The 2×22×2 matrixA=(5−53−3)𝐴=(53−5−3) has two distinct real eigenvalues. 1. Give the characteristic polynomial for A𝐴 in Maple notation in the form t^2 + a*t + b Characteristic polynomial = 2. Find the set of eigenvalues for A𝐴 . Enter your response in Maple set notation, i.e., enclosed in braces { } with the two eigenvalues separated by a comma. A typical answer might look like {-4, 7}. The set of eigenvalues for A𝐴 is 3. Find one eigenvector for each eigenvalue, using Maple notation for vectors, e.g. <5, -6> for the vector (5−6)(5−6) An eigenvector for the smallest eigenvalue for A𝐴 is An eigenvector for the largest eigenvalue for A𝐴 is
Question
The 2×22×2 matrixA=(5−53−3)𝐴=(53−5−3) has two distinct real eigenvalues. 1. Give the characteristic polynomial for A𝐴 in Maple notation in the form t^2 + a*t + b Characteristic polynomial = 2. Find the set of eigenvalues for A𝐴 . Enter your response in Maple set notation, i.e., enclosed in braces { } with the two eigenvalues separated by a comma. A typical answer might look like {-4, 7}. The set of eigenvalues for A𝐴 is 3. Find one eigenvector for each eigenvalue, using Maple notation for vectors, e.g. <5, -6> for the vector (5−6)(5−6) An eigenvector for the smallest eigenvalue for A𝐴 is An eigenvector for the largest eigenvalue for A𝐴 is
Solution
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The characteristic polynomial of a 2x2 matrix A = (a b, c d) is given by t^2 - (a+d)t + (ad - bc). For the matrix A = (5 -5, 3 -3), the characteristic polynomial is t^2 - 2t - 20.
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The eigenvalues of a matrix are the roots of its characteristic polynomial. The roots of the polynomial t^2 - 2t - 20 are -4 and 6. So, the set of eigenvalues for A is {-4, 6}.
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To find the eigenvectors, we substitute each eigenvalue into the equation (A - λI)v = 0 and solve for v.
For λ = -4, the equation becomes:
(5 - (-4) -5) * v1 (3 -3 - (-4)) * v2 = 0
Solving this system of equations gives the eigenvector corresponding to λ = -4 as <1, 1>.
For λ = 6, the equation becomes:
(5 - 6 -5) * v1 (3 -3 - 6) * v2 = 0
Solving this system of equations gives the eigenvector corresponding to λ = 6 as <1, -1>.
So, the eigenvector for the smallest eigenvalue (-4) for A is <1, 1> and the eigenvector for the largest eigenvalue (6) for A is <1, -1>.
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