Practice Final Exam. Solutions.

Transcription

1 MATH Applied Linear Algebra December 6, 8 Practice Final Exam Solutions Find the standard matrix f the linear transfmation T : R R such that T, T, T Solution: Easy to see that the transfmation T can be represented by a matrix A True False If T : R R rotates vects about the igin though an angle π/, then T is a linear transfmation Explain Solution: True, since the transfmation T can be represented by a matrix A cos π sin π sin π cos π Find the eigenvalues and the eigenvects of the matrix Solution: First we find the characteristic polynomial and make it equal to zero, det 5 λ λ 6 7 λ λdet 5 λ 6 λ λλ 7λ + 4 λ[5 λ λ 8] Thus we have three distinct eigenvalues λ, λ 7, and λ To find the eigenvect cresponding to λ, we need to find a nontrivial solution to a homogeneous system A Iv x x x

2 Since we have that x -free and x x x Thus the first eigenvect is x 4 x 4 x x x 4 x x v 6 8 x x To find the second eigenvect cresponding to λ 7, we need to find a nontrivial solution to a homogeneous system A 7Iv Since we have that x and x -free and x x x Thus the second eigenvect is x x x x x v 6 x To find the third eigenvect cresponding to λ, we need to find a nontrivial solution to a homogeneous system A + 4Iv Since we have that x and x -free and x x x x x x x x 9 5 x 6 8

3 Thus finally the third eigenvect is v 4 Let A 5 Find a diagonal matrix D and an invertible matrix P such that A P DP Compute A Solution: First we find the characteristic polynomial λ deta λi det λ5 λ + λ λ 5 λ Thus we have two distinct eigenvalues λ and λ and as a result the cresponding eigenvalues are linearly independent and the matrix A is diagonalizable To find the first eigenvect cresponding to λ, we need to find a nontrivial solution to a homogeneous system 4 Easy to see that the solution is A Iv x x 4 v To find the second eigenvect cresponding to λ, we need to find a nontrivial solution to a homogeneous system A Iv Easy to see that the solution is x x v Thus, and Thus, D which means A 4 4 4, P A P DP A P D P 4, P 4 4 4

4 5 F the following matrices A find the basis f NulA, RowA, ColA What is ranka? A Solution: The matrix A is already in the echelon fm We can see that there are three nonzero rows, hence ranka and right the way we have and RowA span{ ColA span{,,, Thus the dimnula and to find the basis f NulA we need to find all nontrivial solutions to Ax From the matrix A we can see right the way that x, x 4 -free and x 5 Thus, Thus Hence basis f NulA, } } x x x 4 and x x x x + x 4 x x x x 4 x 5 x + x 4 x x 4 x x 4 NulA span{ x, + x 4 Notice that NulA RowA ie any vect in the NulA is thogonal to any vect in RowA 6 If the null space of a 5 6 matrix A is 4-dimensional, a What is the rank of A? b NullA is a subspace of R n, what is n? c ColA is a subspace of R n, what is n? Solution: a ranka 6 dimnula 6 4 b NullA is a subspace of R 6, since A : R 6 R 5 c ColA is a subspace of R 5 }

5 7 Let A be a n-by-n matrix that satisfies A A What can you say about the determinant of A? Solution: Let x deta Since detab detadetb the relation A A implies x x x x xx Hence x x Thus we can conclude that the deta is either equal to 8 Suppose a 4 7 matrix A has four pivot columns Is Col A R 4? Is Nul A R? Explain Solution: Since A has four pivot columns, ranka 4 and dimnula Since ColA is a subset of R 4 and is four dimensional we can conclude indeed that Col A R 4 However Nul A is a subset of R 7 9 Show that the set {u, u, u } is an thogonal set in R Then express a vect x as a linear combination of u s, where 5 u, u, u, x 4 Solution: The thogonality we can check using a dot product Thus, u u u u + 4 u u + 4 Nest we want to find coefficients c, c, c R such that x c u + c u + c u Taking the dot product with u, u, and u and using thogonality we find c u x u u, c u x u u, c u x u u Thus we find and Hence c u x u u c u x u u c u x u u x 4 u + u + u Using the Gram-Schmidt process to produce an thogonal basis f W span{v, v, v }, where 7 v, v 7, v

6 Solution: Using the Gram-Schmidt we produce thogonal vects x,x, and x such that W span{x, x, x } First we take x v Then 7 x v x v x 7 x x Finally, x v x v x x x x v x x x