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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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