# Linear Algebra Final Exam Study Guide Solutions Fall 2012

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1 . Let A = Given that v = Linear Algebra Final Exam Study Guide Solutions Fall 5 explain why it is not possible to diagonalize A. is an eigenvector for A and λ = is an eigenvalue for A diagonalize A or Av = So the eigenvalue corresponding to v is λ =. 7 7 A I = and v = = 5 so the eigenvectors for λ = are v = A is diagonalizable since there are linearly independent eigenvectors. 7 One possible choice is P = and D = 5 7. A discrete dynamical system is defined by ( x = x k+ = Mx k (.. for the transition matrix M =...7 ( M has an eigenvector v = and an eigenvalue λ =.9. Find the eigenvalue corresponding to v and the eigenvector corresponding to λ. Mv = so the eigenvalue for eigenvector v is λ =.. (.. (.. M.9I =.. ( The eigenvector corresponding to λ =.9 is v =. (. Define: p (t = + t p (t = t + t p (t = + t + t

2 (a Is the set B = {p (t p (t p (t} a basis for P? Explain. No. P is isomorphic to R and this set only has vectors. (b Is the vector q(t = + t + t + t in the space spanned by the vectors in B? Explain. Using coordinate vectors we see that so no q is not in the space spanned by the vectors in B.. The matrix C = (a Find a basis for the null space of C. (b Find a basis for the column space of C. 5 is row equivalent to the matrix 5. What is the smallest possible dimension for the null space of a 5 7 matrix? Explain. There are at least free variables so the smallest dimension of the null space is. 6. What is the smallest possible dimension for the null space of a 7 5 matrix? Explain. THere could be no free variables so the dimension of the null space could be zero. 7. A matrix B has rank 8. (a What is the dimension of Col B? 8 (b Col B is a subspace of R n for n = (c What is the dimension of Nul B? 5 (d Nul B is a subspace of R n for n = 8. The dimension of the column space of a 7 5 matrix is 5. Is the column space equal to R 5? Explain. No it s a 5 dimensional subspace of R The dimension of the column space of a 5 7 matrix is 5. Is the column space equal to R 5? Explain. Yes! A 5 dimensional subspace of R 5 must be equal to all of R 5.. Find the inverse of A =. (A I = A =. so. Suppose that the determinant of a matrix C is. Is it possible for Cv = for a non-zero vector v? Explain. No. If determinant is not zero then the matrix is invertible so Cv = has only the trivial solution. (There are no free variables..

3 . Suppose that the determinant of a matrix D is. Is it possible for the transformation x Dx to be onto? Explain. No. If the determinant is zero the matrix is not invertible and so the transformation is not onto.. Suppose that the determinant of a matrix E is 7. Is it possible for the columns of E to be linearly independent? Explain. Yes the columns must be linearly independent since the determinant is non-zero and the matrix is therefore invertible.. Compute the determinant of the matrix B = Show all work! so det B = det Now 5 so det B = ( ( det Now Is the set H = r 5s s r + s r so det B = ( ( det ( r s R a subspace of R 5? Explain. = No. The zero vector is not in H. 7a 5b + c a b + c 6. Is the set H = c a b + c a b c R a subspace of R 6? Explain. a b a 7 5 Yes H = Span and the span of a set of vectors is always a subspace. 7. Is the set H = { p(t = at + at 5 a R } a subspace of P5? Explain. Yes. H = Span{t + t 5 } and the span of a set of vectors is always a subspace Let A = 5 6. Is x = in Nul A? Yes Ax =.

4 9. The matrix A is row equivalent to the matrix. Find a basis for the null space of A. What is the dimension of the null space of A? What is the dimension of the column space of A? Dimension of the null space is. Dimension of the column space is.. Suppose that A B and C are invertible matrices. Solve the following equation for X and simplify. ACXB A = I X = C B. Careful! Order matters!!. Perform the indicated operations or explain why the operation is undefined. ( ( (a + Not possible. Sizes not the same. ( ( ( (b + = ( 6 6 (c = 9 (d ( Not possible. Sizes don t match up right.. Let A be a 6 7 matrix: (a A has 6 rows and 7 columns. (b The columns of A are vectors in R k for k = 6. (c To compute Ax the vector x must be in R m for m = 7. (d Linear combinations of the columns of A are vectors in R n for n = 6. (e The transformation x Ax maps vectors in R m to R n for m = 7 and n = 6. (f The matrix A cannot have a pivot in every (row/column column. (g What does this tell us about the transformation x Ax? The transformation cannot be one-toone since there is not a pivot in every column. (It could be onto or not depending on how many pivots there are. (h i. Can you tell whether the columns of A are linearly independent? Yes. ii. If so are they? No. iii. Explain. No pivot in every column. (i i. The columns of A span a subspace of R k for k = 6. ii. Can you tell whether the columns of A span R k? No iii. If so do they? Maybe. iv. Explain. If there is a pivot in every row then they span R 6.. Suppose that B is a matrix. For each of the following situations write a possible echelon form for B or explain why the situation is impossible.

