Linear Algebra Final
Describe the possible echelon forms of the standard matrix for a linear transformation T where T: ℝ3→ℝ4 is one-to-one.
4 x 3 matrix with pivot in every column and 3!!!! *'s
Let U be a square matrix such that UTU = I. Show that det U = ±1.
Assume that UTU = I. Since the desired result is that det U = ±1, an intermediate step must be found which contains the expression det U. Which of the following can be applied to the assumption UTU = I to achieve the desired result? det (UTU) = det I Simplify the right side of the equation found in the first step. 1 Which property can be used to simplify the left side of the equation found in the first step? Select all that apply. det UT = det U Multiplicative Property Use the properties from the previous step to rewrite the left side of the equation found in the first step. (det U)^2
Let A be an mxn matrix such that ATA is invertible. Show that the columns of A are linearly independent. [Careful: You may not assume that A is invertible; it may not even be square.]
Columns of some matrix A are linearly independent if and only if Ax=0 has only the trivial solution 0. Suppose the nx1 vector x is a solution to Axs=0. To complete the proof, it needs to be shown that x=0. If Ax=0, the matrix expression ATAx is equal to 0. Thus, x must be in Nul ATA. The product ATA is invertible. Therefore, by the Invertible Matrix Theorem, the only vector in the null space of ATA is 0. This implies that the only solution x to Ax = 0 is 0, as required.
Let A be a lower triangular nxn matrix with nonzero entries on the diagonal. Show that A is invertible and A−1 is lower triangular. [Hint: Explain why A can be changed into I using only row replacements and scaling. (Where are the pivots?) Also, explain why the row operations that reduce A to I change I into a lower triangular matrix.]
Consider the augmented matrix [A I]. Remember, an nxn matrix A is invertible if and only if A is row equivalent to In, and in this case, any sequence of elementary row operations that reduces A to the identity matrix also transforms In into the inverse of A. 1. Multiply row 1 by 1 / a 11 2. Add multiples of the first row to the other rows to make their first entries zero. 3. Row reducing the augmented matrix [A I] will result in which augmented matrix? [I A−1] 4. The augmented matrix [A I] has become row reduced and therefore the matrix is invertible. Why was this possible? This is possible because the matrix A is row equivalent to the matrix I and therefore A can be row reduced into I using only row replacements and scaling. 5. Why is the matrix A−1 lower triangular? The row operations performed above only add rows to rows below, so the I on the right of the augmented matrix [A I] changes into a lower triangular matrix. 6. Therefore the matrix A is invertible and the matrix A−1 is lower triangular.
Use the concept of volume to explain why the determinant of a 3 x 3 matrix A is zero if and only if A is not invertible.
Determine what other conditions must be true for a matrix to not be invertible, and determine what the determinant must be under those conditions. Determine what other conditions must be true about matrix A: The columns of matrix A are linearly dependent. What does this mean for the columns in a 3 times ×3 matrix? One of the columns in the plane must be spanned by the other two columns. How does the volume of a 3x3 matrix relate to whether or not it is invertible? If these conditions for a not invertible matrix are true, then the volume of the parallelepiped determined by the columns of the matrix must be zero. What is the volume of a parallelepiped determined by the columns of a 3x3 matrix A? The volume is equal to |det A| How does this prove that the determinant of a 3x3 matrix A is zero if and only if A is not invertible? If the matrix is not invertible, then the volume must be zero, meaning that the determinant of the matrix is zero. If the matrix were invertible, then the volume would not be zero and the determinant could not be zero.
The determinant of A is the product of the pivots in any echelon form U of A, multiplied by (−1)^r, where r is the number of row interchanges made during row reduction from A to U.
False. Reduction to an echelon form may also include scaling a row by a nonzero constant, which can change the value of the determinant.
If Ax=λx for some vector x, then λ is an eigenvalue of A. Choose the correct answer below.
False. The condition that Ax=λx for some vector x is not sufficient to determine if λ is an eigenvalue. The equation Ax=λx must have a nontrivial solution
If Q is a 4x4 matrix and ColQ = ℝ4, what can you say about solutions of equations of the form Qx = b for b in ℝ4?
If ColQ = ℝ4, then the columns of Q span ℝ4. Since Q is square, the Invertible Matrix Theorem shows that Q is invertible and the equation Qx = b has a solution for each b in ℝ4.
Determine if the set is a basis for ℝ3. Justify your answer.
If inconsistent, No, because these vectors do not form the columns of a 3x3 matrix.
Suppose an 8x11 matrix A has 5 pivot columns. Is Col A = ℝ5? Is Nul A = ℝ6? Explain your answers.
Is Col A = ℝ5? No. Since A has 5 pivot columns, dim Col A is 5. But Col A is a five-dimensional subspace of ℝ8, so Col A is not equal to ℝ5. Is Nul A = ℝ6? No, Nul A is not equal to set of real numbers ℝ6. It is true that dim Nul A is equal to 6, but Nul A is a subspace of ℝ11.
Suppose that A is a square matrix such that det A^3 = 0. Explain why A cannot be invertible.
