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Inequality of vector norms with projections
The Next CEO of Stack OverflowVector Spaces and Projectionsshow projection matrix is equal to matrix times its transposeFinding orthogonal projections onto $1$ (co)-dimensional subspaces of $mathbb R^n$How to prove that $P_{w^perp}=I-P_{w}$ for the following condition?Relationship between angle of vector with corresponding projections on orthogonal subspacesFinding orthogonal vectorOrthogonal projection of vector.Does the orthogonal projection of a vector onto a subspace equal the original vector?Proving an inequality about orthogonal projectionsOrthogonal projection and orthogonal complement
$begingroup$
Let $W$ be a vector subspace of $V$, a space with a dot product; $vin V$. Let $p_W(v)$ be the orthogonal projection of $v$ onto $W$ and $win W, wneq p_W(v)$.
How can i prove that $||v-w|| > ||v-p_W(v)||$?
linear-algebra vector-spaces norm orthogonality
asked Mar 16 at 17:42
Noé Duarte GonzálezNoé Duarte González
63
$endgroup$
add a comment |
$begingroup$
Let $W$ be a vector subspace of $V$, a space with a dot product; $vin V$. Let $p_W(v)$ be the orthogonal projection of $v$ onto $W$ and $win W, wneq p_W(v)$.
How can i prove that $||v-w|| > ||v-p_W(v)||$?
linear-algebra vector-spaces norm orthogonality
asked Mar 16 at 17:42
Noé Duarte GonzálezNoé Duarte González
63
$endgroup$
$begingroup$
Pray tell, what is $w'$?
$endgroup$
– Robert Lewis
Mar 16 at 17:49
1
$begingroup$
My bad, I meant $w$
$endgroup$
– Noé Duarte González
Mar 16 at 17:55
$begingroup$
Thanks for the correction!
$endgroup$
– Robert Lewis
Mar 16 at 18:06
add a comment |
$begingroup$
Let $W$ be a vector subspace of $V$, a space with a dot product; $vin V$. Let $p_W(v)$ be the orthogonal projection of $v$ onto $W$ and $win W, wneq p_W(v)$.
How can i prove that $||v-w|| > ||v-p_W(v)||$?
linear-algebra vector-spaces norm orthogonality
asked Mar 16 at 17:42
Noé Duarte GonzálezNoé Duarte González
63
$endgroup$
Let $W$ be a vector subspace of $V$, a space with a dot product; $vin V$. Let $p_W(v)$ be the orthogonal projection of $v$ onto $W$ and $win W, wneq p_W(v)$.
How can i prove that $||v-w|| > ||v-p_W(v)||$?
linear-algebra vector-spaces norm orthogonality
linear-algebra vector-spaces norm orthogonality
asked Mar 16 at 17:42
Noé Duarte GonzálezNoé Duarte González
63
asked Mar 16 at 17:42
Noé Duarte GonzálezNoé Duarte González
63
edited Mar 16 at 17:54
Noé Duarte González
asked Mar 16 at 17:42
Noé Duarte GonzálezNoé Duarte González
63
asked Mar 16 at 17:42
Noé Duarte GonzálezNoé Duarte González
63
asked Mar 16 at 17:42
Noé Duarte GonzálezNoé Duarte González
63
63
$begingroup$
Pray tell, what is $w'$?
$endgroup$
– Robert Lewis
Mar 16 at 17:49
1
$begingroup$
My bad, I meant $w$
$endgroup$
– Noé Duarte González
Mar 16 at 17:55
$begingroup$
Thanks for the correction!
$endgroup$
– Robert Lewis
Mar 16 at 18:06
add a comment |
$begingroup$
Pray tell, what is $w'$?
$endgroup$
– Robert Lewis
Mar 16 at 17:49
1
$begingroup$
My bad, I meant $w$
$endgroup$
– Noé Duarte González
Mar 16 at 17:55
$begingroup$
Thanks for the correction!
$endgroup$
– Robert Lewis
Mar 16 at 18:06
$begingroup$
Pray tell, what is $w'$?
$endgroup$
– Robert Lewis
Mar 16 at 17:49
$begingroup$
Pray tell, what is $w'$?
$endgroup$
– Robert Lewis
Mar 16 at 17:49
1
1
$begingroup$
My bad, I meant $w$
$endgroup$
– Noé Duarte González
Mar 16 at 17:55
$begingroup$
My bad, I meant $w$
$endgroup$
– Noé Duarte González
Mar 16 at 17:55
$begingroup$
Thanks for the correction!
$endgroup$
– Robert Lewis
Mar 16 at 18:06
$begingroup$
Thanks for the correction!
