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Math  /  Algebra

QuestionFind the eigenvalues and eigenvectors associated with the following matrix: A=[35−1−3]A=\left[\begin{array}{cc} 3 & 5 \\ -1 & -3 \end{array}\right]

Studdy Solution

STEP 1

1. We are given a 2×22 \times 2 matrix AA.
2. Eigenvalues λ\lambda are found by solving the characteristic equation det⁡(A−λI)=0\det(A - \lambda I) = 0.
3. Eigenvectors are found by solving (A−λI)v=0(A - \lambda I)\mathbf{v} = \mathbf{0} for each eigenvalue λ\lambda.

STEP 2

1. Set up the characteristic equation.
2. Solve the characteristic equation for eigenvalues.
3. Find eigenvectors for each eigenvalue.

STEP 3

Set up the characteristic equation for matrix AA:
The characteristic equation is given by:
det⁡(A−λI)=0\det(A - \lambda I) = 0
Where II is the identity matrix of the same size as AA. For a 2×22 \times 2 matrix, II is:
I=[1001]I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
Thus, A−λIA - \lambda I is:
A−λI=[35−1−3]−[λ00λ]=[3−λ5−1−3−λ]A - \lambda I = \begin{bmatrix} 3 & 5 \\ -1 & -3 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} = \begin{bmatrix} 3 - \lambda & 5 \\ -1 & -3 - \lambda \end{bmatrix}

STEP 4

Solve the characteristic equation det⁡(A−λI)=0\det(A - \lambda I) = 0:
Calculate the determinant:
det⁡(A−λI)=(3−λ)(−3−λ)−(5)(−1)\det(A - \lambda I) = (3 - \lambda)(-3 - \lambda) - (5)(-1)
Simplify the expression:
=(3−λ)(−3−λ)+5= (3 - \lambda)(-3 - \lambda) + 5
=(9+3λ−3λ+λ2)+5= (9 + 3\lambda - 3\lambda + \lambda^2) + 5
=λ2−9+5= \lambda^2 - 9 + 5
=λ2−4= \lambda^2 - 4
Set the determinant to zero:
λ2−4=0\lambda^2 - 4 = 0
Solve for λ\lambda:
(λ−2)(λ+2)=0(\lambda - 2)(\lambda + 2) = 0
Thus, the eigenvalues are:
λ1=2,λ2=−2\lambda_1 = 2, \quad \lambda_2 = -2

STEP 5

Find eigenvectors for each eigenvalue:
For λ1=2\lambda_1 = 2:
Solve (A−2I)v=0(A - 2I)\mathbf{v} = \mathbf{0}:
A−2I=[15−1−5]A - 2I = \begin{bmatrix} 1 & 5 \\ -1 & -5 \end{bmatrix}
Set up the system of equations:
1v1+5v2=01v_1 + 5v_2 = 0
−v1−5v2=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=[55−1−1]A + 2I = \begin{bmatrix} 5 & 5 \\ -1 & -1 \end{bmatrix}
Set up the system of equations:
5v1+5v2=05v_1 + 5v_2 = 0
−v1−v2=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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