Since the desired conclusion is that A is not invertible, it will be helpful to identify any conditions that are logically equivalent to the condition that A is not invertible. Which of the following conditions is logically equivalent to the condition that A is not invertible? det A = 0 What property can be applied to the left side of the equation det A^3 = 0 in order to yield the desired result? det A^3 = (det A)(det A)(det A) What should be done next to establish that A is not invertible? By the zero factor property, det A = 0
Suppose Ax = λx with x≠0. Let α be a scalar different from the eigenvalues of A, and let B = (A−αI)^−1. Suppose μ is an eigenvalue of B, and that x is a corresponding eigenvector, so that (A−αI)^−1x = μx. Use this equation to find an eigenvalue of A in terms of μ and α. Note that μ ≠ 0 because B is invertible.
The eigenvalue is λ = α+(1/μ)
Suppose A=UDVT where U and V are nxn matrices with the property that UTU=I and VTV=I, and D is a diagonal matrix with positive numbers σ1,...,σn on the diagonal. Show that A is invertible, and find a formula for A−1.
The product of nxn INVERTIBLE matrices is invertible, and the inverse is the product of THEIR INVERSES in the reverse order. Since the matrix U is an nxn matrix with the property that UTU=I , the matrix U is INVERTIBLE and UT = U−1. Similarly, since the matrix V is an nxn matrices with the property that VTV=I, the matrix VT is INVERTIBLE and V= (VT)−1. Since D is a diagonal matrix with positive numbers along the diagonal, it is row equivalent to I and therefore is invertible. This implies that A is invertible because it is the product of nxn INVERTIBLE matrices. Find a formula for A−1. V(D−1)(UT)
If A is diagonalizable, then A is invertible. Choose the correct answer below.
The statement is false because invertibility depends on 0 not being an eigenvalue. A diagonalizable matrix may or may not have 0 as an eigenvalue.
Let A, P, and D be nxn matrices. A is diagonalizable if and only if A has n eigenvalues, counting multiplicities. Choose the correct answer below.
The statement is false because the eigenvalues of A may not produce enough eigenvectors to form a basis of ℝn.
Let A, P, and D be nxn matrices. A is diagonalizable if A = PDP−1 for some matrix D and some invertible matrix P.
The statement is false because the symbol D does not automatically denote a diagonal matrix.
If ℝn has a basis of eigenvectors of A, then A is diagonalizable. Let A, P, and D be nxn matrices.
The statement is true because A is diagonalizable if and only if there are enough eigenvectors to form a basis of ℝn.
Prove the theorem (AB)T=BTAT. [Hint: Consider the ith row of left parenthesis AB right parenthesis Superscript Upper T(AB)T.]
The (i,j)-entry of (AB)T is the (j,i)-entry of AB, which is aj1b1i + ... + ajnbni Complete the second step of the proof by filling in the blank. The entries in row i of BT are b1i, ... bni. Complete the third step of the proof by filling in the blank. The entries in column j of AT are aj1, ..., ajn. Complete the fourth step of the proof by filling in the blank. The (i,j)-entry in BTAT is aj1b1i+...+ajnbni. Write a conclusion by filling in the blank. Therefore, (AB)T = BTAT
A, B, and P are square matrices. Verify that if B = P−1AP and x is an eigenvector of A corresponding to an eigenvalue of λ, then P-1x is an eigenvector of B corresponding also to λ.
To demonstrate that P−1x is an eigenvector of B corresponding to λ, show that BP−1x=λP−1x. If x is an eigenvector of A corresponding to an eigenvalue of λ, then Ax=λx. Use the relationship between A and B to express A in terms of P and B. A=PBP−1 Substitute this expression for A in the matrix equation that shows that x is an eigenvector of A corresponding to λ. PBP−1x = λx How should this equation be manipulated to help show that P-1x is an eigenvector of B corresponding to λ? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. Left-multiply both sides of the equation by P−1. Perform the operation. P−1PBP−1x=P−1λx How can this equation be simplified to obtain the equation that shows that P−1x is an eigenvector of B corresponding to λ? Select the correct choice below and fill in the answer box to complete your choice. Apply the property that states that P−1P = I.
Let R be the triangle with vertices (x1,y1), (x2,y2), (x3,y3). Show that the area of R is given by the formula to the right. 1/2 | det [x1, y1, 1 ]| | [x2, y2, 1 ]| | [x3, y3, 1 ]|
Translate R to a new triangle by subtracting (x3,y3) from each vertex. What are the new vertices? (0,0), (x1−x3, y1−y3), (x2−x3, y2−y3) ***all other answers should match the same pattern of coordinates 1/2 | det [x1, y1, 1 ]| | [x2, y2, 1 ]| | [x3, y3, 1 ]|
A row replacement operation does not affect the determinant of a matrix.
True. If a multiple of one row of a matrix A is added to another to produce a matrix B, then det B equals det A.
If the columns of A are linearly dependent, then det A=0.
True. If the columns of A are linearly dependent, then A is not invertible.
