Linear Discriminant Analysis: Meaning of Negative Eigenvalues?
$begingroup$
In Linear Discriminant Analysis (LDA) we compute two matrices from the data: the between scatter matrix $boldsymbol{S}_b$ and within scatter matrix $boldsymbol{S}_w$. A direction $boldsymbol{w}$ is then considered more discriminative if $(boldsymbol{w}^T boldsymbol{S}_b boldsymbol{w})/(boldsymbol{w}^T boldsymbol{S}_w boldsymbol{w})$ is larger. Therefore, the most $k$ discriminative directions correspond to the top $k$ eigenvectors of the matrix $boldsymbol{S}_b boldsymbol{S}_w^{-1}$.
While the two scatter matrices $boldsymbol{S}_b$ and $boldsymbol{S}_w$ are positive semi-definite, the matrix $boldsymbol{S}_b boldsymbol{S}_w^{-1}$ does not have to be. So I wonder what is the meaning/intuition of negative eigenvalues in this case?
This question arose when I wanted to choose the number $k$ based on what percentage of discriminative information is preserved by the top $k$ eigenvectors. For example, in case of PCA, I could say I want to pick the top $k$ eigenvectors that maintain $90%$ of the variance, i.e. $(sum_{j=1}^k lambda_j)/(sum_{j=1}^n lambda_j) approx 0.9$ ($n$ being the total number of eigenvalues). However, if negative eigenvalues can exist in LDA, how should I modify the PCA idea to achieve similar goal? Should I use $(sum_{j=1}^k lambda_j)/(sum_{j=1}^n lambda_j) approx 0.9$, or $(sum_{j=1}^k lambda_j)/(sum_{j=1}^{n^+} lambda_j) approx 0.9$ ($n^+$ meaning sum over nonnegative eigenvalues only), or $(sum_{j=1}^k lambda_j)/(sum_{j=1}^{n} |lambda_j|) approx 0.9$, or none of these?
Thanks
Golabi
statistics machine-learning
asked Dec 14 '18 at 21:51
GolabiGolabi
727
$endgroup$
add a comment |
$begingroup$
In Linear Discriminant Analysis (LDA) we compute two matrices from the data: the between scatter matrix $boldsymbol{S}_b$ and within scatter matrix $boldsymbol{S}_w$. A direction $boldsymbol{w}$ is then considered more discriminative if $(boldsymbol{w}^T boldsymbol{S}_b boldsymbol{w})/(boldsymbol{w}^T boldsymbol{S}_w boldsymbol{w})$ is larger. Therefore, the most $k$ discriminative directions correspond to the top $k$ eigenvectors of the matrix $boldsymbol{S}_b boldsymbol{S}_w^{-1}$.
While the two scatter matrices $boldsymbol{S}_b$ and $boldsymbol{S}_w$ are positive semi-definite, the matrix $boldsymbol{S}_b boldsymbol{S}_w^{-1}$ does not have to be. So I wonder what is the meaning/intuition of negative eigenvalues in this case?
This question arose when I wanted to choose the number $k$ based on what percentage of discriminative information is preserved by the top $k$ eigenvectors. For example, in case of PCA, I could say I want to pick the top $k$ eigenvectors that maintain $90%$ of the variance, i.e. $(sum_{j=1}^k lambda_j)/(sum_{j=1}^n lambda_j) approx 0.9$ ($n$ being the total number of eigenvalues). However, if negative eigenvalues can exist in LDA, how should I modify the PCA idea to achieve similar goal? Should I use $(sum_{j=1}^k lambda_j)/(sum_{j=1}^n lambda_j) approx 0.9$, or $(sum_{j=1}^k lambda_j)/(sum_{j=1}^{n^+} lambda_j) approx 0.9$ ($n^+$ meaning sum over nonnegative eigenvalues only), or $(sum_{j=1}^k lambda_j)/(sum_{j=1}^{n} |lambda_j|) approx 0.9$, or none of these?
