Mixing
Covariance matrix of data projected onto eigenvectors is diagonal.
$begingroup$
I am reading about PCA and found an exercise that says
Show that when a $N$-dim set of data points $X$ is projected onto the eigenvectors $V = [e_1 e_2...e_n]$ of its covariance matrix $C=XX^T$, the covariance matrix of the projected data $C_p = YY^T$ is diagonal and hence that, in the space of the eigenvector decomposition, the distribution of X is uncorrelated.
What I have so far is
$$Y = V^TX$$
Therefore
$$C_p = YY^T = V^TX(V^TX)^T = V^TXX^TV$$
but there I got stuck. Any advise on how to proceed, moreover, what does "The covariance matrix of the projected data is diagonal" mean?
linear-algebra
asked Jan 18 '14 at 10:18
BRabbit27BRabbit27
352926
$endgroup$
add a comment |
$begingroup$
I am reading about PCA and found an exercise that says
Show that when a $N$-dim set of data points $X$ is projected onto the eigenvectors $V = [e_1 e_2...e_n]$ of its covariance matrix $C=XX^T$, the covariance matrix of the projected data $C_p = YY^T$ is diagonal and hence that, in the space of the eigenvector decomposition, the distribution of X is uncorrelated.
What I have so far is
$$Y = V^TX$$
Therefore
$$C_p = YY^T = V^TX(V^TX)^T = V^TXX^TV$$
but there I got stuck. Any advise on how to proceed, moreover, what does "The covariance matrix of the projected data is diagonal" mean?
linear-algebra
asked Jan 18 '14 at 10:18
BRabbit27BRabbit27
352926
$endgroup$
add a comment |
$begingroup$
I am reading about PCA and found an exercise that says
Show that when a $N$-dim set of data points $X$ is projected onto the eigenvectors $V = [e_1 e_2...e_n]$ of its covariance matrix $C=XX^T$, the covariance matrix of the projected data $C_p = YY^T$ is diagonal and hence that, in the space of the eigenvector decomposition, the distribution of X is uncorrelated.
What I have so far is
$$Y = V^TX$$
Therefore
$$C_p = YY^T = V^TX(V^TX)^T = V^TXX^TV$$
but there I got stuck. Any advise on how to proceed, moreover, what does "The covariance matrix of the projected data is diagonal" mean?
linear-algebra
asked Jan 18 '14 at 10:18
BRabbit27BRabbit27
352926
$endgroup$
I am reading about PCA and found an exercise that says
Show that when a $N$-dim set of data points $X$ is projected onto the eigenvectors $V = [e_1 e_2...e_n]$ of its covariance matrix $C=XX^T$, the covariance matrix of the projected data $C_p = YY^T$ is diagonal and hence that, in the space of the eigenvector decomposition, the distribution of X is uncorrelated.
What I have so far is
$$Y = V^TX$$
Therefore
$$C_p = YY^T = V^TX(V^TX)^T = V^TXX^TV$$
but there I got stuck. Any advise on how to proceed, moreover, what does "The covariance matrix of the projected data is diagonal" mean?
linear-algebra
linear-algebra
asked Jan 18 '14 at 10:18
BRabbit27BRabbit27
352926
asked Jan 18 '14 at 10:18
BRabbit27BRabbit27
352926
asked Jan 18 '14 at 10:18
BRabbit27BRabbit27
352926
asked Jan 18 '14 at 10:18
BRabbit27BRabbit27
352926
asked Jan 18 '14 at 10:18
BRabbit27BRabbit27
352926
352926
add a comment |
add a comment |
1 Answer
1
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$begingroup$
A diagonal matrix is one with zero everywhere and the diagonal entries can be zero or non zero.
Assuming the eigenvectors are normalized in magnitude $|e_i|=1$
$$
begin{align}
because text{ } & V = [e_1 e_2 dots e_n] text{ and } Ce_i =lambda_ie_i \\
therefore text{ } &CV = V
begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\\
because text{ } & C_p = V^T (XX^T) V = V^TCV\\
therefore text{ } & C_p=V^TVbegin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\
because text{ } & V^TV = mathbf{I}\
&text{ as eigenvecotrs of a symmetric matrix are orthogonal}\\
therefore text{ } &C_p= begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\
&text{A diagonal matrix with variance in each eigenvector direction}
end{align}
$$
answered Sep 28 '15 at 2:41
Karim TarabishyKarim Tarabishy
1206
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A diagonal matrix is one with zero everywhere and the diagonal entries can be zero or non zero.
