Question #211219

Find the eigenvalues and eigenvectors of the matrices A =

"

1 2

−8 4#

and B =

"

a b

−b a#

.


1
Expert's answer
2021-06-28T16:32:14-0400

Solution.

1.


A=(1284)A=\begin{pmatrix} 1 & 2\\ -8& 4 \end{pmatrix}

Solve equation for finding eigenvalues:

AλI=0|A-\lambda I|=0


1λ284λ=λ25λ+20.\begin{vmatrix} 1- \lambda& 2 \\ -8 & 4-\lambda \end{vmatrix}=\lambda^2-5\lambda+20.

From here we have eigenvaluesλ1=555i2,λ2=5+55i2.\lambda_1=\frac{5-\sqrt{55}i}{2}, \lambda_2=\frac{5+\sqrt{55}i}{2}.

Solve equation for finding the first eigenvector

(Aλ1I)v1=0.(A-\lambda_1 I)v_1=0.

Hence


(3+55i2283+55i2)(1355i1600)\begin{pmatrix} \frac{-3+\sqrt{55}i}{2} & 2 \\ -8 & \frac{3+\sqrt{55}i}{2} \end{pmatrix}\to \begin{pmatrix} 1 & \frac{-3-\sqrt{55}i}{16}\\ 0& 0 \end{pmatrix}

then x2=x2,x1=3+55i16.x_2=x_2, x_1=\frac{3+\sqrt{55}i}{16}.

So, v1=(3+55i161).v_1=\begin{pmatrix} \frac{3+\sqrt{55}i}{16} \\ 1 \end{pmatrix}.

Solve equation for finding the second eigenvector

(Aλ2I)v2=0.(A-\lambda_2 I)v_2=0.

Hence


(355i228355i2)(13+55i1600)\begin{pmatrix} \frac{-3-\sqrt{55}i}{2} & 2 \\ -8 & \frac{3-\sqrt{55}i}{2} \end{pmatrix}\to \begin{pmatrix} 1 & \frac{-3+\sqrt{55}i}{16}\\ 0& 0 \end{pmatrix}

then x2=x2,x1=355i16.x_2=x_2, x_1=\frac{3-\sqrt{55}i}{16}.

So, v2=(355i161).v_2=\begin{pmatrix} \frac{3-\sqrt{55}i}{16} \\ 1 \end{pmatrix}.

2.


B=(abba)B=\begin{pmatrix} a& b\\ -b& a \end{pmatrix}

Solve equation for finding eigenvalues:

BλI=0|B-\lambda I|=0


aλbbaλ=a2λ2+b2.\begin{vmatrix} a- \lambda& b \\ -b & a-\lambda \end{vmatrix}=a^2-\lambda^2+b^2.

From here we have eigenvaluesλ1=a2+b2,λ2=a2+b2.\lambda_1=\sqrt{a^2+b^2}, \lambda_2=-\sqrt{a^2+b^2}.

Solve equation for finding the first eigenvector

(Bλ1I)v1=0.(B-\lambda_1 I)v_1=0.

Hence

(aa2+b2bbaa2+b2)(baa2+b202ba2+b2)\begin{pmatrix} a- \sqrt{a^2+b^2}& b\\ -b & a-\sqrt{a^2+b^2} \end{pmatrix}\to \begin{pmatrix} b & -a-\sqrt{a^2+b^2}\\ 0& 2b\sqrt{a^2+b^2} \end{pmatrix} then x2=2ba2+b2,x1=2aa2+b2+2(a2+b2).x_2=2b\sqrt{a^2+b^2}, x_1=2a\sqrt{a^2+b^2}+2(a^2+b^2).

So, v1=(2aa2+b2+2(a2+b2)2ba2+b2).v_1=\begin{pmatrix} 2a\sqrt{a^2+b^2}+2(a^2+b^2)\\ 2b\sqrt{a^2+b^2} \end{pmatrix}.

Solve equation for finding the second eigenvector

(Bλ2I)v2=0.(B-\lambda_2 I)v_2=0.

Hence

(a+a2+b2bba+a2+b2)(ba+a2+b202ba2+b2)\begin{pmatrix} a+\sqrt{a^2+b^2}& b\\ -b & a+\sqrt{a^2+b^2} \end{pmatrix}\to \begin{pmatrix} -b & a+\sqrt{a^2+b^2}\\ 0& -2b\sqrt{a^2+b^2} \end{pmatrix} then x2=2ba2+b2,x1=2aa2+b22(a2+b2).x_2=-2b\sqrt{a^2+b^2}, x_1=2a\sqrt{a^2+b^2}-2(a^2+b^2).

So, v2=(2aa2+b22(a2+b2)2ba2+b2).v_2=\begin{pmatrix} 2a\sqrt{a^2+b^2}-2(a^2+b^2)\\ -2b\sqrt{a^2+b^2} \end{pmatrix}.

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