Linear Algebra Exam 3 Review
Identify a nonzero 2×2 matrix that is invertible but not diagonalizable.
1 1 0 1
Find a basis for the eigenspace corresponding to the eigenvalue of A given below. A= matrix (4x4), λ=5
A basis for the eigenspace corresponding to λ = x is... (A - 5I) > subtract 5 times the identity matrix from A > rref the resulting matrix > write individual variable equations > basis = column vectors in terms of each varaible (2)
Find a basis for the eigenspace corresponding to each listed eigenvalue of A below. A = matrix (3x3) λ=4,5,3
A basis for the eigenspace corresponding to λ = x is... eigenvector corresponding to λ=4 > (A - 4I) > subtract 4 times identity matrix from matrix A > (A - 4I)x = 0 so rref above result matrix and write individual variable equations > basis = column vector of solution
Find a basis for the eigenspace corresponding to each listed eigenvalue of A below. A= matrix (2x2) λ=1,2
A basis for the eigenspace corresponding to λ= x is... eigen vector corresponding to λ=1 > (A - 1I) > subtract 1 times identity matrix from matrix A > (A - I)x = 0 so rref above result matrix 1 0 0 1 0 0 becomes x = 0 1 basis = 0 1
Show that if A has n linearly independent eigenvectors, then so does AT. If A has n linearly independent eigenvectors, complete the statements below based on the Diagonalization Theorem. Note that P and P−1 are invertible matrices. What is (P−1)T equal to?
A can be factored as PDP^(-1) The columns of matrix P are n linearly independent eigenvectors of A. D is a diagonal matrix whose diagonal entries are the eigenvalues of A. A^T can be written in the form QDQ^(-1) A^T=(P^(-1))^(T)D^(T)P^(T) (P^(-1))^(T) = (P^(T))^(-1) D^(T) = D Q = (P^(T))^(-1)
Construct an example of a 2×2 matrix with only one distinct eigenvalue.
An example of a 2×2 matrix with only one distinct eigenvalue is 10 01
Let A=PDP−1 and P and D as shown below. Compute A^4.
Do PDP^(-1) on calculator and then do answer ^ 4
A is an n×n matrix. Mark each statement below True or False. Justify each answer. a. If Ax=λx for some vector x, then λ is an eigenvalue of A. Choose the correct answer below. b. A matrix A is not invertible if and only if 0 is an eigenvalue of A. Choose the correct answer below. c. 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. d. 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. e. To find the eigenvalues of A, reduce A to echelon form. 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. 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. 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. 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. False. An echelon form of a matrix A usually does not display the eigenvalues of A.
Show that if A is both diagonalizable and invertible, then so is A^(−1). What does it mean if A is diagonalizable? What does it mean if A is invertible? What is the inverse of A?
If A is diagonalizable, then A=PDP^(−1) for some invertible P and diagonal D. Zero is not an eigenvalue of A, so the diagonal entries in D are not zero, so D is invertible. A^(-1) = PD^(-1)P^(-1)
Show that λ is an eigenvalue of A if and only if λ is an eigenvalue of AT. [Hint: Find out how A−λI and AT−λI are related.] How can this relationship between AT−λI and A−λI be used to determine information about λ? Why does this show that λ is an eigenvalue of A if and only if λ is an eigenvalue of AT?
In order for λ to be an eigenvalue of A and A^T, there must exist nonzero x and v such that Ax=λx and A^(T)v= λv The equations (A−λI)x=0 and (A^(T)-λI)v=0 Matrix A^T−λI is the transpose of matrix A−λI. Since the two matrices are transposes, if either (AT−λI)x=0 or (A−λI)v=0 has at least one nontrivial solution, then all of the statements of the Invertible Matrix Theorem are false for both matrices. If λ is an eigenvalue of either A or AT, then it is an eigenvalue of both A and AT because both (AT−λI)x=0 and (A−λI)v=0 have at least one nontrivial solution.
Is v= vector (3x1) an eigenvector of A= matrix (3x3) If so, find the eigenvalue.
No, v is not an eigenvector of A. multiply matrix A by vector v > Av = vector > find what scalar the original v would need to be multiplied by to become the Av vector > no scalar = no eigenvalue and not an eigenvector
A is a 3×3 matrix with two eigenvalues. Each eigenspace is one-dimensional. Is A diagonalizable? Why?
