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1 V !V such that R(T 1) \N(T 1) = f0gbut V is not a direct sum of R(T 1) and N(T 1) Solution (a)In this case, R(T) = V, so R(T) N(T) is a subspace of V containing V, hence V = R(T) N(T) However, R(T) \N(T) = N(T) 6= f0g, so V is not the direct sum of R(T) and N(T) (b)We take T 1 = Uas de ned in Problem 7 In this case, N(U) = f0g, so RX m s ̕ A ~ ł B T b V A h A G N X e A Ȃǂ̔̔ 烊 t H ̂ k Ⴂ ܂ B 傫 Ȓn } Ō ݒn F @ É x m { s R {48肷 ̎ז ɂȂ Ȃ ߍ ݎ B ̃T j ^ { b N X ̓T C Y 傫 ז ɂȂ Ă ܂ B 萠 ݒu ꍇ A z _ ז ɂȂ 萠 ݂ɂ 肪 ܂ B ̖ ߍ ^ Ƃ 邱 Ƃʼn ܂ B t b V o u { b N X



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B ∈ R2 Clearly T a 0 b 0 = a b , so the range of T is all of R2 Thus the dimension of ran(T) is two ♠ ⋄ Example 22(b) T P1 → P1 defined by T(ax b) = 2bx− a is linear Describe its kernel and range and give the dimension of each T(ax b) = 2bx− a= 0 if, and only if, both a and b are zero Therefore the kernel of T is onlyThe standard matrix T and the matrix of Twith respect to B, T B, are related by TS= ST Band T = ST BS 1 and T B= S 1TS where here S= ~v 1 j j ~v n is the change of basis matrix of the basis In order to understand this relationship better, it is convenient to take it as a de nition and then study it abstractly Given two nT o o s t e t a t e e t a e a t s t t s n n t t t v t t v t d t v v t v t t v t t v t t v v s r l l m a r c i a s t t t t r t v d v v t t t v v t v v t l v d t r t t




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T(~v) B= T B~v Bfor all ~v2V (4) The matrix of T in the basis Band its matrix in the basis Care related by the formula T C= P C BT BP1 C B (5) We see that the matrices of Tin two di erent bases are similar In particular, if V = Rn, Cis the canonical basis of Rn (given by the columns of the n nidentity matrix), T is the matrix0 ' ( 2 3 4 5 4 6 7 8 9 & ' !ZOZOTOWN CALIFORNIA OUTFITTERS i J t H j A A E g t B b ^ Y j ̃s A X i p j i V o n j ȂǖL x Ɏ 葵 t @ b V ʔ̃T C g ł B t v X ^ b Y A h b v ^ C v ̃s A X ȂǁA ԃA C e ŐV g h A C e ܂ŃI C ł w ܂ B V A C e ג I
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