Mixing
eigenvalues of fitted value?
Clash Royale CLAN TAG#URR8PPP
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Let a $mtimes n$ full-column matrix be $A$. Also, denote $mtimes m$ diagonal matrix as $D$.
Define $P(A)D=A(A'A)^-1A'D$ as the orthogonal projection of $D$ onto the column space of $A$, or "fitted value" in regression term. Note that $(A'A)$ is full rank, so $P(A)$ always exists.
I'm wondering how the eigenvalues of $P(A)D$ are determined. Can we use the fact that $D$ is already a diagonal matrix?
linear-algebra matrices eigenvalues-eigenvectors regression projection-matrices
asked Aug 2 at 2:33
user568810
566
add a comment |Â
up vote
1
down vote
favorite
Let a $mtimes n$ full-column matrix be $A$. Also, denote $mtimes m$ diagonal matrix as $D$.
Define $P(A)D=A(A'A)^-1A'D$ as the orthogonal projection of $D$ onto the column space of $A$, or "fitted value" in regression term. Note that $(A'A)$ is full rank, so $P(A)$ always exists.
I'm wondering how the eigenvalues of $P(A)D$ are determined. Can we use the fact that $D$ is already a diagonal matrix?
linear-algebra matrices eigenvalues-eigenvectors regression projection-matrices
asked Aug 2 at 2:33
user568810
566
@philip Can you explain me how my problem is related to the proof about diagonalization?
– user568810
Aug 2 at 2:43
1
What you're doing is looking at square matrices $P$ and $D$, idempotent and diagonal respectively, and asking if there's a shortcut to compute the eigenvalues of $PD$. I'm pretty sure the answer is, no there is not.
– Theo Bendit
Aug 2 at 2:43
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let a $mtimes n$ full-column matrix be $A$. Also, denote $mtimes m$ diagonal matrix as $D$.
Define $P(A)D=A(A'A)^-1A'D$ as the orthogonal projection of $D$ onto the column space of $A$, or "fitted value" in regression term. Note that $(A'A)$ is full rank, so $P(A)$ always exists.
I'm wondering how the eigenvalues of $P(A)D$ are determined. Can we use the fact that $D$ is already a diagonal matrix?
linear-algebra matrices eigenvalues-eigenvectors regression projection-matrices
asked Aug 2 at 2:33
user568810
566
Let a $mtimes n$ full-column matrix be $A$. Also, denote $mtimes m$ diagonal matrix as $D$.
Define $P(A)D=A(A'A)^-1A'D$ as the orthogonal projection of $D$ onto the column space of $A$, or "fitted value" in regression term. Note that $(A'A)$ is full rank, so $P(A)$ always exists.
I'm wondering how the eigenvalues of $P(A)D$ are determined. Can we use the fact that $D$ is already a diagonal matrix?
linear-algebra matrices eigenvalues-eigenvectors regression projection-matrices
asked Aug 2 at 2:33
user568810
566
edited Aug 3 at 3:50
asked Aug 2 at 2:33
user568810
566
asked Aug 2 at 2:33
user568810
566
asked Aug 2 at 2:33
user568810
566
566
@philip Can you explain me how my problem is related to the proof about diagonalization?
– user568810
Aug 2 at 2:43
1
What you're doing is looking at square matrices $P$ and $D$, idempotent and diagonal respectively, and asking if there's a shortcut to compute the eigenvalues of $PD$. I'm pretty sure the answer is, no there is not.
– Theo Bendit
Aug 2 at 2:43
add a comment |Â
@philip Can you explain me how my problem is related to the proof about diagonalization?
– user568810
Aug 2 at 2:43
1
What you're doing is looking at square matrices $P$ and $D$, idempotent and diagonal respectively, and asking if there's a shortcut to compute the eigenvalues of $PD$. I'm pretty sure the answer is, no there is not.
– Theo Bendit
Aug 2 at 2:43
@philip Can you explain me how my problem is related to the proof about diagonalization?
– user568810
Aug 2 at 2:43
@philip Can you explain me how my problem is related to the proof about diagonalization?