5 (a Bx = has only the trivial solution. Not possible there must be a free variable. (b The columns of B span R. (c The columns of B span R. Not possible since the columns are vectors in R not R.. (d The transformation x Bx is one-to-one. Not possible since there can not be a pivot in every column.. x Cx defines a transformation from R to R. For each of the following situations write a possible echelon form for C or explain why the situation is impossible. (a The transformation is onto but not one-to-one. C is a matrix so it is not possible to have a pivot in every row and the transformation cannot be onto. (b The transformation is neither one-to-one nor onto. (c The transformation is one-to-one but not onto. ( 5. Let A = and consider the transformation T (x = Ax. (a Find the image of the vector x = under T. ( Ax = (b Is T (x an onto transformation? Explain. Yes. There is a pivot in every row. (c Is T (x an one-to-one transformation? Explain. No. There is not a pivot in every column. h 6. For what values of h will the columns of A = 5 span R? A h 5 + h As long as h there will be a pivot in every column and the columns will span R. For what values of h will the columns of A be linearly independent? As long as h there will be a pivot in every column and the columns will be linearly independent. 7. Suppose that M is a 5 5 matrix and that there is a non-zero vector b such that Mb =. Could the columns of M span R 5? Explain. No. A non-zero vector b such that Mb = means that there is a free variable and not a pivot in every column. Since M is square this also means that there is not a pivot in every row so the columns cannot span R 5.

6 8. Find the reduced-row echelon form of the augmented matrix for the following linear system of equations and use it to find the solution in parametric vector form (if there is a solution. The augmented matrix is 6 x x = x x + x = 6 x + x = x = 7 7 so the solution is unique. 9. Is it possible to write the vector b = - as a linear combination of v = v = and v = -? 5 Write the solution in parametric vector form (if there is a solution. The augmented matrix is so the equation is inconsistent. There is not solution. 5. The augmented matrix A corresponds to a system of linear equations in x x x x x 5 and x 6. Write the solution to this system of equations in parametric vector form if the reduced row echelon form of A is x = x + x Find conditions involving h and k that determine the number of solutions to the system of equations ( h 6 k ( h 6 h k x + hx = x + 6x = k The system will have exactly one solution if: h The system will have infinitely many solutions if: h = and k = The system will have no solutions if: h = and k

7 . Given u = ( Chapter 6 Questions find a unit vector in the direction of u. u = 5 so a unit vector in the direction of u is (. Are the vectors u = and v = No. u v =. ( Find the orthogonal projection of u onto v. ( u v ( /5 Proj v (u = v = v v 5 v = /5 Find the distance between u and v. The distance between u and v is u v =. ( /5 /5. orthogonal? The orthogonal distance from u to the span of v is u û = /5.. Is the set S = 7 an orthogonal set? Yes. Is S an orthonormal set? No. None of the lengths are one. Normalize the set S Let W = Span. Determine if x = Yes x is orthogonal to each of the three vectors. 6. Apply the Gram-Schmidt process to the following basis for R. {( ( } B = ( v = ( ( v = 5 = ( 6/5 /5 7. Apply the Gram-Schmidt process to the following basis for R. {( ( } 8 B = ( v = ( ( 8 v = 76 7 = ( 6/7 75/7 5 is in W.

8 8. Apply the Gram-Schmidt process to the following basis for R. B = The vectors are already orthogonal! v = v = 9 = v = 9 9 = I m sorry I didn t give you a real Gram-Schmidt example. See section 6. for plenty of more examples. 9. The vectors and span a plane in R. (a Find an orthonormal basis for this subspace. v = v = 5 = 6/5 /5 Normalized basis vectors are (b Find the projection of 7 5 (c Write = 6 5 /5 /5 and 5 onto this subspace. = 5/ / 9/ 6 5 as the sum of a vector in the subspace and a vector orthogonal to the subspace. 5/ / 9/ + (d Find the distance from / 9/ / to this subspace. ( / + (9/ + (/ =. Oh my those were ugly fractions! I promise I won t do that to you on the final!!

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