Mixing
If $A$ is symmetric matrix and $P$ is matrix of orthogonal projection,what is then matrix $Pcdot A$?
$begingroup$
Let $A$ be symmetric matrix, and $P$ be matrix of orthogonal projection on null space of matrix $A$. What can you say about matrix $Pcdot A$ ?
$$ $$
The fact is that symmetric matrices have $n$ different eigenvalues, and $n$ different independent eigenvectors. That means that rank of symmetric matrix is full. Because of that, null space of symmetric matrix will be only zero-vector.I can conclude then that P can't be calculate on null space. Am I right?
$$ $$But if my P was matrix of orthogonal projection on column space of symmetric matrix $A$, I would say that P is identity matrix $(P=A(A^{T}A)^{-1}A^{T})$. So,matrix $PA$ will be only matrix $A$.
If someone wanted to find orthogonal projection of some vector on fundamental subspaces, that vector will be in column space of matrix $A$, because $v_p=Pv$ gives exactly that given vector, and null space of $A^{T}$ will be $v_p=(I-P)v=0$. I would like if someone correct me if I do something wrong..
linear-algebra
asked Feb 2 at 18:16
FiggaroFiggaro
406
$endgroup$
add a comment |
$begingroup$
Let $A$ be symmetric matrix, and $P$ be matrix of orthogonal projection on null space of matrix $A$. What can you say about matrix $Pcdot A$ ?
$$ $$
The fact is that symmetric matrices have $n$ different eigenvalues, and $n$ different independent eigenvectors. That means that rank of symmetric matrix is full. Because of that, null space of symmetric matrix will be only zero-vector.I can conclude then that P can't be calculate on null space. Am I right?
$$ $$But if my P was matrix of orthogonal projection on column space of symmetric matrix $A$, I would say that P is identity matrix $(P=A(A^{T}A)^{-1}A^{T})$. So,matrix $PA$ will be only matrix $A$.
If someone wanted to find orthogonal projection of some vector on fundamental subspaces, that vector will be in column space of matrix $A$, because $v_p=Pv$ gives exactly that given vector, and null space of $A^{T}$ will be $v_p=(I-P)v=0$. I would like if someone correct me if I do something wrong..
linear-algebra
asked Feb 2 at 18:16
FiggaroFiggaro
406
$endgroup$
$begingroup$
The $0$ matrix is symmetric and not of full rank.
$endgroup$
– kimchi lover
Feb 2 at 18:18
$begingroup$
Yes,you're right. But in generally, what can I say about matrix $PA$ ?
$endgroup$
– Figgaro
Feb 2 at 18:40
add a comment |
$begingroup$
Let $A$ be symmetric matrix, and $P$ be matrix of orthogonal projection on null space of matrix $A$. What can you say about matrix $Pcdot A$ ?
$$ $$
The fact is that symmetric matrices have $n$ different eigenvalues, and $n$ different independent eigenvectors. That means that rank of symmetric matrix is full. Because of that, null space of symmetric matrix will be only zero-vector.I can conclude then that P can't be calculate on null space. Am I right?
$$ $$But if my P was matrix of orthogonal projection on column space of symmetric matrix $A$, I would say that P is identity matrix $(P=A(A^{T}A)^{-1}A^{T})$. So,matrix $PA$ will be only matrix $A$.
If someone wanted to find orthogonal projection of some vector on fundamental subspaces, that vector will be in column space of matrix $A$, because $v_p=Pv$ gives exactly that given vector, and null space of $A^{T}$ will be $v_p=(I-P)v=0$. I would like if someone correct me if I do something wrong..
linear-algebra
asked Feb 2 at 18:16
FiggaroFiggaro
406
$endgroup$
Let $A$ be symmetric matrix, and $P$ be matrix of orthogonal projection on null space of matrix $A$. What can you say about matrix $Pcdot A$ ?
$$ $$
The fact is that symmetric matrices have $n$ different eigenvalues, and $n$ different independent eigenvectors. That means that rank of symmetric matrix is full. Because of that, null space of symmetric matrix will be only zero-vector.I can conclude then that P can't be calculate on null space. Am I right?
$$ $$But if my P was matrix of orthogonal projection on column space of symmetric matrix $A$, I would say that P is identity matrix $(P=A(A^{T}A)^{-1}A^{T})$. So,matrix $PA$ will be only matrix $A$.
