Worksheet 3 - Subspaces and InversesFebruary 2, 20101. Determine whether each of the following is a basis for R3.(a)11−2−5−1270−5(b)1−6−73−47−2750892. Mark each statement true or false. Make sure you know why.(a) A subset H of Rnis a subspace if the zero vector is in H.(b) Given vectors ~v1, . . . , ~vpin Rn, the set of all linear combinations of these vectorsis a subspace of Rn.(c) The null space of an m × n matrix is a subspace of Rn.(d) The column space of a matrix A is the set of solutions of A~x =~b.(e) If B is an echelon form of a matrix A, then the pivot columns of B form a basisfor Col A.(f) If B is a basis for the subspace H, then each vector in H can be written in onlyone way as a linear combination of the vectors in B.(g) The dimension of Nul A is the number of variables in the equation A~x =~0.(h) The dimension of the column space of A is rank A.(i) If H is a p-dimensional subspace of Rn, then a linearly independent set of p vectorsin H is a basis for H.3. Here A is a matrix and its echelon form is shown. Find bases for the null space andcolumn space of A.3 −1 7 3 9−2 2 −2 7 5−5 9 3 3 4−2 6 6 3 7∼3 −1 7 0 60 2 4 0 30 0 0 1 10 0 0 0 04. If R is a 6 × 6 matrix and Nul R is not the zero subspace, what can you say about ColR?5. If P is a 5× 5 matrix and Nul P is the zero subspace, what can you say about solutionsof equations of the form P ~x =~b.6. What is the rank of a 4 × 5 matrix whose null space is three-dimensional?1Part 2 - Invertible Matrix Theorem1. Show that if AB is invertible, so if B.2. Suppose A is an n × n matrix with the property that the equation A~x =~0 has only thetrivial solution. Without using the Invertible Matrix Theorem, explain directly whythe equation A~x =~b must have a solution for each~b in Rn.3. Let T (x1, x2) = (6x1−8x2, −5x1+7x2) be a linear transformation. Show T is invertibleand find a formula for T−1.4. Suppose T and U are linear transformations from Rnto Rnsuch that T (U(~x)) = ~xfor all ~x ∈ Rn. Is it true that U(T (~x)) = ~x for all ~x ∈ Rn? Why or why not?5. If n × n matrices E and F have the property that EF = I, then E and F commute.Explain
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