Math Exam Sample Problems Solution Guide October, Note that the following provides a guide to the solutions on the sample problems, but in some cases the complete solution would require more work or justification Determine if the following sets of vectors are or are not linearly independent Explain how you know a set is or is not linearly independent If a set is linearly dependent, then find a linear dependency among the vectors (a 4, 9 (b 4,,, 9 (c 4, 4,, 4 Solution: (a These two vectors are linearly independent To see this we see that if ( ( ( 4 α + β, then 4α + β, and so α Then it follows that β 9 (b These vectors must be linearly dependent To see this, we have 4 vectors in a - dimensional space (where a linearly indpendent set can have no more than elements For a linear dependency we solve a 4 + b + c + d 9 for a, b, c, d This gives us three linear equations a + b c + d 4a + c + d a + 9b + d Solving these equations we find a, b, c, and d 6, is one of many possible solutions
(c These vectors are linearly dependent To see this, take the 4 4 determinant of the matrix whose columns are the four vectors given We find that the determinant is, and so the vectors must be linearly dependent Setting up equations as we did in part (b, we find that one of many dependencies is 4 4 4 + + 4 For (a (c in problem, find bases for the spans of the given sets of vectors {( ( } 4 Solution: (a Since the vectors are linearly independent, we see that, 9 is a basis for the span (b Let V be the span of the four given vectors Since the first two vectors are repeated from part (a and are linearly independent, we see that dim V We calculate det 4 8, 9 so the first three vectors in our set are linearly independent So forms a basis for V (c We wish to find a bais for the column space of the matrix A 4 4 4 {( 4 ( ( 9 },, Applying Gaussian elimination, we find that the reduced row echelon form of A is B Since the number of pivots (leading s of B is, the dimension of the column space of A must be The columns of A corresponding to the three pivots of B will be a {( ( ( } basis for the column space of A: namely 4, 4,
Given v (, v (, S ( 4, 6 find vectors w and w so that S is the transition matrix from {w, w } to {v, v } Solution: In the notation we have used in class S [I] E F, where F {w, w } and E {v, v } That is, for all v R we should have [v] E S[v] F The problem does not say to find w and w with respect to the standarrd basis necessarily, but that is what we will do Let U [I] E ( be the transition matrix from E to the standard basis Let V [I] F (w, w be the transition matrix from F to the standard basis Then Therefore US V Thus V S [I] E F ([I] E [I] F U V ( ( 4 6 ( 4 7 8 Thus we can take w ( 4 and w ( 7 8 {( ( ( } {( 4 Let S,, be the standard basis of R Let A {( ( ( } and B,, (a Show that A is a basis for R Show that B is not a basis for R (, ( }, (b Find a linear equation (or equations whose set of solutions is Span(B ( 4 (c Find the coordinates of with respect to the basis A (You may want to do the next part first (d Find the change of basis matrix which converts coordinates in the S-basis into coordinates in terms of the A-basis Solution: (a We calculate det, and so A is a linearly independent set of vectors Since A consists of vectors, which is also the dimension of R, we see that A must be a basis of R On the other hand, we calculate det,
and so B is a linearly dependent set Therefore, B cannot be a basis ( ( ab (b Let B, and so Span(B is the column space of B A vector is in c the column space of B if and only if the system of equations with augmented matrix a b c is consistent We apply row operations to put the matrix in row echelon form: a a b a b c This is consistent if and only if a b c, which gives the defining equation for the column space of B (d Using the hint, we do part (d first We know that the change of basis matrix that converts coordinates in terms of the A-basis into the S-basis is [I] S A To convert from the standard basis to coordinates with respect to A, we need (c Let v ( 4 In other words, Then [v] A [I] A S 4 which can be checked easily [I] A S ([I] S A S 4, 4 + +, S A Let 4 8 A 4
