Show that any projector onto subspace L is parallel to M
$begingroup$
Assume that $mathbb{R}^n$ is represented as the direct (but not necessarily orthogonal) sum $M_1 dotplus M_2$ of two its subspaces $M_1$ and $M_2$. In particular, every $x in mathbb{R}^n$ can be represented in a unique way as $x = x_1 + x_2$ with some $x_j in M_j$, and the mapping $P_j: x mapsto x_j$ is called the (oblique) projector onto $M_j$ parallel to $M_{3-j}$. It is easy to show that $P_j$ satisfy the following properties: $P_1 + P_2 = I_n$, $P_j^2 = P_j$ and $P_1P_2 = P_2 P_1 = 0$.
Show that any matrix $P$ satisfying the relation $P^2 = P$ is a projector onto some subspace $L$ parallel to $M$, and identify these $L$ and $M$.
inner-product-space orthogonality projection
edited Jan 31 at 21:51
Hanno
2,274628
asked Dec 5 '18 at 20:24
MichaelMichael
1055
$endgroup$
add a comment |
$begingroup$
Assume that $mathbb{R}^n$ is represented as the direct (but not necessarily orthogonal) sum $M_1 dotplus M_2$ of two its subspaces $M_1$ and $M_2$. In particular, every $x in mathbb{R}^n$ can be represented in a unique way as $x = x_1 + x_2$ with some $x_j in M_j$, and the mapping $P_j: x mapsto x_j$ is called the (oblique) projector onto $M_j$ parallel to $M_{3-j}$. It is easy to show that $P_j$ satisfy the following properties: $P_1 + P_2 = I_n$, $P_j^2 = P_j$ and $P_1P_2 = P_2 P_1 = 0$.
Show that any matrix $P$ satisfying the relation $P^2 = P$ is a projector onto some subspace $L$ parallel to $M$, and identify these $L$ and $M$.
inner-product-space orthogonality projection
edited Jan 31 at 21:51
Hanno
2,274628
asked Dec 5 '18 at 20:24
MichaelMichael
1055
$endgroup$
$begingroup$
Formulations like "Show that ..." are a bit oblique, and in most cases orthogonal to the community's readiness to answer a question.
$endgroup$
– Hanno
Jan 31 at 22:03
$begingroup$
"It is easy to show that ... and $P_1P_2 = P_2 P_1 = 0$." is not true for non-orthogonal direct sums. It even characterises orthogonal sums.
$endgroup$
– Hanno
Jan 31 at 22:04
add a comment |
$begingroup$
Assume that $mathbb{R}^n$ is represented as the direct (but not necessarily orthogonal) sum $M_1 dotplus M_2$ of two its subspaces $M_1$ and $M_2$. In particular, every $x in mathbb{R}^n$ can be represented in a unique way as $x = x_1 + x_2$ with some $x_j in M_j$, and the mapping $P_j: x mapsto x_j$ is called the (oblique) projector onto $M_j$ parallel to $M_{3-j}$. It is easy to show that $P_j$ satisfy the following properties: $P_1 + P_2 = I_n$, $P_j^2 = P_j$ and $P_1P_2 = P_2 P_1 = 0$.
Show that any matrix $P$ satisfying the relation $P^2 = P$ is a projector onto some subspace $L$ parallel to $M$, and identify these $L$ and $M$.
inner-product-space orthogonality projection
edited Jan 31 at 21:51
Hanno
2,274628
asked Dec 5 '18 at 20:24
MichaelMichael
1055
$endgroup$
Assume that $mathbb{R}^n$ is represented as the direct (but not necessarily orthogonal) sum $M_1 dotplus M_2$ of two its subspaces $M_1$ and $M_2$. In particular, every $x in mathbb{R}^n$ can be represented in a unique way as $x = x_1 + x_2$ with some $x_j in M_j$, and the mapping $P_j: x mapsto x_j$ is called the (oblique) projector onto $M_j$ parallel to $M_{3-j}$. It is easy to show that $P_j$ satisfy the following properties: $P_1 + P_2 = I_n$, $P_j^2 = P_j$ and $P_1P_2 = P_2 P_1 = 0$.
Show that any matrix $P$ satisfying the relation $P^2 = P$ is a projector onto some subspace $L$ parallel to $M$, and identify these $L$ and $M$.
inner-product-space orthogonality projection
inner-product-space orthogonality projection
edited Jan 31 at 21:51
Hanno
2,274628
asked Dec 5 '18 at 20:24
MichaelMichael
1055
edited Jan 31 at 21:51
Hanno
2,274628
asked Dec 5 '18 at 20:24
MichaelMichael
1055
edited Jan 31 at 21:51
Hanno
2,274628
edited Jan 31 at 21:51
Hanno
2,274628
edited Jan 31 at 21:51
Hanno
2,274628
2,274628
asked Dec 5 '18 at 20:24
MichaelMichael
1055
asked Dec 5 '18 at 20:24
MichaelMichael
1055
asked Dec 5 '18 at 20:24
MichaelMichael
1055
1055
$begingroup$
Formulations like "Show that ..." are a bit oblique, and in most cases orthogonal to the community's readiness to answer a question.
$endgroup$
– Hanno
Jan 31 at 22:03
$begingroup$
"It is easy to show that ... and $P_1P_2 = P_2 P_1 = 0$." is not true for non-orthogonal direct sums. It even characterises orthogonal sums.
$endgroup$
– Hanno
Jan 31 at 22:04
add a comment |
$begingroup$
Formulations like "Show that ..." are a bit oblique, and in most cases orthogonal to the community's readiness to answer a question.
$endgroup$
– Hanno
Jan 31 at 22:03
$begingroup$
"It is easy to show that ... and $P_1P_2 = P_2 P_1 = 0$." is not true for non-orthogonal direct sums. It even characterises orthogonal sums.
$endgroup$
– Hanno
Jan 31 at 22:04
$begingroup$
Formulations like "Show that ..." are a bit oblique, and in most cases orthogonal to the community's readiness to answer a question.
$endgroup$
– Hanno
Jan 31 at 22:03
$begingroup$
Formulations like "Show that ..." are a bit oblique, and in most cases orthogonal to the community's readiness to answer a question.
$endgroup$
– Hanno
Jan 31 at 22:03
$begingroup$
"It is easy to show that ... and $P_1P_2 = P_2 P_1 = 0$." is not true for non-orthogonal direct sums. It even characterises orthogonal sums.
$endgroup$
– Hanno
Jan 31 at 22:04
$begingroup$
"It is easy to show that ... and $P_1P_2 = P_2 P_1 = 0$." is not true for non-orthogonal direct sums. It even characterises orthogonal sums.
$endgroup$
– Hanno
Jan 31 at 22:04
add a comment |
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$begingroup$
Formulations like "Show that ..." are a bit oblique, and in most cases orthogonal to the community's readiness to answer a question.
$endgroup$
– Hanno
Jan 31 at 22:03
$begingroup$
"It is easy to show that ... and $P_1P_2 = P_2 P_1 = 0$." is not true for non-orthogonal direct sums. It even characterises orthogonal sums.
$endgroup$
– Hanno
Jan 31 at 22:04