If someone wanted to find orthogonal projection of some vector on fundamental subspaces, that vector will be in column space of matrix $A$, because $v_p=Pv$ gives exactly that given vector, and null space of $A^{T}$ will be $v_p=(I-P)v=0$. I would like if someone correct me if I do something wrong..
linear-algebra
linear-algebra
asked Feb 2 at 18:16
FiggaroFiggaro
406
asked Feb 2 at 18:16
FiggaroFiggaro
406
asked Feb 2 at 18:16
FiggaroFiggaro
406
asked Feb 2 at 18:16
FiggaroFiggaro
406
asked Feb 2 at 18:16
FiggaroFiggaro
406
406
$begingroup$
The $0$ matrix is symmetric and not of full rank.
$endgroup$
– kimchi lover
Feb 2 at 18:18
$begingroup$
Yes,you're right. But in generally, what can I say about matrix $PA$ ?
$endgroup$
– Figgaro
Feb 2 at 18:40
add a comment |
$begingroup$
The $0$ matrix is symmetric and not of full rank.
$endgroup$
– kimchi lover
Feb 2 at 18:18
$begingroup$
Yes,you're right. But in generally, what can I say about matrix $PA$ ?
$endgroup$
– Figgaro
Feb 2 at 18:40
$begingroup$
The $0$ matrix is symmetric and not of full rank.
$endgroup$
– kimchi lover
Feb 2 at 18:18
$begingroup$
The $0$ matrix is symmetric and not of full rank.
$endgroup$
– kimchi lover
Feb 2 at 18:18
$begingroup$
Yes,you're right. But in generally, what can I say about matrix $PA$ ?
$endgroup$
– Figgaro
Feb 2 at 18:40
$begingroup$
Yes,you're right. But in generally, what can I say about matrix $PA$ ?
$endgroup$
– Figgaro
Feb 2 at 18:40
add a comment |
1 Answer
1
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$begingroup$
The fact is that the symmetric matrices have $n$ linearly independent eigenvectors, but not $n$ distinct eigenvalues. Note that diagonal matrix is a special form of symmetric matrix. To each $i$-th diagonal entry, we can assign value $lambda_i$ to make any set of reals ${lambda_1,lambda_2,cdots,lambda_n}$, counted with multiplicity, as a set of eigenvalues of $A$. Accordingly, $ker A$ (the $0$ eigenspace) does not have to be trivial.
The answer to your problem is quite simple: $PA = O$ whenever $P$ is an orthogonal projection onto $ker A$. To see this, note that for all $y=Ax in mathcal R (A)$ (the range of $A$) and $zin ker A$, it holds that
$$
langle y,zrangle =langle Ax,zrangle = langle x,A^T zrangle = langle x,A zrangle=langle x, 0rangle=0.
$$ This shows $mathcal R(A)perp ker A$, which implies that $Py=PAx=0$ for all $x$. Thus $PA=O$.
answered Feb 2 at 18:40
SongSong
18.6k21651
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The fact is that the symmetric matrices have $n$ linearly independent eigenvectors, but not $n$ distinct eigenvalues. Note that diagonal matrix is a special form of symmetric matrix. To each $i$-th diagonal entry, we can assign value $lambda_i$ to make any set of reals ${lambda_1,lambda_2,cdots,lambda_n}$, counted with multiplicity, as a set of eigenvalues of $A$. Accordingly, $ker A$ (the $0$ eigenspace) does not have to be trivial.
