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Problem 1. Suppose V is a linear space over R, with basis e1 e2 e3 . U, W is two subspace of V. For each of the following problem, Find a basis of U ∩ V (1) U = {ae1 + be2 + ce3|a + b + c = 0} , V = {pe1 + se2 + ce3|3p + 2s + c = 0} (2) U = {ae1 + be2 + ce3|2a + b + 3c = 0} , V = {be1 + ae2 + ce3|4b + 2a + 6c = 0} (Caution: Do some substitution to make sure the scalar in front of each vector of basis are the same) Problem 2. Suppose V is a linear space over R, with basis e1 e2 e3 e4 . U, W is two subspace of V. For each of the following problem, Find a basis of U + V (1) U = {ae1+be2+ce3+de4|a+b+c = 0; a+b = 0} , V = {ae1+be2+ce3|2a+2b+c = 0; c = 0} Problem 3. Suppose V is a linear space over R, with basis e1 e2 e3 e4 ,T : V −→ V is a linear transformation, For each of the following case, determine whether the subspace W is an invariant subspace of T. I’ll show an example over here. Example: T e1 e2 e3 e4 = e1 e2 e3 e4 2 4 5 −9 0 9 5 −12 0 8 −1 −5 1 0 1 0 W = {ae1 + be2 + ce3 + de4|a + b + c + d = 0} We just want to show that TW ⊂ W. We see, we like the matrix form, rewrite W = {ae1 + be2 + ce3 + de4| 1 1 1 1 a b c d = 0} Suppose v = e1 e2 e3 e4 a b c d , then (2) U = {ae1 + be2 + ce3|a + b + 3c = 0} , V = {ae1 + be2 + ce3|4b + 2a + Tv = T e1 e2 e3 e4 a b c d = e1 e2 e3 e4 2 4 5 −9 0 9 5 −12 0 8 −1 −5 1 0 1 0 a b c d We want to see the final vector is in W or not, we discover the coordinate of this final vector is 2 4 5 −11 −2 9 5 −12 0 6 −1 −5 1 0 −1 0 a b c d To see whether this vector is in W or not, we just test left multiply by 1 1 1 1 , and we see 1 1 1 1 2 4 5 −9 0 9 5 −12 0 8 −1 −5 1 0 1 0 a b c d = 2 2 2 2 a b c d = 2 1 1 1 1 a b c d = 2 ∗ 0 = 0 The last equal is true because the vector is already taken in W, and the coordinate of vector in W satisfies this equality. This means Tv ∈ W. so indeed, W is an invariant subspace. Determine whether the following are invariant subspaces of T. (1) T e1 e2 e3 e4 = e1 e2 e3 e4 5 6 3 −9 2 7 0 −12 0 5 0 9 −2 3 0 0 W = {ae1 + be2 + ce3 + de4|3a + 2b + c (2) T e1 e2 e3 e4 = e1 e2 e3 e4 9 0 −10 −9 3 −1 −3 −12 11 −1 −2 9 10 −1 0 0 W = {ae1 + be2 + ce3 + de4|a + 10b + c = 0} 4 Problem 4. Suppose V is 4-dimensional space over R, with chosen basis e1 e2 e3 e4 , T is a linear transformation T : V −→ V , and have the matrix representation: T e1 e2 e3 e4 = e1 e2 e3 e4 1 −2 0 1 1 −2 0 1 0 0 1 0 1 −2 0 1 (1) Find a basis of Im T. (2) We learned in class that image and kernel are all invariant subspace for linear transformation. With your basis for invariant subspace Im T you found above, find the matrix representation of restriction T|Im T T|Im T : Im T −→ Im T (3) Find a basis of Im T2 What is the dimension of Im T2? (4) With your basis for invariant subspace Im T2 , find the matrix representation of restriction of T|Im T2 T|Im T2 : Im T2 −→ Im T2 (5) What is the dimension of Im T3? Does the map T|Im T2 : Im T2 −→ Im T2 an isomorphism? (6) Draw the power-rank curve for T, (means the graph with x axis the power of T, and y axis with Problem 5. Suppose T : V −→ V is a linear map, and dim V = 5 dim Im(T) = 3, T 3 = 0. Without using the result in the class. Analise the following. (1) What is the dimension of Ker(T 2 ) ∩ Im(T)?(Does everything in Image of T killed by T 2?) (2) What is the dimension of Ker(T)? Why?(Using dim(ker) + dim(im) = dim(domain)) (3) What is the dimension of ker(T)∩Im(T) at most be? (it at most be the dimension of kernel of T) (4) In the above, we know the dimension of ker(T) ∩ Im(T) could be 0,1,2. Now we assume if the dimension of ker(T) ∩ Im(T) is 1, then what is the dimension of Im(T 2 )? (5) With assumption in (4), if ker(T) ∩ Im(T) is 1, then what is the dimension of ker(T) ∩ Im(T 2 ) at most be? why? (ker(T) ∩ Im(T 2 ) is a subspace of ker(T) ∩ Im(T)) (6) Continue assumption in (4), if ker(T)∩Im(T 2 ) at most 1, what is the dimension of Im(T 3 ) at least be? why? but we already assumed T 3 = 0, show that then the assumption in (4) is wrong. (7) Analise what will happen if ker(T) ∩ Im(T) = 0?(Hint, then restriction on Im T would be isomorphism. Isomorphism all the way.) (8) With above analysis, we know the only case is dim(Ker(T) ∩ Im(T)) = 2, then find dim(Im(T 2 )) Problem 6. Suppose T : V −→ V is a linear map, and dim V = 6, dim Im(T 3 ) = 3, dim Im(T 5 ) = 2Using the non-increasing and non-accelerating property of the rank curve we learned in class, answer the following question. (1) Draw the power-rank curve, fits other power-rank, what is the rank-stable number? (2) Look at your curve, quick answer the following question. (Sometimes the formula dim(ker)+ dim(Im) = dim(Domain) is super useful!) 1 What is the dimension of ImT ∩ KerT2? 2 What is the dimension of ImT2 ∩ KerT3? 3 What is the dimension of ImT ∩ KerT3? 4 For what number n does the restriction of linear transformation T on ImT n is an isomorphism? Problem 7. Suppose A is a 10 × 10 square matrix, we know rank(A3 ) = 4, A6 = 0, Draw the power-rank curve, fits other ranks, and for each rank say reasons why you fits there. Then find rank(A), rank(A2 ), rank(A4 ), rank(A5 ) 9 Problem 8. Suppose T,R is two commute linear transformation of linear space V, that is, T R = RT, then show that kerR and ImR are invariant subspaces for T.

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