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what is ranks do in Singular value decomposition, if rank = k, others than k first singular values set to 0 or gone?
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hi lets assume we have matrix A with 4 rows x 3 columns, when we input to svd it become:
U = 4x4 , S = 4x3 and VT = 3x3.
after that i specify that rank k=2 so what exactly happen to the other than the first k ranks? is it set to zero or completely gone like
U = 4x2 S=2x2 and VT = 2x3?
both cases resulting the same rows and columns, right? which is 4x3, but does it affect the value of the reduced matrix?
linear-algebra matrices matrix-decomposition svd
asked Jul 30 at 14:25
bernard
102
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up vote
1
down vote
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hi lets assume we have matrix A with 4 rows x 3 columns, when we input to svd it become:
U = 4x4 , S = 4x3 and VT = 3x3.
after that i specify that rank k=2 so what exactly happen to the other than the first k ranks? is it set to zero or completely gone like
U = 4x2 S=2x2 and VT = 2x3?
both cases resulting the same rows and columns, right? which is 4x3, but does it affect the value of the reduced matrix?
linear-algebra matrices matrix-decomposition svd
asked Jul 30 at 14:25
bernard
102
It may be obvious, but in the presence of numerical rounding errors, one expects some spurious nonzero singular values to appear in the decomposition. I'm guessing that these are not of concern to you, but if they are, reply to this or edit the body of the Question, and I'll write up something about "rank revealing SVD".
– hardmath
Jul 30 at 16:13
add a comment |Â
up vote
1
down vote
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up vote
1
down vote
favorite
hi lets assume we have matrix A with 4 rows x 3 columns, when we input to svd it become:
U = 4x4 , S = 4x3 and VT = 3x3.
after that i specify that rank k=2 so what exactly happen to the other than the first k ranks? is it set to zero or completely gone like
U = 4x2 S=2x2 and VT = 2x3?
both cases resulting the same rows and columns, right? which is 4x3, but does it affect the value of the reduced matrix?
linear-algebra matrices matrix-decomposition svd
asked Jul 30 at 14:25
bernard
102
hi lets assume we have matrix A with 4 rows x 3 columns, when we input to svd it become:
U = 4x4 , S = 4x3 and VT = 3x3.
after that i specify that rank k=2 so what exactly happen to the other than the first k ranks? is it set to zero or completely gone like
U = 4x2 S=2x2 and VT = 2x3?
both cases resulting the same rows and columns, right? which is 4x3, but does it affect the value of the reduced matrix?
linear-algebra matrices matrix-decomposition svd
asked Jul 30 at 14:25
bernard
102
asked Jul 30 at 14:25
bernard
102
asked Jul 30 at 14:25
bernard
102
asked Jul 30 at 14:25
bernard
102
102
It may be obvious, but in the presence of numerical rounding errors, one expects some spurious nonzero singular values to appear in the decomposition. I'm guessing that these are not of concern to you, but if they are, reply to this or edit the body of the Question, and I'll write up something about "rank revealing SVD".
– hardmath
Jul 30 at 16:13
add a comment |Â
It may be obvious, but in the presence of numerical rounding errors, one expects some spurious nonzero singular values to appear in the decomposition. I'm guessing that these are not of concern to you, but if they are, reply to this or edit the body of the Question, and I'll write up something about "rank revealing SVD".
– hardmath
Jul 30 at 16:13
It may be obvious, but in the presence of numerical rounding errors, one expects some spurious nonzero singular values to appear in the decomposition. I'm guessing that these are not of concern to you, but if they are, reply to this or edit the body of the Question, and I'll write up something about "rank revealing SVD".
– hardmath
Jul 30 at 16:13
It may be obvious, but in the presence of numerical rounding errors, one expects some spurious nonzero singular values to appear in the decomposition. I'm guessing that these are not of concern to you, but if they are, reply to this or edit the body of the Question, and I'll write up something about "rank revealing SVD".
