DETAILS LARLINALG8 7.R.019. Explain why the matrix is not diagonalizable. 200 A= 1 2 0 0...
DETAILS LARLINALG8 7.2.050. Show that the matrix is not diagonalizable. [ ] : 0 The matrix is not diagonalizable because it only has one linearly independent eigenvector. The matrix is not diagonalizable because it only has one distinct eigenvalue. The matrix is not diagonalizable because [*] is not an eigenvector. The matrix is not diagonalizable because k is not an eigenvalue.
Explain why the matrix is not diagonalizable. 600] A = 1 60 0 0 6 O A is not diagonalizable because it only has one distinct eigenvalue. O A is not diagonalizable because it only has two distinct eigenvalues. O A is not diagonalizable because it only has one linearly independent eigenvector. A is not diagonalizable because it only has two linearly independent eigenvectors
linear algebra Explain why the matrix is not diagonalizable. A= 8 0 0 1 8 0 0 0 8 O A is not diagonalizable because it only has one distinct eigenvalue. O A is not diagonalizable because it only has two distinct eigenvalues. O A is not diagonalizable because it only has one linearly independent eigenvector. O A is not diagonalizable because it only has two linearly independent eigenvectors.
ASK YOUR TEACHER DETAILS LARLINALG8 7.2.023. Find the eigenvalues of the matrix and determine whether there is a sufficient number to guarantee that the matrix is diagonalizable. (Recall that the matrix may be diagonalizable even though it is not guaranteed to be diagonalizable by the theorem shown below.) Sufficient Condition for Diagonalization If an n x n matrix A has a distinct eigenvalues, then the corresponding eigenvectors are linearly independent and A is diagonalizable. Find the eigenvalues. (Enter your answers...
DETAILS LARLINALG8 7.3.033. Show that any two eigenvectors of the symmetric matrix A corresponding to distinct eigenvalues are orthogonal. 3 A = Find the characteristic polynomial of A. |u-A=1 Find the eigenvalues of A. (Enter your answers from smallest to largest.) (14, 12) = Find the general form for every eigenvector corresponding to 1. (Use s as your parameter.) X1 = Find the general form for every eigenvector corresponding to 12. (Use t as your parameter.) X2 = Find x,...
Please do number 2 Assume all matricies are Mmxm(R) unless otherwise specified. 1. (1 point) Prove or disprove that the eigenvalues of A and AT are the same. 2. (2 points) Let A be a matrix with m distinct, non-zero, eigenvalues. Prove that the eigenvectors of A are linearly independent and span R”. Note that this means in this case) that the eigenvectors are distinct and form a base of the space. 3. (1 point) Given that is an eigenvalue...
Let A be the matrix To 1 0] A= -4 4 0 1-2 0 1 (a) Find the eigenvalues and eigenvectors of A. (b) Find the algebraic multiplicity an, and the geometric multiplicity, g, of each eigenvalue. (c) For one of the eigenvalues you should have gi < az. (If not, redo the preceding parts!) Find a generalized eigenvector for this eigenvalue. (d) Verify that the eigenvectors and generalized eigenvectors are all linearly independent. (e) Find a fundamental set of...
Assume all matricies are Mmxm(R) unless otherwise specified. 1. (1 point) Prove or disprove that the eigenvalues of A and AT are the same. 2. (2 points) Let A be a matrix with m distinct, non-zero, eigenvalues. Prove that the eigenvectors of A are linearly independent and span R”. Note that this means in this case) that the eigenvectors are distinct and form a base of the space. 3. (1 point) Given that is an eigenvalue of A associated with...
Exercise 25. Let , be an orthonormal basis of a two-dimensional subspace S of R" and A xyT + (i) Show that x+y and x -y are eigenvectors of A. What are their corresponding eigenvalues? (ii) Show that 0 is an eigenvalue of R" with n - 2 linearly independent eigenvectors. (iii) Explain why A is diagonalizable. Exercise 25. Let , be an orthonormal basis of a two-dimensional subspace S of R" and A xyT + (i) Show that x+y...
linear algebra question 2. (5' each) Give short answers: (a) True or false: If Ai-Adi for some real number λ, then u is an eigenvector of matrix A. If a square matrix is diagonalizable, then it has n distinct real eigenvalues. Two vectors of the same dimension are linearly independent if and only if one is not a multiple of the other. If the span of a set of vectors is R", then that set is a basis of R...