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Inequality of vector norms with projections



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1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 16 at 17:54







Noé Duarte González

















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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    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






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      active

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      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 16 at 18:14









          colt_browningcolt_browning

          768110




          768110























              0












              $begingroup$

              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






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                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






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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






                  share|cite|improve this answer









                  $endgroup$



                  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







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 17 at 19:46









                  Robert LewisRobert Lewis

                  48.5k23167




                  48.5k23167






























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                      Skildring Skildring Tokke komm.svg Norsk nynorsk: Våpen for Tokke kommune i Telemark fylke, godkjent ved kongeleg resolusjon 6. februar 1987. Motiv: På gull grunn ein gåande svart bjørn. [1] Tokke ligg i eit gamalt bjørneområde og bjørnen er eit viktig innslag i fleire lokale historier. Bjørnen kan òg symbolisere naturen og dyrelivet i kommunen. English: Coat of arms of the Norwegian municipality of Tokke, Telemark. Dato 6. februar 1987 Kjelde Original by Stein Davidsen, based upon image from caplex.no. Opphavsperson Eige arbeid Vektorgrafikken vart laga med Inkscape. Løyve (Gjenbruk av denne fila) Public domain Public domain false false This official Norwegian coat of arms is ineligible for copyright, but its usage is restricted according to Norwegian penal law, independent of the copyright status of the depiction shown here. Dette biletet er sett saman av eit flagg, eit våpen, eit sigill eller eit anna offisielt insignium . Bruken av slike symbol er avgrensa i fleire land. Desse r
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