Assuming the eigenvectors are normalized in magnitude $|e_i|=1$
$$
begin{align}
because text{ } & V = [e_1 e_2 dots e_n] text{ and } Ce_i =lambda_ie_i \\
therefore text{ } &CV = V
begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\\
because text{ } & C_p = V^T (XX^T) V = V^TCV\\
therefore text{ } & C_p=V^TVbegin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\
because text{ } & V^TV = mathbf{I}\
&text{ as eigenvecotrs of a symmetric matrix are orthogonal}\\
therefore text{ } &C_p= begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\
&text{A diagonal matrix with variance in each eigenvector direction}
end{align}
$$
answered Sep 28 '15 at 2:41
Karim TarabishyKarim Tarabishy
1206
$endgroup$
add a comment |
$begingroup$
A diagonal matrix is one with zero everywhere and the diagonal entries can be zero or non zero.
Assuming the eigenvectors are normalized in magnitude $|e_i|=1$
$$
begin{align}
because text{ } & V = [e_1 e_2 dots e_n] text{ and } Ce_i =lambda_ie_i \\
therefore text{ } &CV = V
begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\\
because text{ } & C_p = V^T (XX^T) V = V^TCV\\
therefore text{ } & C_p=V^TVbegin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\
because text{ } & V^TV = mathbf{I}\
&text{ as eigenvecotrs of a symmetric matrix are orthogonal}\\
therefore text{ } &C_p= begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\
&text{A diagonal matrix with variance in each eigenvector direction}
end{align}
$$
answered Sep 28 '15 at 2:41
Karim TarabishyKarim Tarabishy
1206
$endgroup$
add a comment |
$begingroup$
A diagonal matrix is one with zero everywhere and the diagonal entries can be zero or non zero.
Assuming the eigenvectors are normalized in magnitude $|e_i|=1$
$$
begin{align}
because text{ } & V = [e_1 e_2 dots e_n] text{ and } Ce_i =lambda_ie_i \\
therefore text{ } &CV = V
begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\\
because text{ } & C_p = V^T (XX^T) V = V^TCV\\
therefore text{ } & C_p=V^TVbegin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\
because text{ } & V^TV = mathbf{I}\
&text{ as eigenvecotrs of a symmetric matrix are orthogonal}\\
therefore text{ } &C_p= begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\
&text{A diagonal matrix with variance in each eigenvector direction}
end{align}
$$
answered Sep 28 '15 at 2:41
Karim TarabishyKarim Tarabishy
1206
$endgroup$
A diagonal matrix is one with zero everywhere and the diagonal entries can be zero or non zero.
Assuming the eigenvectors are normalized in magnitude $|e_i|=1$
$$
begin{align}
because text{ } & V = [e_1 e_2 dots e_n] text{ and } Ce_i =lambda_ie_i \\
therefore text{ } &CV = V
begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\\
because text{ } & C_p = V^T (XX^T) V = V^TCV\\
therefore text{ } & C_p=V^TVbegin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\
because text{ } & V^TV = mathbf{I}\
&text{ as eigenvecotrs of a symmetric matrix are orthogonal}\\
therefore text{ } &C_p= begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_n
end{bmatrix}\
&text{A diagonal matrix with variance in each eigenvector direction}
end{align}
$$
answered Sep 28 '15 at 2:41
Karim TarabishyKarim Tarabishy
1206
answered Sep 28 '15 at 2:41
Karim TarabishyKarim Tarabishy
1206
answered Sep 28 '15 at 2:41
Karim TarabishyKarim Tarabishy
1206
answered Sep 28 '15 at 2:41
Karim TarabishyKarim Tarabishy
1206
1206
add a comment |
add a comment |
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