Questions on PCA The Next CEO of Stack OverflowThoroughly understand the concepts and formulas in Linear AlgebraUnderstanding singular value decompositionSolving differential equations in linear algebraChecking connectivity of adjacency matrixDetermine matrix of linear transformation of square into parrallogramPrincipal component analysis (PCA) results in sinusoids, what is the underlying cause?Decide if each is a basis for $P_2$. (a) $(x^2 + x - 1, 2x + 1, 2x - 1)$If $lambda = i$ is an eigenvalue of $A in mathbb R^n times n$, explain why the rank of $A^3 + A$ is less than $n$What does it mean when the rank (number of non-zero rows) of a reduced matrix is MORE than the number of variables?One variable equal to 0 in linear equation system.
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Questions on PCA
The Next CEO of Stack OverflowThoroughly understand the concepts and formulas in Linear AlgebraUnderstanding singular value decompositionSolving differential equations in linear algebraChecking connectivity of adjacency matrixDetermine matrix of linear transformation of square into parrallogramPrincipal component analysis (PCA) results in sinusoids, what is the underlying cause?Decide if each is a basis for $P_2$. (a) $(x^2 + x - 1, 2x + 1, 2x - 1)$If $lambda = i$ is an eigenvalue of $A in mathbb R^n times n$, explain why the rank of $A^3 + A$ is less than $n$What does it mean when the rank (number of non-zero rows) of a reduced matrix is MORE than the number of variables?One variable equal to 0 in linear equation system.
$begingroup$
I have a hard time understand the following statements below about PCA (normed or not normed).
a) the matrix to diagonalize is the matrix of linear correlations of original variables.
b) An illustrative variable is well represented on a factorial axis if its contribution is high on this axis.
It would be really helpful if someone could explain.
linear-algebra matrix-decomposition
asked Jan 4 '16 at 15:29
Saul GarciaSaul Garcia
1052
$endgroup$
add a comment |
$begingroup$
I have a hard time understand the following statements below about PCA (normed or not normed).
a) the matrix to diagonalize is the matrix of linear correlations of original variables.
b) An illustrative variable is well represented on a factorial axis if its contribution is high on this axis.
It would be really helpful if someone could explain.
linear-algebra matrix-decomposition
asked Jan 4 '16 at 15:29
Saul GarciaSaul Garcia
1052
$endgroup$
add a comment |
$begingroup$
I have a hard time understand the following statements below about PCA (normed or not normed).
a) the matrix to diagonalize is the matrix of linear correlations of original variables.
b) An illustrative variable is well represented on a factorial axis if its contribution is high on this axis.
It would be really helpful if someone could explain.
linear-algebra matrix-decomposition
asked Jan 4 '16 at 15:29
Saul GarciaSaul Garcia
1052
$endgroup$
I have a hard time understand the following statements below about PCA (normed or not normed).
a) the matrix to diagonalize is the matrix of linear correlations of original variables.
b) An illustrative variable is well represented on a factorial axis if its contribution is high on this axis.
It would be really helpful if someone could explain.
linear-algebra matrix-decomposition
linear-algebra matrix-decomposition
asked Jan 4 '16 at 15:29
Saul GarciaSaul Garcia
1052
asked Jan 4 '16 at 15:29
Saul GarciaSaul Garcia
1052
asked Jan 4 '16 at 15:29
Saul GarciaSaul Garcia
1052
asked Jan 4 '16 at 15:29
Saul GarciaSaul Garcia
1052
asked Jan 4 '16 at 15:29
Saul GarciaSaul Garcia
1052
1052
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I suggest you read the chapter for the Singular Value Decomposition (SVD) in the book 'Linear Algebra' of Gilbert Strang. There's no other book that explains it more intuitively and clearly. The SVD and PCA have the same purpose in practical applications: Identifying the directions in which the data varies the most. In the end one tries to describe (approximate) the data with fewer dimensions by leaving out the dimensions that don't vary a lot.
I can't explain the whole SVD in detail. But here's the big picture: We'd like to decompose $A$ into $A=USigma V^T$, where $U$ and $V$ have orthogonal columns and $Sigma$ is just a diagonal matrix.
