Linear Algebra

396 Chapter Five. Similarityis (x − 3) 2 and so T has only the single eigenvalue 3. Thus for( )−1 −1T − 3I =1 1the only eigenvalue is 0 and T −3I is nilpotent. Finding the null spaces is routine;to ease this computation we take T to represent a transformation t: C 2 → C 2with respect to the standard basis (we shall do this for the rest of the chapter).( )−y ∣∣N (t − 3) = { y ∈ C} N ((t − 3) 2 ) = C 2yThe dimensions of these null spaces show that the action of the map t − 3 on astring basis is ⃗β 1 ↦→ ⃗β 2 ↦→ ⃗0. Thus, here is the canonical form for t − 3 with onechoice for a string basis.( )0 0Rep B,B (t − 3) = N =1 0By Lemma 2.3, T is similar to this matrix.( )3 0Rep B,B (t) = N + 3I =1 3( ) ( )1 −2B = 〈 , 〉1 2We can produce the similarity computation. Recall how to find the change ofbasis matrices P and P −1 to express N as P(T − 3I)P −1 . The similarity diagramC 2 wrt E 2⏐id↓Pt−3−−−−→T −3IC 2 wrt E 2⏐id↓PC 2 wrt Bt−3−−−−→NC 2 wrt Bdescribes that to move from the lower left to the upper left we multiply byP −1 = ( Rep E2 ,B(id) ) ( )−1 1 −2= RepB,E2 (id) =1 2and to move from the upper right to the lower right we multiply by this matrix.( ) −1 ( )1 −2 1/2 1/2P ==1 2 −1/4 1/4So this equation expresses the similarity.( ) ( ) (3 0 1/2 1/2 2 −1=1 3 −1/4 1/4 1 4) ()1 −21 2