Show that if A is both diagonalizable and invertible, then so is A−1.
What does it mean if A is diagonalizable? If A is diagonalizable, then A=PDP-1 for some invertible P and diagonal D. What does it mean if A is invertible? Zero is not an eigenvalue of A, so the diagonal entries in D are not zero, so D is invertible. What is the inverse of A? PD−1P−1
If A is an invertible nxn matrix, then the inverse of matrix A is A-1 = (1/det(A)) * adjA. If A = [a b][c d] and ad−bc≠0, then A is invertible and the inverse is (1/ad−bc) [d -b][-c a]. Show that if A is 2x2, then the first theorem gives the same formula for A−1 as that given by the second theorem.
What must be done to prove the theorems are equivalent? Evaluate the inverse of matrix A =[a b][c d] using A−1=(1/det A) * adj A Set the 2 x 2 matrix A equal to [a b][c d]. Evaluate the adjugate of this matrix. adj A=[d -b][-c a] Find the determinant of matrix A = [a b][c d]. det A = ad−bc What is the final step in proving the two theorems are equivalent? Substitute the values for det A and adj A into the first theorem to show that it is now equivalent to the second theorem.
Find formulas for X, Y, and Z in terms of A and B. It may be necessary to make assumptions about the size of a matrix in order to produce a formula. [Hint: Compute the product on the left, and set it equal to the right side.] [A Z] [X 0 0] [0 0] = [1 0] [Y 0 1] [B 1] [0 1]
X = A-1 Y = -BA-1 Z = 0
Let S be the tetrahedron in ℝ3, with vertices at vectors 0, e1, e2, and e3, and let S′ be the tetrahedron with vertices at vectors 0, v1, v2, and v3. See the figures to the right. Complete parts (a) and (b) below Find a formula for the volume of the tetrahedron S′ using the fact that {volume of S} = 1/3{area of base} x {height}.
volume = 1/6 * |det[v1 v2 v3]|
Give a formula for (ABx)T, where x is a vector and A and B are matrices of appropriate size.
xTBTAT
If the nxn matrices E and F have the property that EF = I, then E and F commute. Explain why.
According the Invertible Matrix Theorem, E and F must be invertible and inverses. So FE = I and I = EF. Thus, E and F commute.
To find the eigenvalues of A, reduce A to echelon form. Choose the correct answer below.
False. An echelon form of a matrix A usually does not display the eigenvalues of A.
det(A+B) = detA + detB
False. If A = [ 1 0 ] [ 0 1 ] and B = [−1 0 ] [ 0 −1 ], then det(A+B)=0 and detA + detB = 2.
Is it possible that all solutions of a homogeneous system of thirteen linear equations in seventeen variables are multiples of one fixed nonzero solution? Discuss. Consider the system as Ax=0, where A is a 13x17 matrix. Choose the correct answer below.
No. Since A has at most 13 pivot positions, rank A≤13. By the Rank Theorem, dim Nul A= 17 −rankA ≥4. Thus, it is impossible to find a single vector in Nul A that spans Nul A.
A number c is an eigenvalue of A if and only if the equation (A−cI)x=0 has a nontrivial solution. Choose the correct answer below.
True. A number c is an eigenvalue of A if and only if the equation Ax=cx has nontrivial solutions, and Ax=cx and (A−cI)x=0 are equivalent equations.
Finding an eigenvector of A may be difficult, but checking whether a given vector u is in fact an eigenvector is easy. Choose the correct answer below.
True. Checking whether a given vector u is in fact an eigenvector is easy because it only requires checking that u is a nonzero vector and finding if Au is a scalar multiple of u.
A matrix A is not invertible if and only if 0 is an eigenvalue of A. Choose the correct answer below.
True. If 0 is an eigenvalue of A, then there are nontrivial solutions to the equation Ax=0x. The equation Ax=0x is equivalent to the equation Ax=0, and Ax=0 has nontrivial solutions if and only if A is not invertible.
A linearly independent set {v1, ... , vk} in ℝn can be expanded to a basis for ℝn. One way to do this is to create A = [v1•••vke1•••en] with e1, ... , en the columns of the identity matrix; the pivot columns of A form a basis for ℝn. Complete parts (a) and (b) below.
Use the method described to extend the following vectors to a basis for ℝ5. Choose the correct answer below. ***Row reduce augmented v1 v2 v3 In and see which columns make a basis for ColA b. Explain why the method works in general. Why are the original vectors v1, ... , vk included in the basis found for Col A? The original vectors are the first k columns of A. Since the set of original vectors is assumed to BE linearly independent, these columns of A will be PIVOT columns and the ORIGINAL SET OF VECTORS will be included in the basis. Why is Col A =ℝn? Since all of the columns of the nxn identity matrix are COLUMNS of A, every vector in RN is in COL A. Since every column of A is in ℝn, every vector in COL A is in RN. This shows that Col A and ℝn are equivalent.
Give integers p and q such that Nul A is a subspace of ℝp and Col A is a subspace of ℝq.
p is the number of columns (Nul A) q Is the number of rows (Col A)