We know that $A^top A$ is symmetric and thus diagonalizable. Since
$$
A^top A = (USigma V^top)^top USigma V^top
=VSigma U^top USigma V^top = VSigma^2 V^T,
$$
by diagonalising $A^top A$ and orthonormalising the eigenvectors of the decomposition we can get $V$ and $Sigma$ from this process. The eigenvalues of $Sigma$ are usually sorted in decreasing order. We then just use $A=USigma V^top Longleftrightarrow AV=USigma$ in order to determine $U$.
See also:
https://en.wikipedia.org/wiki/Singular_value_decomposition (Has some good pictures)
https://en.wikipedia.org/wiki/Principal_component_analysis
answered Jan 4 '16 at 15:30
ndrizzandrizza
493312
$endgroup$
add a comment |
$begingroup$
PCA is kind of order reduction. You simply categorize the data to fewer dimensions in a way you can visualized such data. Imagine that you have list of 1000 kinds of win to pick one or two among such a list regards/ based on 5 or 6 or 7.... features.
PCA analysis tool is a vital tool in such a case. PCA allows you to have a clear look on wins based on 2D dimensions (2 factors / 2 PCA's). Such factors/ PCAs take in accounts all the other features(factors) with various
To get more of the physical meaning of PCA, I highly recommend t check the following youtube:
https://www.youtube.com/watch?v=g-Hb26agBFg
Reagrds to the statements you mentioned above:
a) the matrix to diagonalize is the matrix of linear correlations of original variables.
[U,D]=eigs(A'*A) A'*A is the covariance matrix which is more or less (correlation).
b) An illustrative variable is well represented on a factorial axis if its contribution is high on this axis.
This means that high illustrative variable has high PCA.
Best regards
answered Mar 28 at 1:37
Mohamed AbugammarMohamed Abugammar
9
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I suggest you read the chapter for the Singular Value Decomposition (SVD) in the book 'Linear Algebra' of Gilbert Strang. There's no other book that explains it more intuitively and clearly. The SVD and PCA have the same purpose in practical applications: Identifying the directions in which the data varies the most. In the end one tries to describe (approximate) the data with fewer dimensions by leaving out the dimensions that don't vary a lot.
I can't explain the whole SVD in detail. But here's the big picture: We'd like to decompose $A$ into $A=USigma V^T$, where $U$ and $V$ have orthogonal columns and $Sigma$ is just a diagonal matrix.
We know that $A^top A$ is symmetric and thus diagonalizable. Since
$$
A^top A = (USigma V^top)^top USigma V^top
=VSigma U^top USigma V^top = VSigma^2 V^T,
$$
by diagonalising $A^top A$ and orthonormalising the eigenvectors of the decomposition we can get $V$ and $Sigma$ from this process. The eigenvalues of $Sigma$ are usually sorted in decreasing order. We then just use $A=USigma V^top Longleftrightarrow AV=USigma$ in order to determine $U$.
See also:
https://en.wikipedia.org/wiki/Singular_value_decomposition (Has some good pictures)
https://en.wikipedia.org/wiki/Principal_component_analysis
answered Jan 4 '16 at 15:30
ndrizzandrizza
493312
$endgroup$
add a comment |
$begingroup$
I suggest you read the chapter for the Singular Value Decomposition (SVD) in the book 'Linear Algebra' of Gilbert Strang. There's no other book that explains it more intuitively and clearly. The SVD and PCA have the same purpose in practical applications: Identifying the directions in which the data varies the most. In the end one tries to describe (approximate) the data with fewer dimensions by leaving out the dimensions that don't vary a lot.
I can't explain the whole SVD in detail. But here's the big picture: We'd like to decompose $A$ into $A=USigma V^T$, where $U$ and $V$ have orthogonal columns and $Sigma$ is just a diagonal matrix.
