document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=15391>Question 15391</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose V is an n dimensional vector space and that T is an operator on V. T has n distinct eigenvalues. Also suppose that S is also an operator on V which has the same eigenvectors as T ( not necessarily with the same eigenvalues).<BR>\n" );
document.write( "Prove that ST = TS </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #7649 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=7649\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  Suppose V is an n dimensional vector space and that T is an operator on V. T has n distinct eigenvalues. Also suppose that S is also an operator on V which has the same eigenvectors as T ( not necessarily with the same eigenvalues).<BR>\n" );
document.write( "Prove that ST = TS\r<BR>\n" );
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document.write( "Proof: If T has n distinct eigenvalues  {&#955;i| i=1,2,..,n}(i.e. <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=lambda%5Bi%5D ALIGN=MIDDLE  ALT=\"lambda%5Bi%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" > ) with corresponding eigenvectors {vi|i=1,2,..n}, <BR>\n" );
document.write( "then {vi|i=1,2,..n} is linear independent and so forms a basis of V. <BR>\n" );
document.write( "For each i, vi is an eigenvector of S means there exists scalar &#946;i <BR>\n" );
document.write( "such that Svi = &#946;i vi.<BR>\n" );
document.write( "Consider ST (vi) = S(&#955;i vi ) = &#955;i S(vi) = &#955;i&#946;i vi and <BR>\n" );
document.write( "TS (vi ) = TS (&#946;i vi ) = &#946;i T(vi) =&#955;i&#946;i vi for every i=1,2,..n.<BR>\n" );
document.write( "This shows ST = TS (since {vi|i=1,2,..n} is a basis of V).  \r<BR>\n" );
document.write( "\n" );
document.write( "Fact. The eigenvectors {vi|i=1,2,..,n} of T is linearly independent <BR>\n" );
document.write( " Proof: If <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=SIGMA ALIGN=MIDDLE  ALT=\"SIGMA\"   DISABLED_style= \"max-width:90%; height:auto;\" > <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=c%5Bi%5D+v%5Bi%5D+=+0+ ALIGN=MIDDLE  ALT=\"c%5Bi%5D+v%5Bi%5D+=+0+\"   DISABLED_style= \"max-width:90%; height:auto;\" > for some scalars ci, <BR>\n" );
document.write( "set L to be the product operator<BR>\n" );
document.write( " II(T –&#955;j I) (j=1,2,..,n-1 ). Then, we have L(&#931;ci vi) =&#931;ci L(vi) <BR>\n" );
document.write( "= &#931;ci (&#955;i vi -&#955;i vi) (i=1,2,..,n-1) + cn II(&#955;n -&#955;i )vn<BR>\n" );
document.write( "= cn II(&#955;n -&#955;i )vn  = 0 .<BR>\n" );
document.write( "Since all {&#955;i} are distinct, II(&#955;n -&#955;i ) <> 0. (i=1,2,..,n-1) <BR>\n" );
document.write( "We obtain cn = 0. Similarly, we can prove that all other scalars ci are zeros. This shows that {vi} are independent. \r<BR>\n" );
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document.write( " Kenny<BR>\n" );
document.write( " PS. I don't quite believe that some questions like this would be posted<BR>\n" );
document.write( " here. </SPAN>\n" );
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