Finding rank/trace for projection involving pseudo-inverse?
$begingroup$
Define the projection matrix $H = X (X^T X)^- X^T$, where $(X^T X)^-$ is a pseudo/generalised inverse (ie. $A A^- A = A$).
How can I find the rank of $H$?
I know the following two facts:
$H$ is idempontent
Since $H$ is idempotent, $r(H) = tr(H)$
So the problem reduces to finding the trace of $H$. But here I am stuck. If $(X^T X)^-$ was a regular inverse, we could use cyclic communative property of trace to show that
$$tr(X(X^T X)^{-1} X^T) = tr((X^T X)^{-1} X^T X) = tr (I)$$
But since we have a pseudo-inverse, not a regular inverse, this approach doesn't work.
linear-algebra matrix-rank
asked Jan 15 at 1:49
XiaomiXiaomi
1,081115
$endgroup$
add a comment |
$begingroup$
Define the projection matrix $H = X (X^T X)^- X^T$, where $(X^T X)^-$ is a pseudo/generalised inverse (ie. $A A^- A = A$).
How can I find the rank of $H$?
I know the following two facts:
$H$ is idempontent
Since $H$ is idempotent, $r(H) = tr(H)$
So the problem reduces to finding the trace of $H$. But here I am stuck. If $(X^T X)^-$ was a regular inverse, we could use cyclic communative property of trace to show that
$$tr(X(X^T X)^{-1} X^T) = tr((X^T X)^{-1} X^T X) = tr (I)$$
But since we have a pseudo-inverse, not a regular inverse, this approach doesn't work.
linear-algebra matrix-rank
asked Jan 15 at 1:49
XiaomiXiaomi
1,081115
$endgroup$
2
$begingroup$
Note that $ker[(X^TX)^-] = ker(X^TX) = ker(X)$. This is enough to determine that $ker(H) = ker(X)$, so that $H$ and $X$ have the same rank.
$endgroup$
– Omnomnomnom
Jan 15 at 2:00
add a comment |
$begingroup$
Define the projection matrix $H = X (X^T X)^- X^T$, where $(X^T X)^-$ is a pseudo/generalised inverse (ie. $A A^- A = A$).
How can I find the rank of $H$?
I know the following two facts:
$H$ is idempontent
Since $H$ is idempotent, $r(H) = tr(H)$
So the problem reduces to finding the trace of $H$. But here I am stuck. If $(X^T X)^-$ was a regular inverse, we could use cyclic communative property of trace to show that
$$tr(X(X^T X)^{-1} X^T) = tr((X^T X)^{-1} X^T X) = tr (I)$$
But since we have a pseudo-inverse, not a regular inverse, this approach doesn't work.
linear-algebra matrix-rank
asked Jan 15 at 1:49
XiaomiXiaomi
1,081115
$endgroup$
Define the projection matrix $H = X (X^T X)^- X^T$, where $(X^T X)^-$ is a pseudo/generalised inverse (ie. $A A^- A = A$).
How can I find the rank of $H$?
I know the following two facts:
$H$ is idempontent
Since $H$ is idempotent, $r(H) = tr(H)$
So the problem reduces to finding the trace of $H$. But here I am stuck. If $(X^T X)^-$ was a regular inverse, we could use cyclic communative property of trace to show that
$$tr(X(X^T X)^{-1} X^T) = tr((X^T X)^{-1} X^T X) = tr (I)$$
But since we have a pseudo-inverse, not a regular inverse, this approach doesn't work.
linear-algebra matrix-rank
linear-algebra matrix-rank
asked Jan 15 at 1:49
XiaomiXiaomi
1,081115
asked Jan 15 at 1:49
XiaomiXiaomi
1,081115
asked Jan 15 at 1:49
XiaomiXiaomi
1,081115
asked Jan 15 at 1:49
XiaomiXiaomi
1,081115
asked Jan 15 at 1:49
XiaomiXiaomi
1,081115
1,081115
2
$begingroup$
Note that $ker[(X^TX)^-] = ker(X^TX) = ker(X)$. This is enough to determine that $ker(H) = ker(X)$, so that $H$ and $X$ have the same rank.
$endgroup$
– Omnomnomnom
Jan 15 at 2:00
add a comment |
2
$begingroup$
Note that $ker[(X^TX)^-] = ker(X^TX) = ker(X)$. This is enough to determine that $ker(H) = ker(X)$, so that $H$ and $X$ have the same rank.
$endgroup$
– Omnomnomnom
Jan 15 at 2:00
2
2
$begingroup$
Note that $ker[(X^TX)^-] = ker(X^TX) = ker(X)$. This is enough to determine that $ker(H) = ker(X)$, so that $H$ and $X$ have the same rank.
$endgroup$
– Omnomnomnom
Jan 15 at 2:00
$begingroup$
Note that $ker[(X^TX)^-] = ker(X^TX) = ker(X)$. This is enough to determine that $ker(H) = ker(X)$, so that $H$ and $X$ have the same rank.
$endgroup$
– Omnomnomnom
Jan 15 at 2:00
add a comment |
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$begingroup$
Note that $ker[(X^TX)^-] = ker(X^TX) = ker(X)$. This is enough to determine that $ker(H) = ker(X)$, so that $H$ and $X$ have the same rank.
$endgroup$
– Omnomnomnom
Jan 15 at 2:00