document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=27384>Question 27384</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let V = P3(Q) and W = { f(x) 2 V | f(x) = f(&#8722;x + 1) }. Find a basis for W.\r<BR>\n" );
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document.write( "Let V and W be same as above. Find a basis for V that contains a basis for W.\r<BR>\n" );
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document.write( "I am totally confused.  I really need some help here.\r<BR>\n" );
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document.write( "Thanks. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #15002 by </B><A HREF=/tutors/aboutme.mpl?userid=khwang><B>khwang(438)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=khwang><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=15002\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let V = P3(Q) and W = { f(x) 2 V | f(x) = f(&#8722;x + 1) }. Find a basis for W. <BR>\n" );
document.write( "Let V and W be same as above. Find a basis for V that contains a basis for W. <BR>\n" );
document.write( "[Usually, P3(Q) means the vector space of polynomials of deg <= 3 in Q[x])\r<BR>\n" );
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document.write( "Sol: If f(x) =<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ax%5E3%2Bbx%5E2%2Bcx+%2B+d ALIGN=MIDDLE  ALT=\"ax%5E3%2Bbx%5E2%2Bcx+%2B+d\"   DISABLED_style= \"max-width:90%; height:auto;\" > belongs to W , then f(x) = f(-x+1) implies <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ax%5E3%2Bbx%5E2%2Bcx+%2B+d+=+a%28-x%2B1%29%5E3%2Bb%28-x%2B1%29%5E2%2Bc%28-x%2B1%29+%2B+d+ ALIGN=MIDDLE  ALT=\"ax%5E3%2Bbx%5E2%2Bcx+%2B+d+=+a%28-x%2B1%29%5E3%2Bb%28-x%2B1%29%5E2%2Bc%28-x%2B1%29+%2B+d+\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ax%5E3%2Bbx%5E2%2Bcx+%2B+d+ ALIGN=MIDDLE  ALT=\"ax%5E3%2Bbx%5E2%2Bcx+%2B+d+\"   DISABLED_style= \"max-width:90%; height:auto;\" >= <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%28-x%5E3%2B+3x%5E2+-+3x+-1%29%2Bb%28x%5E2-2x%2B1%29%2B+c%28-x%2B1%29+%2B+d ALIGN=MIDDLE  ALT=\"a%28-x%5E3%2B+3x%5E2+-+3x+-1%29%2Bb%28x%5E2-2x%2B1%29%2B+c%28-x%2B1%29+%2B+d\"   DISABLED_style= \"max-width:90%; height:auto;\" >= <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-ax%5E3%2B+%283a%2Bb%29x%5E2%2B%28-3a-2b-c%29x+-+a+%2B+b+%2B+c+%2B+d+ ALIGN=MIDDLE  ALT=\"-ax%5E3%2B+%283a%2Bb%29x%5E2%2B%28-3a-2b-c%29x+-+a+%2B+b+%2B+c+%2B+d+\"   DISABLED_style= \"max-width:90%; height:auto;\" > <BR>\n" );
document.write( "    By comparing the coefficients, we have a=-a,  3a+b = b, -3a-2b-c=c ,    d= –a+b+c+d. Hence, a= 0, b+c = 0. We see that W = { <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%28x%5E2-+x%29+%2B+d+ ALIGN=MIDDLE  ALT=\"b%28x%5E2-+x%29+%2B+d+\"   DISABLED_style= \"max-width:90%; height:auto;\" > | b,d in Q} (generated by <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2-x ALIGN=MIDDLE  ALT=\"x%5E2-x\"   DISABLED_style= \"max-width:90%; height:auto;\" > & 1and so dim W = 2. By choosing B’ = {1, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2-x ALIGN=MIDDLE  ALT=\"x%5E2-x\"   DISABLED_style= \"max-width:90%; height:auto;\" > } we get a basis of W. The adjoining two independent vectors <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E3+ ALIGN=MIDDLE  ALT=\"x%5E3+\"   DISABLED_style= \"max-width:90%; height:auto;\" >and x (not in W) to B', then we can get a basis B= {1, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2-x ALIGN=MIDDLE  ALT=\"x%5E2-x\"   DISABLED_style= \"max-width:90%; height:auto;\" >, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E3 ALIGN=MIDDLE  ALT=\"x%5E3\"   DISABLED_style= \"max-width:90%; height:auto;\" >, x} of V, which contains a basis B’ for W.\r<BR>\n" );
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document.write( " Kenny </SPAN>\n" );
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