Thanks
Golabi
statistics machine-learning
asked Dec 14 '18 at 21:51
GolabiGolabi
727
$endgroup$
add a comment |
$begingroup$
In Linear Discriminant Analysis (LDA) we compute two matrices from the data: the between scatter matrix $boldsymbol{S}_b$ and within scatter matrix $boldsymbol{S}_w$. A direction $boldsymbol{w}$ is then considered more discriminative if $(boldsymbol{w}^T boldsymbol{S}_b boldsymbol{w})/(boldsymbol{w}^T boldsymbol{S}_w boldsymbol{w})$ is larger. Therefore, the most $k$ discriminative directions correspond to the top $k$ eigenvectors of the matrix $boldsymbol{S}_b boldsymbol{S}_w^{-1}$.
While the two scatter matrices $boldsymbol{S}_b$ and $boldsymbol{S}_w$ are positive semi-definite, the matrix $boldsymbol{S}_b boldsymbol{S}_w^{-1}$ does not have to be. So I wonder what is the meaning/intuition of negative eigenvalues in this case?
This question arose when I wanted to choose the number $k$ based on what percentage of discriminative information is preserved by the top $k$ eigenvectors. For example, in case of PCA, I could say I want to pick the top $k$ eigenvectors that maintain $90%$ of the variance, i.e. $(sum_{j=1}^k lambda_j)/(sum_{j=1}^n lambda_j) approx 0.9$ ($n$ being the total number of eigenvalues). However, if negative eigenvalues can exist in LDA, how should I modify the PCA idea to achieve similar goal? Should I use $(sum_{j=1}^k lambda_j)/(sum_{j=1}^n lambda_j) approx 0.9$, or $(sum_{j=1}^k lambda_j)/(sum_{j=1}^{n^+} lambda_j) approx 0.9$ ($n^+$ meaning sum over nonnegative eigenvalues only), or $(sum_{j=1}^k lambda_j)/(sum_{j=1}^{n} |lambda_j|) approx 0.9$, or none of these?
Thanks
Golabi
statistics machine-learning
asked Dec 14 '18 at 21:51
GolabiGolabi
727
$endgroup$
In Linear Discriminant Analysis (LDA) we compute two matrices from the data: the between scatter matrix $boldsymbol{S}_b$ and within scatter matrix $boldsymbol{S}_w$. A direction $boldsymbol{w}$ is then considered more discriminative if $(boldsymbol{w}^T boldsymbol{S}_b boldsymbol{w})/(boldsymbol{w}^T boldsymbol{S}_w boldsymbol{w})$ is larger. Therefore, the most $k$ discriminative directions correspond to the top $k$ eigenvectors of the matrix $boldsymbol{S}_b boldsymbol{S}_w^{-1}$.
While the two scatter matrices $boldsymbol{S}_b$ and $boldsymbol{S}_w$ are positive semi-definite, the matrix $boldsymbol{S}_b boldsymbol{S}_w^{-1}$ does not have to be. So I wonder what is the meaning/intuition of negative eigenvalues in this case?
This question arose when I wanted to choose the number $k$ based on what percentage of discriminative information is preserved by the top $k$ eigenvectors. For example, in case of PCA, I could say I want to pick the top $k$ eigenvectors that maintain $90%$ of the variance, i.e. $(sum_{j=1}^k lambda_j)/(sum_{j=1}^n lambda_j) approx 0.9$ ($n$ being the total number of eigenvalues). However, if negative eigenvalues can exist in LDA, how should I modify the PCA idea to achieve similar goal? Should I use $(sum_{j=1}^k lambda_j)/(sum_{j=1}^n lambda_j) approx 0.9$, or $(sum_{j=1}^k lambda_j)/(sum_{j=1}^{n^+} lambda_j) approx 0.9$ ($n^+$ meaning sum over nonnegative eigenvalues only), or $(sum_{j=1}^k lambda_j)/(sum_{j=1}^{n} |lambda_j|) approx 0.9$, or none of these?
Thanks
Golabi
statistics machine-learning
statistics machine-learning
asked Dec 14 '18 at 21:51
GolabiGolabi
727
asked Dec 14 '18 at 21:51
GolabiGolabi
727
edited Dec 14 '18 at 22:33
Golabi
asked Dec 14 '18 at 21:51
GolabiGolabi
727
asked Dec 14 '18 at 21:51
GolabiGolabi
727
asked Dec 14 '18 at 21:51
GolabiGolabi
727
727
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