– hardmath
Jul 30 at 16:13
add a comment |Â
1 Answer
1
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0
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In your case, $A = USV^T$, where
$$U = [u_1 u_2 u_3 u_4]$$
$$V = [v_1 v_2 v_3]$$
where $u_i$ is the $i^th$ column of $U$. (Same for $V$). Now, $S$ is a diagonal rectangular containing the Singular values of $A$, i.e.
$$S = beginbmatrix
sigma_1 & 0 & 0 \
0 & sigma_2 & 0 \
0 & 0 & sigma_3 \
0 & 0 & 0 \
endbmatrix$$
Notice that the number of non-zero singular values indicates the rank of $A$, hence if you have rank$A$ = 2, then (assuming the singular values are sorted $sigma_1 geq sigma_2 geq sigma_3$), we will have $sigma_3 = 0$
Hence
$$S = beginbmatrix
sigma_1 & 0 & 0 \
0 & sigma_2 & 0 \
0 & 0 & 0\
0 & 0 & 0 \
endbmatrix$$
Hence
$$A = USV^T = sigma_1 u_1v_1^T + sigma_2 u_2v_2^T = barUbarSbarV^T$$, where
$$barU = [u_1 u_2] in mathbbR^4times2$$
$$barV = [v_1 v_2 ] in mathbbR^3times2 $$
and
$$barS = beginbmatrix
sigma_1 & 0 \
0 & sigma_2 \
endbmatrix in mathbbR^2times2$$
answered Jul 30 at 14:41
Ahmad Bazzi
2,2551417
ahh i see, now i get it. thank youu
– bernard
Jul 31 at 7:45
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
In your case, $A = USV^T$, where
$$U = [u_1 u_2 u_3 u_4]$$
$$V = [v_1 v_2 v_3]$$
where $u_i$ is the $i^th$ column of $U$. (Same for $V$). Now, $S$ is a diagonal rectangular containing the Singular values of $A$, i.e.
$$S = beginbmatrix
sigma_1 & 0 & 0 \
0 & sigma_2 & 0 \
0 & 0 & sigma_3 \
0 & 0 & 0 \
endbmatrix$$
Notice that the number of non-zero singular values indicates the rank of $A$, hence if you have rank$A$ = 2, then (assuming the singular values are sorted $sigma_1 geq sigma_2 geq sigma_3$), we will have $sigma_3 = 0$
Hence
$$S = beginbmatrix
sigma_1 & 0 & 0 \
0 & sigma_2 & 0 \
0 & 0 & 0\
0 & 0 & 0 \
endbmatrix$$
Hence
$$A = USV^T = sigma_1 u_1v_1^T + sigma_2 u_2v_2^T = barUbarSbarV^T$$, where
$$barU = [u_1 u_2] in mathbbR^4times2$$
$$barV = [v_1 v_2 ] in mathbbR^3times2 $$
and
$$barS = beginbmatrix
sigma_1 & 0 \
0 & sigma_2 \
endbmatrix in mathbbR^2times2$$
answered Jul 30 at 14:41
Ahmad Bazzi
2,2551417
ahh i see, now i get it. thank youu
– bernard
Jul 31 at 7:45
add a comment |Â
up vote
0
down vote
accepted
In your case, $A = USV^T$, where
$$U = [u_1 u_2 u_3 u_4]$$
$$V = [v_1 v_2 v_3]$$
where $u_i$ is the $i^th$ column of $U$. (Same for $V$). Now, $S$ is a diagonal rectangular containing the Singular values of $A$, i.e.
$$S = beginbmatrix
sigma_1 & 0 & 0 \
0 & sigma_2 & 0 \
0 & 0 & sigma_3 \
0 & 0 & 0 \
endbmatrix$$
Notice that the number of non-zero singular values indicates the rank of $A$, hence if you have rank$A$ = 2, then (assuming the singular values are sorted $sigma_1 geq sigma_2 geq sigma_3$), we will have $sigma_3 = 0$
Hence
$$S = beginbmatrix
sigma_1 & 0 & 0 \
0 & sigma_2 & 0 \
0 & 0 & 0\
0 & 0 & 0 \
endbmatrix$$
Hence
$$A = USV^T = sigma_1 u_1v_1^T + sigma_2 u_2v_2^T = barUbarSbarV^T$$, where
$$barU = [u_1 u_2] in mathbbR^4times2$$
$$barV = [v_1 v_2 ] in mathbbR^3times2 $$
and
$$barS = beginbmatrix
sigma_1 & 0 \
0 & sigma_2 \
endbmatrix in mathbbR^2times2$$
answered Jul 30 at 14:41
Ahmad Bazzi
2,2551417
ahh i see, now i get it. thank youu
– bernard
Jul 31 at 7:45
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
In your case, $A = USV^T$, where
$$U = [u_1 u_2 u_3 u_4]$$
$$V = [v_1 v_2 v_3]$$
where $u_i$ is the $i^th$ column of $U$. (Same for $V$). Now, $S$ is a diagonal rectangular containing the Singular values of $A$, i.e.
