eigenvalue and eigenvector

Exercise:

Let’s find the eigenvalue and eigenvector of this vector

(1) $\begin{align*}M = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix}\end{align*}$

Solution:

Step 1: Characteristic Polynomial

For the definition of eigenvalue and eigenvector, we got eigenvector $x \not = 0$ to satisfy $Ax = \lambda x$ , which can be rewritten as $(A- \lambda I)x=0$ .
Because $x \not = 0$ , $(A-\lambda I)$ is invertible. Its determinant is thus 0, $det(A-\lambda I) = 0$ .

Step 2: Eigenvalue

(2) $\begin{align*}(\begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix}) \times x = 0 \\\begin{bmatrix} 4- \lambda & 2 \\ 1 & 3 - \lambda \end{bmatrix} \times x = 0\\\end{align*}$

Step 3: Eigenvector and eigenspace

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