Consider the matrix A=[2213]
. The eigen decomposition of A
 is given by 

[121−1][4001][132313−13]

⎡⎣13√23√12√−12√⎤⎦[4001]⎡⎣13√22√313√−2√3⎤⎦

⎡⎣13√23√12√−12√⎤⎦[2001]⎡⎣13√22√313√−2√3⎤⎦

⎡⎣13√23√12√−12√⎤⎦[2001]⎡⎣13√12√23√−12√⎤⎦

It seems there are some formatting issues in the problem you provided. Let me clarify the scenario and walk you through the eigen decomposition of the given matrix.

Given a matrix \( A \), the eigen decomposition of \( A \) is typically expressed as:

\[
A = PDP^{-1}
\]

where:
- \( P \) is the matrix whose columns are the eigenvectors of \( A \),
- \( D \) is a diagonal matrix with the eigenvalues of \( A \) on the diagonal.

Let's assume the matrix \( A \) you provided is:

\[
A = \begin{pmatrix} 2 & 2 \\ 1 & 3 \end{pmatrix}
\]

You provided several expressions that seem to be eigen decompositions of different matrices. Let's focus on the correct eigen decomposition for matrix \( A = \begin{pmatrix} 2 & 2 \\ 1 & 3 \end{pmatrix} \).

### Step 1: Finding Eigenvalues
The eigenvalues \( \lambda \) are found by solving the characteristic equation:

\[
\text{det}(A - \lambda I) = 0
\]

\[
\begin{vmatrix} 2 - \lambda & 2 \\ 1 & 3 - \lambda \end{vmatrix} = 0
\]

\[
(2 - \lambda)(3 - \lambda) - 2 \times 1 = \lambda^2 - 5\lambda + 4 = 0
\]

Solving this quadratic equation:

\[
\lambda = \frac{5 \pm \sqrt{25 - 16}}{2} = \frac{5 \pm 3}{2}
\]

\[
\lambda_1 = 4, \quad \lambda_2 = 1

Explore the eigen decomposition of matrix A = [2 2; 1 3] with detailed steps and solutions. Learn about eigenvalues, eigenvectors, and the decomposition formula A = PDP^{-1}. Suitable for advanced undergraduate students studying matrix algebra and eigenvalue theorems.

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