$$S = beginbmatrix
sigma_1 & 0 & 0 \
0 & sigma_2 & 0 \
0 & 0 & sigma_3 \
0 & 0 & 0 \
endbmatrix$$
Notice that the number of non-zero singular values indicates the rank of $A$, hence if you have rank$A$ = 2, then (assuming the singular values are sorted $sigma_1 geq sigma_2 geq sigma_3$), we will have $sigma_3 = 0$
Hence
$$S = beginbmatrix
sigma_1 & 0 & 0 \
0 & sigma_2 & 0 \
0 & 0 & 0\
0 & 0 & 0 \
endbmatrix$$
Hence
$$A = USV^T = sigma_1 u_1v_1^T + sigma_2 u_2v_2^T = barUbarSbarV^T$$, where
$$barU = [u_1 u_2] in mathbbR^4times2$$
$$barV = [v_1 v_2 ] in mathbbR^3times2 $$
and
$$barS = beginbmatrix
sigma_1 & 0 \
0 & sigma_2 \
endbmatrix in mathbbR^2times2$$
answered Jul 30 at 14:41
Ahmad Bazzi
2,2551417
In your case, $A = USV^T$, where
$$U = [u_1 u_2 u_3 u_4]$$
$$V = [v_1 v_2 v_3]$$
where $u_i$ is the $i^th$ column of $U$. (Same for $V$). Now, $S$ is a diagonal rectangular containing the Singular values of $A$, i.e.
$$S = beginbmatrix
sigma_1 & 0 & 0 \
0 & sigma_2 & 0 \
0 & 0 & sigma_3 \
0 & 0 & 0 \
endbmatrix$$
Notice that the number of non-zero singular values indicates the rank of $A$, hence if you have rank$A$ = 2, then (assuming the singular values are sorted $sigma_1 geq sigma_2 geq sigma_3$), we will have $sigma_3 = 0$
Hence
$$S = beginbmatrix
sigma_1 & 0 & 0 \
0 & sigma_2 & 0 \
0 & 0 & 0\
0 & 0 & 0 \
endbmatrix$$
Hence
$$A = USV^T = sigma_1 u_1v_1^T + sigma_2 u_2v_2^T = barUbarSbarV^T$$, where
$$barU = [u_1 u_2] in mathbbR^4times2$$
$$barV = [v_1 v_2 ] in mathbbR^3times2 $$
and
$$barS = beginbmatrix
sigma_1 & 0 \
0 & sigma_2 \
endbmatrix in mathbbR^2times2$$
answered Jul 30 at 14:41
Ahmad Bazzi
2,2551417
answered Jul 30 at 14:41
Ahmad Bazzi
2,2551417
answered Jul 30 at 14:41
Ahmad Bazzi
2,2551417
answered Jul 30 at 14:41
Ahmad Bazzi
2,2551417
2,2551417
ahh i see, now i get it. thank youu
– bernard
Jul 31 at 7:45
add a comment |Â
ahh i see, now i get it. thank youu
– bernard
Jul 31 at 7:45
ahh i see, now i get it. thank youu
– bernard
Jul 31 at 7:45
ahh i see, now i get it. thank youu
– bernard
Jul 31 at 7:45
add a comment |Â
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It may be obvious, but in the presence of numerical rounding errors, one expects some spurious nonzero singular values to appear in the decomposition. I'm guessing that these are not of concern to you, but if they are, reply to this or edit the body of the Question, and I'll write up something about "rank revealing SVD".
– hardmath
Jul 30 at 16:13