We know that $A^top A$ is symmetric and thus diagonalizable. Since
$$
A^top A = (USigma V^top)^top USigma V^top
=VSigma U^top USigma V^top = VSigma^2 V^T,
$$
by diagonalising $A^top A$ and orthonormalising the eigenvectors of the decomposition we can get $V$ and $Sigma$ from this process. The eigenvalues of $Sigma$ are usually sorted in decreasing order. We then just use $A=USigma V^top Longleftrightarrow AV=USigma$ in order to determine $U$.
See also:
https://en.wikipedia.org/wiki/Singular_value_decomposition (Has some good pictures)
https://en.wikipedia.org/wiki/Principal_component_analysis
answered Jan 4 '16 at 15:30
ndrizzandrizza
493312
$endgroup$
add a comment |
$begingroup$
I suggest you read the chapter for the Singular Value Decomposition (SVD) in the book 'Linear Algebra' of Gilbert Strang. There's no other book that explains it more intuitively and clearly. The SVD and PCA have the same purpose in practical applications: Identifying the directions in which the data varies the most. In the end one tries to describe (approximate) the data with fewer dimensions by leaving out the dimensions that don't vary a lot.
I can't explain the whole SVD in detail. But here's the big picture: We'd like to decompose $A$ into $A=USigma V^T$, where $U$ and $V$ have orthogonal columns and $Sigma$ is just a diagonal matrix.
We know that $A^top A$ is symmetric and thus diagonalizable. Since
$$
A^top A = (USigma V^top)^top USigma V^top
=VSigma U^top USigma V^top = VSigma^2 V^T,
$$
by diagonalising $A^top A$ and orthonormalising the eigenvectors of the decomposition we can get $V$ and $Sigma$ from this process. The eigenvalues of $Sigma$ are usually sorted in decreasing order. We then just use $A=USigma V^top Longleftrightarrow AV=USigma$ in order to determine $U$.
See also:
https://en.wikipedia.org/wiki/Singular_value_decomposition (Has some good pictures)
https://en.wikipedia.org/wiki/Principal_component_analysis
answered Jan 4 '16 at 15:30
ndrizzandrizza
493312
$endgroup$
I suggest you read the chapter for the Singular Value Decomposition (SVD) in the book 'Linear Algebra' of Gilbert Strang. There's no other book that explains it more intuitively and clearly. The SVD and PCA have the same purpose in practical applications: Identifying the directions in which the data varies the most. In the end one tries to describe (approximate) the data with fewer dimensions by leaving out the dimensions that don't vary a lot.
I can't explain the whole SVD in detail. But here's the big picture: We'd like to decompose $A$ into $A=USigma V^T$, where $U$ and $V$ have orthogonal columns and $Sigma$ is just a diagonal matrix.
We know that $A^top A$ is symmetric and thus diagonalizable. Since
$$
A^top A = (USigma V^top)^top USigma V^top
=VSigma U^top USigma V^top = VSigma^2 V^T,
$$
by diagonalising $A^top A$ and orthonormalising the eigenvectors of the decomposition we can get $V$ and $Sigma$ from this process. The eigenvalues of $Sigma$ are usually sorted in decreasing order. We then just use $A=USigma V^top Longleftrightarrow AV=USigma$ in order to determine $U$.
See also:
https://en.wikipedia.org/wiki/Singular_value_decomposition (Has some good pictures)
https://en.wikipedia.org/wiki/Principal_component_analysis
answered Jan 4 '16 at 15:30
ndrizzandrizza
493312
edited Jan 4 '16 at 15:45
answered Jan 4 '16 at 15:30
ndrizzandrizza
493312
answered Jan 4 '16 at 15:30
ndrizzandrizza
493312
answered Jan 4 '16 at 15:30
ndrizzandrizza
493312
493312
add a comment |
add a comment |
$begingroup$
PCA is kind of order reduction. You simply categorize the data to fewer dimensions in a way you can visualized such data. Imagine that you have list of 1000 kinds of win to pick one or two among such a list regards/ based on 5 or 6 or 7.... features.
