document.write( "Question 8159: is there any linear transformation from R^3 to R^2 defined by
\n" ); document.write( "T(1,-1,1)=(1,0)
\n" ); document.write( "T(1,1,1)=(0,1) \n" ); document.write( "
Algebra.Com's Answer #4491 by khwang(438)\"\" \"About 
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Yes, since (1,-1,1) and (1,1,1) are linearly independent in R^3
\n" ); document.write( " Also, (1,0) and (0,1) are linearly independent in R^2 can form a basis.\r
\n" ); document.write( "\n" ); document.write( " We know that any linear transformation is uniquely determined by
\n" ); document.write( " the the values on the basis.
\n" ); document.write( " Set another vector v in R^3, which is independent of (1,-1,1) and (1,1,1),
\n" ); document.write( " say (0, 0,1) , then define T(0,0,1)= (0,0) [or choose v = (1,-1,1)x(1,1,1)\r
\n" ); document.write( "\n" ); document.write( " We obtain a linear transformation from R^3 to R^2 generated by
\n" ); document.write( " T(1,-1,1)=(1,0)
\n" ); document.write( " T(1,1,1)=(0,1) with Kernel(T) = <(0,0,1)>
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\n" ); document.write( " More precisely, T(a(1,-1,1)+b(1,1,1)+c(0,0,1))
\n" ); document.write( " = a(1,0)+ b(0,1)= (a,b) for all real a,b,c\r
\n" ); document.write( "\n" ); document.write( " Kenny
\n" ); document.write( " PS: Important notice:
\n" ); document.write( " I will not solve your or other student's questions with repeated posting. \r
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