document.write( "Question 15128: Let T be a linear operator on V.\r
\n" ); document.write( "\n" ); document.write( "Tv1= -v1-v2 and Tv2= v1-3v2 where (v1,v2) is a basis for V.
\n" ); document.write( "a) Find the characteristic and minimal polynomial.
\n" ); document.write( "b) Find all the linearly independent eigenvectors.
\n" ); document.write( "c) Find a jordan basis for T and exhibit the jordan form. \n" ); document.write( "
Algebra.Com's Answer #7524 by khwang(438)\"\" \"About 
You can put this solution on YOUR website!
I really cannot believe that anybody would ask questions about Jordan form,
\n" ); document.write( " since I know that nodbody knows much about linear algebra here.\r
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\n" ); document.write( "Sol: a) T(1, 0) = (-1,-1), T(0,1) = (1,-3)
\n" ); document.write( "(-v1 –v2) (-1 ,–1) (v1)
\n" ); document.write( "(v1 –3v2) (1 ,- 3) (v2)\r
\n" ); document.write( "\n" ); document.write( " The matrix representation [T}B as (where B = {v1,v2} )
\n" ); document.write( " (-1 ,1)
\n" ); document.write( " (1, -3)\r
\n" ); document.write( "\n" ); document.write( "char(T) = det(xI –T) = (x+1)(x+3) + 1 = \"x%5E2+%2B4x+%2B4+=%28x%2B2%29%5E2+\",
\n" ); document.write( "there is one double root: -2
\n" ); document.write( "So, the only eigenvalue of T is -2.
\n" ); document.write( "Since T + 2I =
\n" ); document.write( "(1 ,-1)
\n" ); document.write( "(1, -1) is not the zero matrix.
\n" ); document.write( "And we know that the minimal polynomial min(T) is a factor of char(T) so,
\n" ); document.write( "min(T) = char(T) = \"%28x%2B2%29%5E2+\". \r
\n" ); document.write( "\n" ); document.write( "b) To find the eigenvectors(eigenspace) for the eigenvalue -2,we have to solve Tx = -2 x for nonzero vectors x.
\n" ); document.write( "That is, for the null space N(T+2I) of T+2I as:
\n" ); document.write( "Solve equations:
\n" ); document.write( "(1 –1) (x) (0)
\n" ); document.write( "(1 –1) (y) = (0), we see that x - y = 0 .\r
\n" ); document.write( "\n" ); document.write( " Hence, the eigenspace for the unique egenvalue–2 is <(1,1)>= {c(1,1) | c in R} (one dim. subspace generated by (1,1)). That is, the eigenspace has a basis {(1,1)}\r
\n" ); document.write( "\n" ); document.write( "c) Since {(1,1)} is a basis of the N(T+2I) and T+2I is not the zero vector.
\n" ); document.write( "Find a vector (x,y) such that (T + 2 I) (x, y) = (1,1). That is, x-y =1. Say, choose (1,0). Then, {(1,0), (1,1)} can form a Jordan basis for T and the corresponding Jordan form as
\n" ); document.write( "(-2 0)
\n" ); document.write( "(1 –2).
\n" ); document.write( "[Check: T(1,0) = (-2,1) = -2(1,0) + (0,1) and T(1,1) = -2(0,1) correct.] \r
\n" ); document.write( "\n" ); document.write( " Try to read carefully about every details and study hard.
\n" ); document.write( " I won't do too much for lazy students.\r
\n" ); document.write( "\n" ); document.write( " Kenny
\n" ); document.write( " PS.
\n" ); document.write( " By the way, since min(T) = \"%28x%2B2%29%5E2+\", we claim that {v, (T+2I)v} can be a Jordan basis for T if v is a vector such that (T+2I)v being non-zero.(i.e v is not in Null(T+2T)) (Why ?)
\n" ); document.write( "This shows there are (infinitely) many Jordan basis for T.
\n" ); document.write( "In fact, for this example, any v = (a,b) with a <> b ,
\n" ); document.write( "then {v, (1,1)} can be a Jordan basis for T.
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