The answer to your problem is quite simple: $PA = O$ whenever $P$ is an orthogonal projection onto $ker A$. To see this, note that for all $y=Ax in mathcal R (A)$ (the range of $A$) and $zin ker A$, it holds that
$$
langle y,zrangle =langle Ax,zrangle = langle x,A^T zrangle = langle x,A zrangle=langle x, 0rangle=0.
$$ This shows $mathcal R(A)perp ker A$, which implies that $Py=PAx=0$ for all $x$. Thus $PA=O$.
answered Feb 2 at 18:40
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
The fact is that the symmetric matrices have $n$ linearly independent eigenvectors, but not $n$ distinct eigenvalues. Note that diagonal matrix is a special form of symmetric matrix. To each $i$-th diagonal entry, we can assign value $lambda_i$ to make any set of reals ${lambda_1,lambda_2,cdots,lambda_n}$, counted with multiplicity, as a set of eigenvalues of $A$. Accordingly, $ker A$ (the $0$ eigenspace) does not have to be trivial.
The answer to your problem is quite simple: $PA = O$ whenever $P$ is an orthogonal projection onto $ker A$. To see this, note that for all $y=Ax in mathcal R (A)$ (the range of $A$) and $zin ker A$, it holds that
$$
langle y,zrangle =langle Ax,zrangle = langle x,A^T zrangle = langle x,A zrangle=langle x, 0rangle=0.
$$ This shows $mathcal R(A)perp ker A$, which implies that $Py=PAx=0$ for all $x$. Thus $PA=O$.
answered Feb 2 at 18:40
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
The fact is that the symmetric matrices have $n$ linearly independent eigenvectors, but not $n$ distinct eigenvalues. Note that diagonal matrix is a special form of symmetric matrix. To each $i$-th diagonal entry, we can assign value $lambda_i$ to make any set of reals ${lambda_1,lambda_2,cdots,lambda_n}$, counted with multiplicity, as a set of eigenvalues of $A$. Accordingly, $ker A$ (the $0$ eigenspace) does not have to be trivial.
The answer to your problem is quite simple: $PA = O$ whenever $P$ is an orthogonal projection onto $ker A$. To see this, note that for all $y=Ax in mathcal R (A)$ (the range of $A$) and $zin ker A$, it holds that
$$
langle y,zrangle =langle Ax,zrangle = langle x,A^T zrangle = langle x,A zrangle=langle x, 0rangle=0.
$$ This shows $mathcal R(A)perp ker A$, which implies that $Py=PAx=0$ for all $x$. Thus $PA=O$.
answered Feb 2 at 18:40
SongSong
18.6k21651
$endgroup$
The fact is that the symmetric matrices have $n$ linearly independent eigenvectors, but not $n$ distinct eigenvalues. Note that diagonal matrix is a special form of symmetric matrix. To each $i$-th diagonal entry, we can assign value $lambda_i$ to make any set of reals ${lambda_1,lambda_2,cdots,lambda_n}$, counted with multiplicity, as a set of eigenvalues of $A$. Accordingly, $ker A$ (the $0$ eigenspace) does not have to be trivial.
The answer to your problem is quite simple: $PA = O$ whenever $P$ is an orthogonal projection onto $ker A$. To see this, note that for all $y=Ax in mathcal R (A)$ (the range of $A$) and $zin ker A$, it holds that
$$
langle y,zrangle =langle Ax,zrangle = langle x,A^T zrangle = langle x,A zrangle=langle x, 0rangle=0.
$$ This shows $mathcal R(A)perp ker A$, which implies that $Py=PAx=0$ for all $x$. Thus $PA=O$.
answered Feb 2 at 18:40
SongSong
18.6k21651
answered Feb 2 at 18:40
SongSong
18.6k21651
answered Feb 2 at 18:40
SongSong
18.6k21651
answered Feb 2 at 18:40
SongSong
18.6k21651
18.6k21651
add a comment |
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$begingroup$
The $0$ matrix is symmetric and not of full rank.
$endgroup$
– kimchi lover
Feb 2 at 18:18
$begingroup$
Yes,you're right. But in generally, what can I say about matrix $PA$ ?
$endgroup$
– Figgaro
Feb 2 at 18:40