PCA analysis tool is a vital tool in such a case. PCA allows you to have a clear look on wins based on 2D dimensions (2 factors / 2 PCA's). Such factors/ PCAs take in accounts all the other features(factors) with various
To get more of the physical meaning of PCA, I highly recommend t check the following youtube:
https://www.youtube.com/watch?v=g-Hb26agBFg
Reagrds to the statements you mentioned above:
a) the matrix to diagonalize is the matrix of linear correlations of original variables.
[U,D]=eigs(A'*A) A'*A is the covariance matrix which is more or less (correlation).
b) An illustrative variable is well represented on a factorial axis if its contribution is high on this axis.
This means that high illustrative variable has high PCA.
Best regards
answered Mar 28 at 1:37
Mohamed AbugammarMohamed Abugammar
9
$endgroup$
add a comment |
$begingroup$
PCA is kind of order reduction. You simply categorize the data to fewer dimensions in a way you can visualized such data. Imagine that you have list of 1000 kinds of win to pick one or two among such a list regards/ based on 5 or 6 or 7.... features.
PCA analysis tool is a vital tool in such a case. PCA allows you to have a clear look on wins based on 2D dimensions (2 factors / 2 PCA's). Such factors/ PCAs take in accounts all the other features(factors) with various
To get more of the physical meaning of PCA, I highly recommend t check the following youtube:
https://www.youtube.com/watch?v=g-Hb26agBFg
Reagrds to the statements you mentioned above:
a) the matrix to diagonalize is the matrix of linear correlations of original variables.
[U,D]=eigs(A'*A) A'*A is the covariance matrix which is more or less (correlation).
b) An illustrative variable is well represented on a factorial axis if its contribution is high on this axis.
This means that high illustrative variable has high PCA.
Best regards
answered Mar 28 at 1:37
Mohamed AbugammarMohamed Abugammar
9
$endgroup$
add a comment |
$begingroup$
PCA is kind of order reduction. You simply categorize the data to fewer dimensions in a way you can visualized such data. Imagine that you have list of 1000 kinds of win to pick one or two among such a list regards/ based on 5 or 6 or 7.... features.
PCA analysis tool is a vital tool in such a case. PCA allows you to have a clear look on wins based on 2D dimensions (2 factors / 2 PCA's). Such factors/ PCAs take in accounts all the other features(factors) with various
To get more of the physical meaning of PCA, I highly recommend t check the following youtube:
https://www.youtube.com/watch?v=g-Hb26agBFg
Reagrds to the statements you mentioned above:
a) the matrix to diagonalize is the matrix of linear correlations of original variables.
[U,D]=eigs(A'*A) A'*A is the covariance matrix which is more or less (correlation).
b) An illustrative variable is well represented on a factorial axis if its contribution is high on this axis.
This means that high illustrative variable has high PCA.
Best regards
answered Mar 28 at 1:37
Mohamed AbugammarMohamed Abugammar
9
$endgroup$
PCA is kind of order reduction. You simply categorize the data to fewer dimensions in a way you can visualized such data. Imagine that you have list of 1000 kinds of win to pick one or two among such a list regards/ based on 5 or 6 or 7.... features.
PCA analysis tool is a vital tool in such a case. PCA allows you to have a clear look on wins based on 2D dimensions (2 factors / 2 PCA's). Such factors/ PCAs take in accounts all the other features(factors) with various
To get more of the physical meaning of PCA, I highly recommend t check the following youtube:
https://www.youtube.com/watch?v=g-Hb26agBFg
Reagrds to the statements you mentioned above:
a) the matrix to diagonalize is the matrix of linear correlations of original variables.
[U,D]=eigs(A'*A) A'*A is the covariance matrix which is more or less (correlation).
b) An illustrative variable is well represented on a factorial axis if its contribution is high on this axis.
This means that high illustrative variable has high PCA.
Best regards
answered Mar 28 at 1:37
Mohamed AbugammarMohamed Abugammar
9
answered Mar 28 at 1:37
Mohamed AbugammarMohamed Abugammar
9
answered Mar 28 at 1:37
Mohamed AbugammarMohamed Abugammar
9
answered Mar 28 at 1:37
Mohamed AbugammarMohamed Abugammar
9
9
add a comment |
add a comment |
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