Determine whether or not the 5x5 matrix 𝐴=(3 0 -5 -6 2/ 6 2 -12 -9 3/ 6 0 -10 -9 3/ -5 0 7 5 -1/ 1 0 -5 -6 4) is diagonalizable. You may use the fact that the characteristic polynomial is 𝑐𝐴(𝑡)=(𝑡−2)^3(𝑡+1)^2.
Question
Determine whether or not the 5x5 matrix 𝐴=(3 0 -5 -6 2/ 6 2 -12 -9 3/ 6 0 -10 -9 3/ -5 0 7 5 -1/ 1 0 -5 -6 4) is diagonalizable. You may use the fact that the characteristic polynomial is 𝑐𝐴(𝑡)=(𝑡−2)^3(𝑡+1)^2.
Solution
To determine if a matrix is diagonalizable, we need to check if it has enough eigenvectors to form a basis for the space. In other words, the algebraic multiplicity of each eigenvalue must equal its geometric multiplicity.
From the given characteristic polynomial, we can see that the eigenvalues of the matrix A are 2 and -1. The algebraic multiplicity of 2 is 3 and of -1 is 2.
The geometric multiplicity of an eigenvalue is the dimension of the eigenspace corresponding to that eigenvalue. It is equal to the number of linearly independent eigenvectors corresponding to that eigenvalue.
Since we are dealing with a 5x5 matrix, we need 5 linearly independent eigenvectors for the matrix to be diagonalizable.
Given that the algebraic multiplicities of the eigenvalues are 3 and 2, we can infer that there are at most 3 linearly independent eigenvectors for the eigenvalue 2 and at most 2 linearly independent eigenvectors for the eigenvalue -1.
Therefore, the matrix A does not have enough eigenvectors to form a basis for the space, and hence it is not diagonalizable.
Similar Questions
The matrix A𝐴 has eigenvalues −5−5, −1−1, and 22 with corresponding eigenvectors ⎡⎣⎢07−6⎤⎦⎥[07−6], ⎡⎣⎢−13−1⎤⎦⎥[−13−1] and ⎡⎣⎢−764⎤⎦⎥[−764]. Which of the following is a valid diagonalisation? Select all that apply.This matrix cannot be diagonalisedA=V−1DV𝐴=𝑉−1𝐷𝑉 with V=𝑉= ⎡⎣⎢07−6−13−1−764⎤⎦⎥[0−1−7736−6−14] and D=𝐷= ⎡⎣⎢−5000−10002⎤⎦⎥[−5000−10002]A=V−1DV𝐴=𝑉−1𝐷𝑉 with V=𝑉= ⎡⎣⎢0−1−7736−6−14⎤⎦⎥[07−6−13−1−764] and D=𝐷= ⎡⎣⎢−5000−10002⎤⎦⎥[−5000−10002]A=VDV−1𝐴=𝑉𝐷𝑉−1 with V=𝑉= ⎡⎣⎢0−1−7736−6−14⎤⎦⎥[07−6−13−1−764] and D=𝐷= ⎡⎣⎢−5000−10002⎤⎦⎥[−5000−10002]A=VDV−1𝐴=𝑉𝐷𝑉−1 with V=𝑉= ⎡⎣⎢07−6−13−1−764⎤⎦⎥[0−1−7736−6−14] and D=𝐷= ⎡⎣⎢−5000−10002⎤⎦⎥
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