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is"></A><FONT=20
color=3D#ff0000>Theorem: If <I>A</I> is an m x n matrix, =
then there is=20
an orthonormal basis of <IMG height=3D17 =
alt=3D[Graphics:svdgr22.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr22.gif" =
width=3D19=20
align=3DABSCENTER>, <IMG height=3D17 =
alt=3D[Graphics:svdgr23.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr23.gif" =
width=3D61=20
align=3DABSCENTER> so that for all <IMG height=3D17=20
alt=3D[Graphics:svdgr24.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr24.gif" =
width=3D28=20
align=3DABSCENTER>, <IMG height=3D19 =
alt=3D[Graphics:svdgr25.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr25.gif" =
width=3D73=20
align=3DABSCENTER>.</FONT></H3>
<H3><FONT color=3D#3333ff>Proof 1: For <IMG height=3D17=20
alt=3D[Graphics:svdgr26.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr26.gif" =
width=3D28=20
align=3DABSCENTER> matrices</FONT></H3>You can get 2D perpframes=20
using <IMG height=3D32 alt=3D[Graphics:svdgr27.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr27.gif" =
width=3D189=20
align=3DABSCENTER> and specifying an angle <IMG height=3D29=20
alt=3D[Graphics:svdgr28.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr28.gif" =
width=3D56=20
align=3DABSCENTER>. =20
<P>You'd like to show that for any m x 2 matrix there's an angle t =
so=20
that <BR><IMG height=3D17 alt=3D[Graphics:svdgr29.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr29.gif" =
width=3D143=20
align=3DABSCENTER> is zero. =20
<P>How do you know there is always such a <IMG height=3D17=20
alt=3D[Graphics:svdgr30.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr30.gif" =
width=3D11=20
align=3DABSCENTER>? =20
<P><B><FONT color=3D#3333ff>Answer</FONT></B>: Go with the =
example=20
matrix <IMG height=3D28 alt=3D[Graphics:svdgr31.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr31.gif" =
width=3D77=20
align=3DABSCENTER> and look at the following plot that shows the=20
perpframe <IMG height=3D17 alt=3D[Graphics:svdgr32.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr32.gif" =
width=3D27=20
align=3DABSCENTER> and <IMG height=3D17 =
alt=3D[Graphics:svdgr33.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr33.gif" =
width=3D27=20
align=3DABSCENTER> before and after the hit with A. =20
<CENTER>
<H4><IMG height=3D135=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/tofrom.gif"=20
width=3D288></H4></CENTER>Notice that when <IMG height=3D17=20
alt=3D[Graphics:svdgr34.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr34.gif" =
width=3D32=20
align=3DABSCENTER>, the angle between <IMG height=3D17=20
alt=3D[Graphics:svdgr35.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr35.gif" =
width=3D48=20
align=3DABSCENTER>and <IMG height=3D17 =
alt=3D[Graphics:svdgr36.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr36.gif" =
width=3D48=20
align=3DABSCENTER> is greater than <IMG height=3D30=20
alt=3D[Graphics:svdgr37.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr37.gif" =
width=3D26=20
align=3DABSCENTER>. =20
<P>By the time <IMG height=3D30 alt=3D[Graphics:svdgr38.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr38.gif" =
width=3D89=20
align=3DABSCENTER>, the angle is less than <IMG height=3D30=20
alt=3D[Graphics:svdgr39.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr39.gif" =
width=3D26=20
align=3DABSCENTER>. =20
<P>That's enough to guarantee that somewhere between <IMG =
height=3D17=20
alt=3D[Graphics:svdgr40.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr40.gif" =
width=3D32=20
align=3DABSCENTER> and <IMG height=3D30 =
alt=3D[Graphics:svdgr41.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr41.gif" =
width=3D47=20
align=3DABSCENTER> there's an angle where <IMG height=3D17=20
alt=3D[Graphics:svdgr42.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr42.gif" =
width=3D48=20
align=3DABSCENTER>and <IMG height=3D17 =
alt=3D[Graphics:svdgr43.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr43.gif" =
width=3D48=20
align=3DABSCENTER>are perpendicular. =20
<P>Most reasonable folks would accept the evidence given in the =
plots, but=20
the doubters might want to look at the function: =20
<CENTER><IMG height=3D32 alt=3D[Graphics:svdgr44.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr44.gif" =
width=3D188=20
align=3DABSCENTER> </CENTER>
<CENTER> </CENTER>
<P>Then <IMG height=3D32 alt=3D[Graphics:svdgr45.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr45.gif" =
width=3D276=20
align=3DABSCENTER> =20
<P>(1) The definition of f. =20
<P>(2) Linearity of matrix hits. =20
<P>(3) The definition of f. <BR> =20
<P>This equation says that <IMG height=3D17 =
alt=3D[Graphics:svdgr46.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr46.gif" =
width=3D32=20
align=3DABSCENTER>and <IMG height=3D28 =
alt=3D[Graphics:svdgr47.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr47.gif" =
width=3D41=20
