Why in spectral clustering number of eigen vectors is same as number of clusters that we want?
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In spectral clustering we take eigenvector corresponding to K smallest eigenvalues. Then we do K means clustering on these eigenvector to get final clusters. What will happen if we take different number of eigenvectors than number of clusters we want ?
clustering
asked 2 days ago
sneh guptasneh gupta
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In spectral clustering we take eigenvector corresponding to K smallest eigenvalues. Then we do K means clustering on these eigenvector to get final clusters. What will happen if we take different number of eigenvectors than number of clusters we want ?
clustering
asked 2 days ago
sneh guptasneh gupta
1
New contributor
sneh gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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In spectral clustering we take eigenvector corresponding to K smallest eigenvalues. Then we do K means clustering on these eigenvector to get final clusters. What will happen if we take different number of eigenvectors than number of clusters we want ?
clustering
asked 2 days ago
sneh guptasneh gupta
1
New contributor
sneh gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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In spectral clustering we take eigenvector corresponding to K smallest eigenvalues. Then we do K means clustering on these eigenvector to get final clusters. What will happen if we take different number of eigenvectors than number of clusters we want ?
clustering
clustering
asked 2 days ago
sneh guptasneh gupta
1
New contributor
sneh gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
sneh guptasneh gupta
1
New contributor
sneh gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
sneh guptasneh gupta
1
New contributor
sneh gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
sneh guptasneh gupta
1
asked 2 days ago
sneh guptasneh gupta
1
1
New contributor
sneh gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
sneh gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Theoretically, log2(k) components could be enough.
But usually clusters are not that well balanced, that you could get all 2^l combinations of eigenvectors stable, usually one will mask the other.
If you go back to the theory of spectral clustering, you'd have one eigenvector for each connected component. Now the clusters are unfortunately connected, so we don't completely get this ideal situation, but you'd expect one eigenvector for each "almost component", too.
answered 2 days ago
Anony-MousseAnony-Mousse
4,856624
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Can you please share the related resource or give intuition behind sqrt(k) components?
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– sneh gupta
yesterday
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Sorry. Log2(k) of course. Enough to separate k components.
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– Anony-Mousse
yesterday
add a comment |
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1 Answer
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$begingroup$
Theoretically, log2(k) components could be enough.
But usually clusters are not that well balanced, that you could get all 2^l combinations of eigenvectors stable, usually one will mask the other.
If you go back to the theory of spectral clustering, you'd have one eigenvector for each connected component. Now the clusters are unfortunately connected, so we don't completely get this ideal situation, but you'd expect one eigenvector for each "almost component", too.
answered 2 days ago
Anony-MousseAnony-Mousse
4,856624
$endgroup$
$begingroup$
Can you please share the related resource or give intuition behind sqrt(k) components?
$endgroup$
– sneh gupta
yesterday
$begingroup$
Sorry. Log2(k) of course. Enough to separate k components.
$endgroup$
– Anony-Mousse
yesterday
add a comment |
$begingroup$
Theoretically, log2(k) components could be enough.
But usually clusters are not that well balanced, that you could get all 2^l combinations of eigenvectors stable, usually one will mask the other.
If you go back to the theory of spectral clustering, you'd have one eigenvector for each connected component. Now the clusters are unfortunately connected, so we don't completely get this ideal situation, but you'd expect one eigenvector for each "almost component", too.
answered 2 days ago
Anony-MousseAnony-Mousse
4,856624
$endgroup$
$begingroup$
Can you please share the related resource or give intuition behind sqrt(k) components?
$endgroup$
– sneh gupta
yesterday
$begingroup$
Sorry. Log2(k) of course. Enough to separate k components.
$endgroup$
– Anony-Mousse
yesterday
add a comment |
$begingroup$
Theoretically, log2(k) components could be enough.
But usually clusters are not that well balanced, that you could get all 2^l combinations of eigenvectors stable, usually one will mask the other.
If you go back to the theory of spectral clustering, you'd have one eigenvector for each connected component. Now the clusters are unfortunately connected, so we don't completely get this ideal situation, but you'd expect one eigenvector for each "almost component", too.
answered 2 days ago
Anony-MousseAnony-Mousse
4,856624
$endgroup$
Theoretically, log2(k) components could be enough.
But usually clusters are not that well balanced, that you could get all 2^l combinations of eigenvectors stable, usually one will mask the other.
If you go back to the theory of spectral clustering, you'd have one eigenvector for each connected component. Now the clusters are unfortunately connected, so we don't completely get this ideal situation, but you'd expect one eigenvector for each "almost component", too.
answered 2 days ago
Anony-MousseAnony-Mousse
4,856624
edited 23 hours ago
answered 2 days ago
Anony-MousseAnony-Mousse
4,856624
answered 2 days ago
Anony-MousseAnony-Mousse
4,856624
answered 2 days ago
Anony-MousseAnony-Mousse
4,856624
4,856624
$begingroup$
Can you please share the related resource or give intuition behind sqrt(k) components?
$endgroup$
– sneh gupta
yesterday
$begingroup$
Sorry. Log2(k) of course. Enough to separate k components.
$endgroup$
– Anony-Mousse
yesterday
add a comment |
$begingroup$
Can you please share the related resource or give intuition behind sqrt(k) components?
$endgroup$
– sneh gupta
yesterday
$begingroup$
Sorry. Log2(k) of course. Enough to separate k components.
$endgroup$
– Anony-Mousse
yesterday
$begingroup$
Can you please share the related resource or give intuition behind sqrt(k) components?
$endgroup$
– sneh gupta
yesterday
$begingroup$
Can you please share the related resource or give intuition behind sqrt(k) components?
$endgroup$
– sneh gupta
yesterday
$begingroup$
Sorry. Log2(k) of course. Enough to separate k components.
$endgroup$
– Anony-Mousse
yesterday
$begingroup$
Sorry. Log2(k) of course. Enough to separate k components.
$endgroup$
– Anony-Mousse
yesterday
add a comment |
sneh gupta is a new contributor. Be nice, and check out our Code of Conduct.
sneh gupta is a new contributor. Be nice, and check out our Code of Conduct.
sneh gupta is a new contributor. Be nice, and check out our Code of Conduct.
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