(PCA) Effect of holding the sum of squared coefficients to 1
$begingroup$
In principal-component analysis, the variables $x1, x2 ... xp$ are replaced by a more convenient set $u1, u2 ... up$ where the new variables are linear combinations of the old.
To hold the new vectors to the same scale as the original set, the sum of the squared coefficients is fixed to unity:
How exactly does fixing the squared coefficients to 1 maintain the same scale? And why the squared coefficients not the coefficients themselves?
linear-algebra
asked Dec 2 '18 at 16:17
hlineehlinee
755
$endgroup$
add a comment |
$begingroup$
In principal-component analysis, the variables $x1, x2 ... xp$ are replaced by a more convenient set $u1, u2 ... up$ where the new variables are linear combinations of the old.
To hold the new vectors to the same scale as the original set, the sum of the squared coefficients is fixed to unity:
How exactly does fixing the squared coefficients to 1 maintain the same scale? And why the squared coefficients not the coefficients themselves?
linear-algebra
asked Dec 2 '18 at 16:17
hlineehlinee
755
$endgroup$
add a comment |
$begingroup$
In principal-component analysis, the variables $x1, x2 ... xp$ are replaced by a more convenient set $u1, u2 ... up$ where the new variables are linear combinations of the old.
To hold the new vectors to the same scale as the original set, the sum of the squared coefficients is fixed to unity:
How exactly does fixing the squared coefficients to 1 maintain the same scale? And why the squared coefficients not the coefficients themselves?
linear-algebra
asked Dec 2 '18 at 16:17
hlineehlinee
755
$endgroup$
In principal-component analysis, the variables $x1, x2 ... xp$ are replaced by a more convenient set $u1, u2 ... up$ where the new variables are linear combinations of the old.
To hold the new vectors to the same scale as the original set, the sum of the squared coefficients is fixed to unity:
How exactly does fixing the squared coefficients to 1 maintain the same scale? And why the squared coefficients not the coefficients themselves?
linear-algebra
linear-algebra
asked Dec 2 '18 at 16:17
hlineehlinee
755
asked Dec 2 '18 at 16:17
hlineehlinee
755
asked Dec 2 '18 at 16:17
hlineehlinee
755
asked Dec 2 '18 at 16:17
hlineehlinee
755
asked Dec 2 '18 at 16:17
hlineehlinee
755
755
add a comment |
add a comment |
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