Math  /  Algebra

QuestionFind the eigenvalues and eigenvectors associated with the following matrix: A=[3513]A=\left[\begin{array}{cc} 3 & 5 \\ -1 & -3 \end{array}\right]

Studdy Solution
Find eigenvectors for each eigenvalue:
For λ1=2\lambda_1 = 2:
Solve (A2I)v=0(A - 2I)\mathbf{v} = \mathbf{0}:
A2I=[1515]A - 2I = \begin{bmatrix} 1 & 5 \\ -1 & -5 \end{bmatrix}
Set up the system of equations:
1v1+5v2=01v_1 + 5v_2 = 0
v15v2=0-v_1 - 5v_2 = 0
Both equations are equivalent, so we can solve:
v1=5v2v_1 = -5v_2
Let v2=tv_2 = t, then v1=5tv_1 = -5t.
Thus, an eigenvector is:
v1=[51]\mathbf{v_1} = \begin{bmatrix} -5 \\ 1 \end{bmatrix}
For λ2=2\lambda_2 = -2:
Solve (A+2I)v=0(A + 2I)\mathbf{v} = \mathbf{0}:
A+2I=[5511]A + 2I = \begin{bmatrix} 5 & 5 \\ -1 & -1 \end{bmatrix}
Set up the system of equations:
5v1+5v2=05v_1 + 5v_2 = 0
v1v2=0-v_1 - v_2 = 0
Both equations are equivalent, so we can solve:
v1=v2v_1 = -v_2
Let v2=sv_2 = s, then v1=sv_1 = -s.
Thus, an eigenvector is:
v2=[11]\mathbf{v_2} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}
The eigenvalues are λ1=2\lambda_1 = 2 and λ2=2\lambda_2 = -2. The corresponding eigenvectors are v1=[51]\mathbf{v_1} = \begin{bmatrix} -5 \\ 1 \end{bmatrix} and v2=[11]\mathbf{v_2} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}.

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