$endgroup$
– Robert Lewis
Mar 16 at 18:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $v = v_| + v_bot $, where $v_| = p_W(v)$, $v_bot in W^bot$.
$ |v-w|^2 = (v-w)^2 = v^2 + w^2 - 2vw = v_bot^2 + v_|^2 + w^2 - 2v_|w = v_bot^2 + (v_|-w)^2$
$ |v-v_||^2 = v_bot^2 < v_bot^2 + (v_|-w)^2 $, QED.
answered Mar 16 at 18:14
colt_browningcolt_browning
768110
$endgroup$
add a comment |
$begingroup$
Since $W subset V$ is a fixed subspace, I am going to drop the subscript "$W"$ from $P$; in this answer, $P = P_W$.
Recall that an orthogonal projection $P$ satisfies
$P^2 = P = P^T; tag 1$
thus for any
$x, y in V tag 2$
we have
$langle x, Py rangle = langle P^Tx, y rangle = langle Px, y rangle; tag 3$
now consider the three vectors $v$, $Pv$, and $w$;
$Pv, w in W Longrightarrow Pv - w in W, tag 4$
$langle v - Pv, w rangle = langle v, w rangle - langle Pv, w rangle = langle v, w rangle - langle v, Pw rangle = langle v, w rangle - langle v, w rangle = 0, tag 5$
where we have used (3) and the fact that $Pw = w$ for $w in W$; since this holds for every $w in W$, we have established that
$v - Pv in W^bot; tag 6$
we write
$v - w = (v - Pv) + (Pv - w), tag 7$
and by virtue of (4) and (6)
$langle v - Pv, Pv - w rangle = 0, tag 8$
whence
$Vert v - w Vert^2 = langle v - w, v - w rangle = langle (v - Pv) + (Pv - w), (v - Pv) + (Pv - w) rangle$
$= langle v - Pv, v - Pv rangle -2langle v - Pv, Pv - w rangle + langle Pv - w, Pv - w rangle$
$= Vert v - Pv Vert^2 + Vert Pv - w Vert^2;tag 9$
since each term on the right is non-negative, we find that
$Vert v - w Vert^2 ge Vert v - Pv Vert^2, tag{10}$
whence
$Vert v - w Vert ge Vert v - Pv Vert, ; forall w in W, tag{11}$
with equality holding precisely when
$Vert w - Pv Vert = 0 Longleftrightarrow w = Pv; tag{12}$
thus
$w ne Pv Longrightarrow Vert v - w Vert > Vert v - Pv Vert. tag{13}$
$OEDelta$.
Inequality of vector norms with projections
answered Mar 17 at 19:46
Robert LewisRobert Lewis
48.5k23167
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $v = v_| + v_bot $, where $v_| = p_W(v)$, $v_bot in W^bot$.
$ |v-w|^2 = (v-w)^2 = v^2 + w^2 - 2vw = v_bot^2 + v_|^2 + w^2 - 2v_|w = v_bot^2 + (v_|-w)^2$
$ |v-v_||^2 = v_bot^2 < v_bot^2 + (v_|-w)^2 $, QED.
answered Mar 16 at 18:14
colt_browningcolt_browning
768110
$endgroup$
add a comment |
$begingroup$
Let $v = v_| + v_bot $, where $v_| = p_W(v)$, $v_bot in W^bot$.
$ |v-w|^2 = (v-w)^2 = v^2 + w^2 - 2vw = v_bot^2 + v_|^2 + w^2 - 2v_|w = v_bot^2 + (v_|-w)^2$
$ |v-v_||^2 = v_bot^2 < v_bot^2 + (v_|-w)^2 $, QED.
answered Mar 16 at 18:14
colt_browningcolt_browning
768110
$endgroup$
add a comment |
$begingroup$
Let $v = v_| + v_bot $, where $v_| = p_W(v)$, $v_bot in W^bot$.
$ |v-w|^2 = (v-w)^2 = v^2 + w^2 - 2vw = v_bot^2 + v_|^2 + w^2 - 2v_|w = v_bot^2 + (v_|-w)^2$
$ |v-v_||^2 = v_bot^2 < v_bot^2 + (v_|-w)^2 $, QED.
answered Mar 16 at 18:14
colt_browningcolt_browning
768110
$endgroup$
Let $v = v_| + v_bot $, where $v_| = p_W(v)$, $v_bot in W^bot$.
