PDQ, but enter Q as row vectors instead of column, this is so when calculating the pseudoinverse we can do A+ = P(T)D+Q(T)
$endgroup$
– Mitul Suchak
Jan 13 at 16:01
$begingroup$
IM SO SILLY, my P was wrong, no idea how I got 5 0 for the first row, dont think i divided by anything
$endgroup$
– Mitul Suchak
Jan 13 at 16:26
$begingroup$
IM SO SILLY, my P was wrong, no idea how I got 5 0 for the first row, dont think i divided by anything
$endgroup$
– Mitul Suchak
Jan 13 at 16:26
add a comment |
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$begingroup$
You need to calculate $PDQ^T$. Also, $P=I_2$.
$endgroup$
– obareey
Jan 13 at 9:57
$begingroup$
See the way they teach it to me at uni is PDQ, but enter Q as row vectors instead of column, this is so when calculating the pseudoinverse we can do A+ = P(T)D+Q(T)
$endgroup$
– Mitul Suchak
Jan 13 at 16:01
$begingroup$
IM SO SILLY, my P was wrong, no idea how I got 5 0 for the first row, dont think i divided by anything
$endgroup$
– Mitul Suchak
Jan 13 at 16:26