– user568810
Aug 2 at 2:43
1
1
What you're doing is looking at square matrices $P$ and $D$, idempotent and diagonal respectively, and asking if there's a shortcut to compute the eigenvalues of $PD$. I'm pretty sure the answer is, no there is not.
– Theo Bendit
Aug 2 at 2:43
What you're doing is looking at square matrices $P$ and $D$, idempotent and diagonal respectively, and asking if there's a shortcut to compute the eigenvalues of $PD$. I'm pretty sure the answer is, no there is not.
– Theo Bendit
Aug 2 at 2:43
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
It's not true that the positive eigenvalues of $P(A)D$ are diagonal elements of $D$. Consider e.g.
$$P(A) = pmatrix8/9 & -2/9 & -2/9cr
-2/9 & 5/9 & -4/9cr
-2/9 & -4/9 & 5/9cr, D = pmatrixa & 0 & 0cr 0 & b & 0cr 0 & 0 & c$$
Then $P(A)D$ has characteristic polynomial $$t^3- left(frac89 a +frac59 b + frac59 cright)t^2+ frac4ac+bc+4ab9 t
$$
and none of $a,b,c$ are roots of this.
answered Aug 2 at 3:04
Robert Israel
303k22200440
Thank you! If I ask you more, when the eigenvalues of $P(A)D$ becomes the distinct elements of $D$? Now I see from you answer that it's generally not always the case, but I found some cases where it holds.
– user568810
Aug 2 at 3:07
Although I found some cases where it holds, the problem is I don't know what condition of $A$ will lead this property and WHY it holds. :(
– user568810
Aug 2 at 3:15
1
If $P(A)$ and $D$ commute, they can be jointly diagonalized, and then it's true.
– Robert Israel
Aug 2 at 4:33
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It's not true that the positive eigenvalues of $P(A)D$ are diagonal elements of $D$. Consider e.g.
$$P(A) = pmatrix8/9 & -2/9 & -2/9cr
-2/9 & 5/9 & -4/9cr
-2/9 & -4/9 & 5/9cr, D = pmatrixa & 0 & 0cr 0 & b & 0cr 0 & 0 & c$$
Then $P(A)D$ has characteristic polynomial $$t^3- left(frac89 a +frac59 b + frac59 cright)t^2+ frac4ac+bc+4ab9 t
$$
and none of $a,b,c$ are roots of this.
answered Aug 2 at 3:04
Robert Israel
303k22200440
Thank you! If I ask you more, when the eigenvalues of $P(A)D$ becomes the distinct elements of $D$? Now I see from you answer that it's generally not always the case, but I found some cases where it holds.
– user568810
Aug 2 at 3:07
Although I found some cases where it holds, the problem is I don't know what condition of $A$ will lead this property and WHY it holds. :(
– user568810
Aug 2 at 3:15
1
If $P(A)$ and $D$ commute, they can be jointly diagonalized, and then it's true.
– Robert Israel
Aug 2 at 4:33
add a comment |Â
up vote
2
down vote
accepted
It's not true that the positive eigenvalues of $P(A)D$ are diagonal elements of $D$. Consider e.g.
$$P(A) = pmatrix8/9 & -2/9 & -2/9cr
-2/9 & 5/9 & -4/9cr
-2/9 & -4/9 & 5/9cr, D = pmatrixa & 0 & 0cr 0 & b & 0cr 0 & 0 & c$$
Then $P(A)D$ has characteristic polynomial $$t^3- left(frac89 a +frac59 b + frac59 cright)t^2+ frac4ac+bc+4ab9 t
$$
and none of $a,b,c$ are roots of this.
answered Aug 2 at 3:04
Robert Israel
303k22200440
Thank you! If I ask you more, when the eigenvalues of $P(A)D$ becomes the distinct elements of $D$? Now I see from you answer that it's generally not always the case, but I found some cases where it holds.
– user568810
Aug 2 at 3:07
Although I found some cases where it holds, the problem is I don't know what condition of $A$ will lead this property and WHY it holds. :(
– user568810
Aug 2 at 3:15
1
If $P(A)$ and $D$ commute, they can be jointly diagonalized, and then it's true.