$ |v-w|^2 = (v-w)^2 = v^2 + w^2 - 2vw = v_bot^2 + v_|^2 + w^2 - 2v_|w = v_bot^2 + (v_|-w)^2$
$ |v-v_||^2 = v_bot^2 < v_bot^2 + (v_|-w)^2 $, QED.
answered Mar 16 at 18:14
colt_browningcolt_browning
768110
answered Mar 16 at 18:14
colt_browningcolt_browning
768110
answered Mar 16 at 18:14
colt_browningcolt_browning
768110
answered Mar 16 at 18:14
colt_browningcolt_browning
768110
768110
add a comment |
add a comment |
$begingroup$
Since $W subset V$ is a fixed subspace, I am going to drop the subscript "$W"$ from $P$; in this answer, $P = P_W$.
Recall that an orthogonal projection $P$ satisfies
$P^2 = P = P^T; tag 1$
thus for any
$x, y in V tag 2$
we have
$langle x, Py rangle = langle P^Tx, y rangle = langle Px, y rangle; tag 3$
now consider the three vectors $v$, $Pv$, and $w$;
$Pv, w in W Longrightarrow Pv - w in W, tag 4$
$langle v - Pv, w rangle = langle v, w rangle - langle Pv, w rangle = langle v, w rangle - langle v, Pw rangle = langle v, w rangle - langle v, w rangle = 0, tag 5$
where we have used (3) and the fact that $Pw = w$ for $w in W$; since this holds for every $w in W$, we have established that
$v - Pv in W^bot; tag 6$
we write
$v - w = (v - Pv) + (Pv - w), tag 7$
and by virtue of (4) and (6)
$langle v - Pv, Pv - w rangle = 0, tag 8$
whence
$Vert v - w Vert^2 = langle v - w, v - w rangle = langle (v - Pv) + (Pv - w), (v - Pv) + (Pv - w) rangle$
$= langle v - Pv, v - Pv rangle -2langle v - Pv, Pv - w rangle + langle Pv - w, Pv - w rangle$
$= Vert v - Pv Vert^2 + Vert Pv - w Vert^2;tag 9$
since each term on the right is non-negative, we find that
$Vert v - w Vert^2 ge Vert v - Pv Vert^2, tag{10}$
whence
$Vert v - w Vert ge Vert v - Pv Vert, ; forall w in W, tag{11}$
with equality holding precisely when
$Vert w - Pv Vert = 0 Longleftrightarrow w = Pv; tag{12}$
thus
$w ne Pv Longrightarrow Vert v - w Vert > Vert v - Pv Vert. tag{13}$
$OEDelta$.
Inequality of vector norms with projections
answered Mar 17 at 19:46
Robert LewisRobert Lewis
48.5k23167
$endgroup$
add a comment |
$begingroup$
Since $W subset V$ is a fixed subspace, I am going to drop the subscript "$W"$ from $P$; in this answer, $P = P_W$.
Recall that an orthogonal projection $P$ satisfies
$P^2 = P = P^T; tag 1$
thus for any
$x, y in V tag 2$
we have
$langle x, Py rangle = langle P^Tx, y rangle = langle Px, y rangle; tag 3$
now consider the three vectors $v$, $Pv$, and $w$;
$Pv, w in W Longrightarrow Pv - w in W, tag 4$
$langle v - Pv, w rangle = langle v, w rangle - langle Pv, w rangle = langle v, w rangle - langle v, Pw rangle = langle v, w rangle - langle v, w rangle = 0, tag 5$
where we have used (3) and the fact that $Pw = w$ for $w in W$; since this holds for every $w in W$, we have established that
$v - Pv in W^bot; tag 6$
we write
$v - w = (v - Pv) + (Pv - w), tag 7$
and by virtue of (4) and (6)
$langle v - Pv, Pv - w rangle = 0, tag 8$
whence
$Vert v - w Vert^2 = langle v - w, v - w rangle = langle (v - Pv) + (Pv - w), (v - Pv) + (Pv - w) rangle$
$= langle v - Pv, v - Pv rangle -2langle v - Pv, Pv - w rangle + langle Pv - w, Pv - w rangle$
$= Vert v - Pv Vert^2 + Vert Pv - w Vert^2;tag 9$
since each term on the right is non-negative, we find that
$Vert v - w Vert^2 ge Vert v - Pv Vert^2, tag{10}$
whence
$Vert v - w Vert ge Vert v - Pv Vert, ; forall w in W, tag{11}$
with equality holding precisely when
$Vert w - Pv Vert = 0 Longleftrightarrow w = Pv; tag{12}$
thus
$w ne Pv Longrightarrow Vert v - w Vert > Vert v - Pv Vert. tag{13}$
$OEDelta$.
Inequality of vector norms with projections
answered Mar 17 at 19:46
Robert LewisRobert Lewis
48.5k23167
$endgroup$
add a comment |
$begingroup$
Since $W subset V$ is a fixed subspace, I am going to drop the subscript "$W"$ from $P$; in this answer, $P = P_W$.
