Find the eigenvalues and an eigenvector per eigenvalue of the matrix
A= [ 0 1 1]
[ 3 -4 -3]
[ -5 7 6]
Is A is diagonalisable ? Give reason your answer.
1
Expert's answer
2021-07-26T09:10:38-0400
Given the matrix A to find the Eigenvalues we have the following:det(A−λI)=0⟹∣∣−λ3−51−λ−471−36−λ∣∣=0Add column 2 multiplied by λ to column 1:C1=C1+λC2∣∣−λ3−51−λ−471−36−λ∣∣=∣∣0−λ(λ+4)+37λ−51−λ−471−36−λ∣∣Subtract column 2 from column 3: C3=C3−C2∣∣0−λ(λ+4)+37λ−51−λ−471−36−λ∣∣=∣∣0−λ(λ+4)+37λ−51−λ−470λ+1−λ−1∣∣Expand along row 1:∣∣0−λ(λ+4)+37λ−51−λ−470λ+1−λ−1∣∣=(0)(−1)1+1∣∣−λ−47λ+1−λ−1∣∣+(1)(−1)1+2∣∣−λ(λ+4)+37λ−5λ+1−λ−1∣∣+(0)(−1)1+3∣∣−λ(λ+4)+37λ−5−λ−47∣∣=−∣∣−λ(λ+4)+37λ−5λ+1−λ−1∣∣Note that: ∣∣−λ(λ+4)+37λ−5λ+1−λ−1∣∣=(−λ(λ+4)+3)⋅(−λ−1)−(λ+1)⋅(7λ−5)=λ3−2λ2−λ+2Finally we have the following: (−1)⋅(λ3−2λ2−λ+2)=−(λ−2)(λ−1)(λ+1).⟹det(A−λI)=−(λ−2)(λ−1)(λ+1)=0Solving this we have that: λ=−1,1,2.∴The eigenvalues of A are λ=−1,λ=1,λ=2We now fin the eigenvectors of the matrix. 1.For λ=2:⎣⎡−λ3−51−λ−471−36−λ⎦⎤=⎣⎡−23−51−671−34⎦⎤The nullspace of this matrix is ⎣⎡31−311⎦⎤which is the eigenvector.2.For λ=1:⎣⎡−λ3−51−λ−471−36−λ⎦⎤=⎣⎡−13−51−571−35⎦⎤The nullspace for this matrix is: ⎣⎡101⎦⎤and this is the eigenvector3.For λ=−1:⎣⎡−λ3−51−λ−471−36−λ⎦⎤=⎣⎡13−51−371−37⎦⎤The nullspace of this matrix is ⎣⎡0−11⎦⎤which is the eigenvector.A is diagonalisable since each of the eigenvectors of A are linearly independent and they are distinct.
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#339914 on Dec 2023
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