(a Find bases for the row space and the column space of A (b What is the rank of A? (c What is the nullity of A? ( c8 (d Let b For what value or values of c will Ax b be consistent? 6 Solution: (a We put A in reduced row echelon form and find 4 U From this we see that {(,,, 4, (,,, } is a basis for the row space of A, {( ( 4 } and, is a basis for the column space of A (b The rank is the dimension of the row space of A, which is (c Since rank(a + nullity(a 4 the number of columns of A, we see that the nullity of A is also (d The system of equations Ax b is consistent if and only if b is in the column space of A After setting up the equations, we see that this happens only if c : 4 8 6 6 Let A and B be n n matrices, and let O be the n n zero matrix (a Prove that AB O if and only if the column space of B is a subspace of the null space of A (b Show that if AB O, then rank(a + rank(b n Solution: (a ( : Assume that AB O Let v be in the column space of B Since AB O, if we multiply A by any column of B we get Therefore, A times any linear combination of columns of B will be, and so Av Thus, v N(A ( : Suppose that the column space of B is contained in the null space of A Then in particular A times any column of B is Therefoer AB O (b Suppose that AB O By part (a we see that the column space of B is contained in the null space of A We know that rank(a + dim N(A n
The column space of B is contained in the null space of A, and the rank of B is the dimension of its column space Thus rank(b dim N(A From the equation above, we see then that rank(a + rank(b n 7 Consider the function L : R R defined by a ( L b a + 6b c 9a c (a Is L a linear operator? Explain (b Find a matrix A so that L(x Ax for x R (c What is the dimension of the kernel and image of L Explain Solution: (a Yes The reason is that a a ( L b + b (a + a + 6(b + b (c c c + c 9(a + a ( ( a + 6b a + 6b + L c 9a c 9a a b c + L a b c (b We know that we can take A ( L(e, L(e, L(e ( 6 9 (c The dimension of the kernel of L is equal to the nullity of A The dimension of the image of L is equal to the dimension of the column space of A, which is the same as the rank of A It is easy to see that the second and third columns of A are linearly independent, so the rank of A is Therefore dim Im(L Since rank(a + nullity(a, the nullity of A is, and so dim ker(l 8 Let T : R R be defined by T (x ( x,
where x R is written with respect to the standard basis Let B {(, ( } What is the matrix representing T with respect to the basis B? (Ie, what is [T ] B B? Solution: We know that the transition matrix that converts vectors in coordinates with respect to B into the standard basis S is ( [I] S B Therefore [T ] B B ([I] S B [T ] S S [I] S B ( ( ( ( 8 9 9 Suppose that A and B are similar n n matrices Let c R, and let I denote the n n identity matrix Show that A ci and B ci are similar matrices Solution: Since A and B are similar matrices, there exists an invertible n n matrix S so that B S AS Note then that S (B cis S BS S (cis A ci Therefore, A ci and B ci are similar For each of the following matrices, find the eigenvalues and determine the corresponding eigenspaces ( (a 4 (b Solution: (a Let A be the matrix given Then ( λ det(a λi det ( λ( λ λ λ 4 λ We solve λ λ (λ (λ +,
and so the eigenvalues of A are λ, We look for eigenvectors for each of these eigenvalues For λ we solve ( ( ( a a, 4 b b which amounts to the system a + b a and 4a + b b homogeneous system This becomes the 4a + b 4a b The second equation is redundant, and so ( a b is an eigenvector with eigenvalue if and only if a 4 b Thus v ( 4 is one, and the space is -dimensional, so the eigenspace with eigenvalue is R ( 4 Similarly, for λ we solve which leads to the system of equations ( ( ( a a, 4 b b a + b 4a + 4b Thus we need a b, and so w ( is an eigenvector with eigenvalue, and the corresponding eigenspace is R ( (b Let B be the matrix given in the problem Then λ det(b λi λ x + x + x λ We solve x x x + (x (x, and find that the eigenvalues of B are λ, ± In a similar manner as in part (a, ( we find that the eigenspace of eigenvalue λ is spanned by The eigenspace of eigenvalue λ is spanned by, and the eigenspace of eigenvalue λ is spanned by ( ( + +