Recall that an orthogonal projection $P$ satisfies
$P^2 = P = P^T; tag 1$
thus for any
$x, y in V tag 2$
we have
$langle x, Py rangle = langle P^Tx, y rangle = langle Px, y rangle; tag 3$
now consider the three vectors $v$, $Pv$, and $w$;
$Pv, w in W Longrightarrow Pv - w in W, tag 4$
$langle v - Pv, w rangle = langle v, w rangle - langle Pv, w rangle = langle v, w rangle - langle v, Pw rangle = langle v, w rangle - langle v, w rangle = 0, tag 5$
where we have used (3) and the fact that $Pw = w$ for $w in W$; since this holds for every $w in W$, we have established that
$v - Pv in W^bot; tag 6$
we write
$v - w = (v - Pv) + (Pv - w), tag 7$
and by virtue of (4) and (6)
$langle v - Pv, Pv - w rangle = 0, tag 8$
whence
$Vert v - w Vert^2 = langle v - w, v - w rangle = langle (v - Pv) + (Pv - w), (v - Pv) + (Pv - w) rangle$
$= langle v - Pv, v - Pv rangle -2langle v - Pv, Pv - w rangle + langle Pv - w, Pv - w rangle$
$= Vert v - Pv Vert^2 + Vert Pv - w Vert^2;tag 9$
since each term on the right is non-negative, we find that
$Vert v - w Vert^2 ge Vert v - Pv Vert^2, tag{10}$
whence
$Vert v - w Vert ge Vert v - Pv Vert, ; forall w in W, tag{11}$
with equality holding precisely when
$Vert w - Pv Vert = 0 Longleftrightarrow w = Pv; tag{12}$
thus
$w ne Pv Longrightarrow Vert v - w Vert > Vert v - Pv Vert. tag{13}$
$OEDelta$.
Inequality of vector norms with projections
answered Mar 17 at 19:46
Robert LewisRobert Lewis
48.5k23167
$endgroup$
Since $W subset V$ is a fixed subspace, I am going to drop the subscript "$W"$ from $P$; in this answer, $P = P_W$.
Recall that an orthogonal projection $P$ satisfies
$P^2 = P = P^T; tag 1$
thus for any
$x, y in V tag 2$
we have
$langle x, Py rangle = langle P^Tx, y rangle = langle Px, y rangle; tag 3$
now consider the three vectors $v$, $Pv$, and $w$;
$Pv, w in W Longrightarrow Pv - w in W, tag 4$
$langle v - Pv, w rangle = langle v, w rangle - langle Pv, w rangle = langle v, w rangle - langle v, Pw rangle = langle v, w rangle - langle v, w rangle = 0, tag 5$
where we have used (3) and the fact that $Pw = w$ for $w in W$; since this holds for every $w in W$, we have established that
$v - Pv in W^bot; tag 6$
we write
$v - w = (v - Pv) + (Pv - w), tag 7$
and by virtue of (4) and (6)
$langle v - Pv, Pv - w rangle = 0, tag 8$
whence
$Vert v - w Vert^2 = langle v - w, v - w rangle = langle (v - Pv) + (Pv - w), (v - Pv) + (Pv - w) rangle$
$= langle v - Pv, v - Pv rangle -2langle v - Pv, Pv - w rangle + langle Pv - w, Pv - w rangle$
$= Vert v - Pv Vert^2 + Vert Pv - w Vert^2;tag 9$
since each term on the right is non-negative, we find that
$Vert v - w Vert^2 ge Vert v - Pv Vert^2, tag{10}$
whence
$Vert v - w Vert ge Vert v - Pv Vert, ; forall w in W, tag{11}$
with equality holding precisely when
$Vert w - Pv Vert = 0 Longleftrightarrow w = Pv; tag{12}$
thus
$w ne Pv Longrightarrow Vert v - w Vert > Vert v - Pv Vert. tag{13}$
$OEDelta$.
Inequality of vector norms with projections
answered Mar 17 at 19:46
Robert LewisRobert Lewis
48.5k23167
answered Mar 17 at 19:46
Robert LewisRobert Lewis
48.5k23167
answered Mar 17 at 19:46
Robert LewisRobert Lewis
48.5k23167
answered Mar 17 at 19:46
Robert LewisRobert Lewis
48.5k23167
48.5k23167
add a comment |
add a comment |
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$begingroup$
Pray tell, what is $w'$?
$endgroup$
– Robert Lewis
Mar 16 at 17:49
1
$begingroup$
My bad, I meant $w$
$endgroup$
– Noé Duarte González
Mar 16 at 17:55
$begingroup$
Thanks for the correction!
$endgroup$
– Robert Lewis
Mar 16 at 18:06