Question #235230

Find the eigenvalues and corresponding eigenvectors of matrix G given below.

G= [2 -3]

[4 -5]


1
Expert's answer
2021-09-10T14:35:02-0400

G=[2345]G= \begin{bmatrix} 2 & -3 \\ 4 & -5 \end{bmatrix}

Finding the eigenvalues:

det((2345)λ(1001))\det \left(\begin{pmatrix}2&-3\\ 4&-5\end{pmatrix}-λ\begin{pmatrix}1&0\\ 0&1\end{pmatrix}\right)

=(2λ(3)040(5)λ)=\begin{pmatrix}2-λ&\left(-3\right)-0\\ 4-0&\left(-5\right)-λ\end{pmatrix}

(2λ)(5λ)(3)4\left(2-λ\right)\left(-5-λ\right)-\left(-3\right)\cdot \:4

λ2+3λ+2=0λ^2+3λ+2=0

Solving this λ2+3λ+2=0λ^2+3λ+2=0

λ=1,λ=2λ=-1,\:λ=-2

Therefore eigenvalues are: -1 and -2

Calculating Eigenvectors for :λ=1: λ=-1

=(2345)(1)(1001)= \begin{pmatrix}2&-3\\ 4&-5\end{pmatrix}-\left(-1\right)\begin{pmatrix}1&0\\ 0&1\end{pmatrix}

=(2345)(1001)=\begin{pmatrix}2&-3\\ 4&-5\end{pmatrix}-\begin{pmatrix}-1&0\\ 0&-1\end{pmatrix}

=(3344)=\begin{pmatrix}3&-3\\ 4&-4\end{pmatrix}

To solve (3344)(xy)=(00),\begin{pmatrix}3&-3\\ 4&-4\end{pmatrix} \begin{pmatrix} x\\y\end{pmatrix} = \begin{pmatrix} 0\\0\end{pmatrix}, reduce the matrix

=(11)= \begin{pmatrix} 1\\1 \end{pmatrix} as the eigenvectors for λ=1λ=-1

Calculating Eigenvectors for :λ=2: λ=-2

=(2345)(2)(1001)= \begin{pmatrix}2&-3\\ 4&-5\end{pmatrix}-\left(-2\right)\begin{pmatrix}1&0\\ 0&1\end{pmatrix}

=(4343)=\begin{pmatrix}4&-3\\ 4&-3\end{pmatrix}

To solve (4343)(xy)=(00),\begin{pmatrix}4&-3\\ 4&-3\end{pmatrix} \begin{pmatrix} x\\y\end{pmatrix} = \begin{pmatrix} 0\\0\end{pmatrix}, reduce the matrix

=(34)= \begin{pmatrix} 3\\4 \end{pmatrix} as the eigenvectors for λ=2λ=-2

The Eigenvectors for (2345)\begin{pmatrix}2&-3 \\ 4 & -5\end{pmatrix} becomes =(11),(34)=\begin{pmatrix}1\\ 1\end{pmatrix},\:\begin{pmatrix}3\\ 4\end{pmatrix}


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