No. The sum of the dimensions of the eigenspaces equals 2 and the matrix has 3 columns. The sum of the dimensions of the eigenspace and the number of columns must be equal.
Let λ be an eigenvalue of an invertible matrix A. Show that λ−1 is an eigenvalue of A−1. [Hint: Suppose a nonzero x satisfies Ax=λx.] Suppose a nonzero x satisfies Ax=λx. What is the first operation that should be performed on Ax=λx so that an equation similar to the one in the previous step can be obtained? Perform the operation and simplify. Why does this show that λ−1 is defined? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. How does this show that λ^(−1) is an eigenvalue of A^(−1)? Select the correct choice below and fill in the answer box to complete your choice.
Note that A−1 exists. In order for λ−1 to be an eigenvalue of A−1, there must exist a nonzero x such that A^(−1)x=λ^(−1)x. Left-multiply both sides of Ax=λx by A^(−1). Perform the operation and simplify. x = A^(−1)(λx) By definition, x is nonzero and A is invertible. So, the previous equation cannot be satisfied if λ=0. Both sides of the equation can be multiplied by λ^(−1) and one side can be simplified to obtain λ^(−1)x=A^(−1)x.
For A= matrix, find one eigenvalue, with no calculation. Justify your answer.
One eigenvalue of A is λ=0. This is because the columns of A are linearly dependent, so the matrix is not invertible.
Use a property of determinants to show that A and AT have the same characteristic polynomial.
Start with det(A^(T)−λI)=det(A^(T)−λI^(T))=det(A−λI)T. Then use the formula det A^(T)=det A.
Find the characteristic polynomial of the matrix, using either a cofactor expansion or the special formula for 3×3 determinants [Note: Finding the characteristic polynomial of a 3×3 matrix is not easy to do with just row operations, because the variable λ is involved.]
The characteristic polynomial is (A - λI) > do matrix a minus the identity matrix with alphas on the diagonal > for 3x3 matrix copy the (A - λI) matrix and repeat the first two columns again at the end. D's are multiplying down the diagonals and U's are multiplying up diagonals. det = D1+D2+D3-U1-U2-U3 > simplified = polynomial
For the matrix, list the eigenvalues, repeated according to their multiplicities. -2 0 0 0 -7 -2 0 0 0 -8 5 0 -9 -1 9 -7
The eigenvalues are −2,−2,5,−7. the diagonals of a triangular matrix
Diagonalize the following matrix.
The matrix cannot be diagonalized since the diagonals are the same value
For the matrix, list the real eigenvalues, repeated according to their multiplicities. 4 -3 0 4 0 6 5 1 0 0 8 5 0 0 0 4
The real eigenvalues are 4,4,6,8. the diagonals of a triangular matrix
Start with detAT−λI)=detAT−λIT)=det(A−λI)T. Then use the formula det AT=det A. a. The determinant of A is the product of the diagonal entries in A. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. b. An elementary row operation on A does not change the determinant. Choose the correct answer below. c. (det A)(det B)=detAB. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. d. If λ+5 is a factor of the characteristic polynomial of A, then 5 is an eigenvalue of A. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
The statement is false because the determinant of the 2×2 matrix A = 11 11 is not equal to the product of the entries on the main diagonal of A. The statement is false because scaling a row also scales the determinant by the same scalar factor. The statement is true because it is the Multiplicative Property of determinants. The statement is false because in order for 5 to be an eigenvalue of A, the characteristic polynomial would need to have a factor of lambda minus λ−5.
Let A, P, and D be n×n matrices. Mark each statement true or false. Justify each answer. Complete parts (a) through (d) below. a. A is diagonalizable if A=PDP−1 for some matrix D and some invertible matrix P. b. If ℝn has a basis of eigenvectors of A, then A is diagonalizable. c. A is diagonalizable if and only if A has n eigenvalues, counting multiplicities. d. If A is diagonalizable, then A is invertible.
The statement is false because the symbol D does not automatically denote a diagonal matrix. The statement is true because A is diagonalizable if and only if there are enough eigenvectors to form a basis of ℝn. The statement is false because the eigenvalues of A may not produce enough eigenvectors to form a basis of ℝn. 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.