– Robert Israel
Aug 2 at 4:33
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It's not true that the positive eigenvalues of $P(A)D$ are diagonal elements of $D$. Consider e.g.
$$P(A) = pmatrix8/9 & -2/9 & -2/9cr
-2/9 & 5/9 & -4/9cr
-2/9 & -4/9 & 5/9cr, D = pmatrixa & 0 & 0cr 0 & b & 0cr 0 & 0 & c$$
Then $P(A)D$ has characteristic polynomial $$t^3- left(frac89 a +frac59 b + frac59 cright)t^2+ frac4ac+bc+4ab9 t
$$
and none of $a,b,c$ are roots of this.
answered Aug 2 at 3:04
Robert Israel
303k22200440
It's not true that the positive eigenvalues of $P(A)D$ are diagonal elements of $D$. Consider e.g.
$$P(A) = pmatrix8/9 & -2/9 & -2/9cr
-2/9 & 5/9 & -4/9cr
-2/9 & -4/9 & 5/9cr, D = pmatrixa & 0 & 0cr 0 & b & 0cr 0 & 0 & c$$
Then $P(A)D$ has characteristic polynomial $$t^3- left(frac89 a +frac59 b + frac59 cright)t^2+ frac4ac+bc+4ab9 t
$$
and none of $a,b,c$ are roots of this.
answered Aug 2 at 3:04
Robert Israel
303k22200440
answered Aug 2 at 3:04
Robert Israel
303k22200440
answered Aug 2 at 3:04
Robert Israel
303k22200440
answered Aug 2 at 3:04
Robert Israel
303k22200440
303k22200440
Thank you! If I ask you more, when the eigenvalues of $P(A)D$ becomes the distinct elements of $D$? Now I see from you answer that it's generally not always the case, but I found some cases where it holds.
– user568810
Aug 2 at 3:07
Although I found some cases where it holds, the problem is I don't know what condition of $A$ will lead this property and WHY it holds. :(
– user568810
Aug 2 at 3:15
1
If $P(A)$ and $D$ commute, they can be jointly diagonalized, and then it's true.
– Robert Israel
Aug 2 at 4:33
add a comment |Â
Thank you! If I ask you more, when the eigenvalues of $P(A)D$ becomes the distinct elements of $D$? Now I see from you answer that it's generally not always the case, but I found some cases where it holds.
– user568810
Aug 2 at 3:07
Although I found some cases where it holds, the problem is I don't know what condition of $A$ will lead this property and WHY it holds. :(
– user568810
Aug 2 at 3:15
1
If $P(A)$ and $D$ commute, they can be jointly diagonalized, and then it's true.
– Robert Israel
Aug 2 at 4:33
Thank you! If I ask you more, when the eigenvalues of $P(A)D$ becomes the distinct elements of $D$? Now I see from you answer that it's generally not always the case, but I found some cases where it holds.
– user568810
Aug 2 at 3:07
Thank you! If I ask you more, when the eigenvalues of $P(A)D$ becomes the distinct elements of $D$? Now I see from you answer that it's generally not always the case, but I found some cases where it holds.
– user568810
Aug 2 at 3:07
Although I found some cases where it holds, the problem is I don't know what condition of $A$ will lead this property and WHY it holds. :(
– user568810
Aug 2 at 3:15
Although I found some cases where it holds, the problem is I don't know what condition of $A$ will lead this property and WHY it holds. :(
– user568810
Aug 2 at 3:15
1
1
If $P(A)$ and $D$ commute, they can be jointly diagonalized, and then it's true.
– Robert Israel
Aug 2 at 4:33
If $P(A)$ and $D$ commute, they can be jointly diagonalized, and then it's true.
– Robert Israel
Aug 2 at 4:33
add a comment |Â
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@philip Can you explain me how my problem is related to the proof about diagonalization?
– user568810
Aug 2 at 2:43
1
What you're doing is looking at square matrices $P$ and $D$, idempotent and diagonal respectively, and asking if there's a shortcut to compute the eigenvalues of $PD$. I'm pretty sure the answer is, no there is not.
– Theo Bendit
Aug 2 at 2:43