align=3DABSCENTER>have opposite signs. This should convince even =
the=20
doubters that there's a <IMG height=3D17 =
alt=3D[Graphics:svdgr48.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr48.gif" =
width=3D11=20
align=3DABSCENTER> between <IMG height=3D17 =
alt=3D[Graphics:svdgr49.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr49.gif" =
width=3D11=20
align=3DABSCENTER> and <IMG height=3D28 =
alt=3D[Graphics:svdgr50.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr50.gif" =
width=3D20=20
align=3DABSCENTER> that makes <IMG height=3D17 =
alt=3D[Graphics:svdgr51.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr51.gif" =
width=3D32=20
align=3DABSCENTER> zero. =20
<P>The same argument holds for any m by 2 matrix. =20
<P>
<HR width=3D"100%">
<H3><FONT color=3D#3333ff>Proof 2: General case</FONT></H3>The =
proof above=20
works well enough for m x 2 matrices, but this proof shows that =
for all m=20
x n matrices there's is a perpendicular frame <IMG =
height=3D17=20
alt=3D[Graphics:svdgr52.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr52.gif" =
width=3D143=20
align=3DABSCENTER> of unit vectors so that <IMG height=3D18=20
alt=3D[Graphics:svdgr53.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr53.gif" =
width=3D82=20
align=3DABSCENTER> for <IMG height=3D17 =
alt=3D[Graphics:svdgr54.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr54.gif" =
width=3D32=20
align=3DABSCENTER>. =20
<P>Here's one way to get such a perpframe. =20
<P>Start with a m x n matrix <I>A</I>. =20
<P>Step 1: Let <IMG height=3D17 alt=3D[Graphics:svdgr55.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr55.gif" =
width=3D17=20
align=3DABSCENTER> be a unit vector maximizing <IMG =
height=3D17=20
alt=3D[Graphics:svdgr56.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr56.gif" =
width=3D49=20
align=3DABSCENTER>. =20
<P>Step 2: Let <IMG height=3D17 alt=3D[Graphics:svdgr57.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr57.gif" =
width=3D17=20
align=3DABSCENTER> be the space perpendicular to <IMG =
height=3D17=20
alt=3D[Graphics:svdgr58.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr58.gif" =
width=3D62=20
align=3DABSCENTER> <BR>Let <IMG height=3D17=20
alt=3D[Graphics:svdgr59.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr59.gif" =
width=3D17=20
align=3DABSCENTER> be a unit vector in <IMG height=3D17=20
alt=3D[Graphics:svdgr60.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr60.gif" =
width=3D17=20
align=3DABSCENTER> maximizing <IMG height=3D17 =
alt=3D[Graphics:svdgr61.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr61.gif" =
width=3D49=20
align=3DABSCENTER>. =20
<P>Step 3: Let <IMG height=3D17 alt=3D[Graphics:svdgr62.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr62.gif" =
width=3D17=20
align=3DABSCENTER> be the space perpendicular to <IMG =
height=3D17=20
alt=3D[Graphics:svdgr63.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr63.gif" =
width=3D86=20
align=3DABSCENTER> <BR>Let <IMG height=3D17=20
alt=3D[Graphics:svdgr64.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr64.gif" =
width=3D17=20
align=3DABSCENTER> be a unit vector in <IMG height=3D17=20
alt=3D[Graphics:svdgr65.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr65.gif" =
width=3D17=20
align=3DABSCENTER> maximizing <IMG height=3D17 =
alt=3D[Graphics:svdgr66.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr66.gif" =
width=3D49=20
align=3DABSCENTER>. <BR> . <BR> . =20
<BR> . <BR>Step n: Let <IMG height=3D17=20
alt=3D[Graphics:svdgr67.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr67.gif" =
width=3D17=20
align=3DABSCENTER> be the space perpendicular to <IMG =
height=3D17=20
alt=3D[Graphics:svdgr68.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr68.gif" =
width=3D129=20
align=3DABSCENTER> <BR>Let <IMG height=3D17=20
alt=3D[Graphics:svdgr69.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr69.gif" =
width=3D17=20
align=3DABSCENTER> be a unit vector in <IMG height=3D17=20
alt=3D[Graphics:svdgr70.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr70.gif" =
width=3D17=20
align=3DABSCENTER> maximizing <IMG height=3D17 =
alt=3D[Graphics:svdgr71.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr71.gif" =
width=3D49=20
align=3DABSCENTER>. <BR> <BR> =20
<P>This gets you a perpendicular frame of unit vectors <IMG =
height=3D17=20
alt=3D[Graphics:svdgr72.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr72.gif" =
width=3D143=20
align=3DABSCENTER>, but how do you know that <IMG height=3D18 =
alt=3D[Graphics:svdgr73.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr73.gif" =
width=3D82=20
align=3DABSCENTER> for <IMG height=3D17 =
alt=3D[Graphics:svdgr74.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr74.gif" =
width=3D32=20
align=3DABSCENTER>? <BR> =20
<P>Look at <IMG height=3D18 alt=3D[Graphics:svdgr75.gif]=20
src=3D"http://www.uwlax.edu/faculty/will/svd/svd/svdgr75.gif" =
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