Linear Algebra Final

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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 ​(UT​U) = 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)


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