Assume A, B, P, and D are n×n matrices. Determine whether the folowing statements are true or false. Justify each answer. a. A matrix A is diagonalizable if A has n eigenvectors. b. If A is diagonalizable, then A has n distinct eigenvalues. c. If AP=PD, with D diagonal, then the nonzero columns of P must be eigenvectors of A. d. If A is invertible, then A is diagonalizable.
The statement is false. A diagonalizable matrix must have n linearly independent eigenvectors. The statement is false. A diagonalizable matrix can have fewer than n eigenvalues and still have n linearly independent eigenvectors. The statement is true. Let v be a nonzero column in P and let λ be the corresponding diagonal element in D. Then AP=PD implies that Av=λv, which means that v is an eigenvector of A. The statement is false. An invertible matrix may have fewer than n linearly independent eigenvectors, making it not diagonalizable.
Find the eigenvalues of the matrix. -6 0 0 0 0 0 -1 0 -7
The eigenvalue(s) of the matrix is/are: -6,0,-7 diagonals of the triangular matrix
Find the eigenvalues of the matrix. 0 0 0 0 -9 3 0 0 7
The eigenvalue(s) of the matrix is/are: 0,-9,7 diagonals of the triangular matrix
Is v = vector (2x1) an eigenvector of A = matrix (2x2)? If so, find the eigenvalue.
Yes, v is an eigenvector of A. The eigenvalue is λ=2. multiply matrix A by vector v > Av = vector > find what scalar the original v would need to be multiplied by to become the Av vector > scalar = eigenvalue
Is λ=2 an eigenvalue of A = matrix? Why or why not?
Yes, λ is an eigenvalue of A because Ax=λx has a nontrivial solution. Ax = 2x > (A - 2I)x = 0 > subtract 2 times identify matrix from matrix A > resulting matrix: either column can be formed by multiplying the other by a scalar so linearly dependent meaning it has non trivial solution and λ=2 an eigenvalue of A
Is λ=3 an eigenvalue of matrix? If so, find one corresponding eigenvector.
Yes, λ=3 is an eigenvalue of matrix. One corresponding eigenvector is vector. Ax = λx > (A - 3I)x = 0 > subtract 3 times the identity vector from matrix A > since it is linearly independent λ=3 is an eigenvalue > solve (A - 3I)x = 0 > rref A - 3I matrix > write individual equations for each variable > One corresponding eigenvector is *column vector of solution coefficients*
Find the characteristic polynomial and the eigenvalues of the matrix. (2x2) The characteristic polynomial is Select the correct choice below and, if necessary, fill in the answer box within your choice.
characteristic polynomial (A - λI) > do matrix a minus the identity matrix with alphas on the diagonal > for 2x2 matrix abcd det = ad - bc > multiply out and simplify ad - bc The real eigenvalue(s) of the matrix is/are factor the polynomial or use quadratic formula > ex (λ-13)(λ-9) means 13 and 9 are the eigenvalues If you cant factor or quadratic is negative no eigenvalues
Diagonalize the following matrix, if possible.
det(A - λI) = det(original matrix minus λ's on the diagonal) > multiply the diagonal terms > solve det = 0 for λ values > substitute the solved values for λ into the (A - λI) matrix > simplify and set the matrix = 0 by making augmented matrix > determine general solution by writing equations > repeat for other value of λ > construct matrix P from the 2 vector solutions > construct matrix D corresponding to the columns of P where the diagonals are the values of λ
Matrix A is factored in the form PDP−1. Use the Diagonalization Theorem to find the eigenvalues of A and a basis for each eigenspace.
look at the second matrix to the right of the second equal sign and pull the eigenvalues from the diagonal. the basis for the eigenspace corresponding to an eigenvalue is the column(s) of the first matrix to the right of the second equals where the eigenvalue occured (column 1 eigenvalue = column 1 as basis)
Diagonalize the following matrix. The real eigenvalues are given to the right of the matrix.
substitute λ = a given x into A - λI and simplify > now augment the resulting matrix to solve (A - #I)x = 0 > rref > write general solution in terms of a variable > a basis for the eigenspace corresponding to λ = # is general solution vector > repeat for all values of λ > matrix P = combination of vectors and D = matrix of diagonal eigenvalues that were given
Use the factorization A=PDP−1 to compute Ak, where k represents an arbitrary integer.
take the matrix on the left side of the equation and raise all a's and b's to the k