4. Find the eigenvalues and a basis for each eigenspace in C². Show all of the work (no calculator). 3 -2 3

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### Problem 4: Finding Eigenvalues and Eigenvectors

**Problem Statement:**
Given the matrix 

\[ 
\begin{pmatrix} 
3 & -2 \\ 
2 & 3 
\end{pmatrix}
\]

find the eigenvalues and a basis for each eigenspace in \(\mathbb{C}^2\). Show all of the work (no calculator).

**Step-by-Step Solution:**

**Step 1: Find the Eigenvalues**

The eigenvalues \(\lambda\) of a matrix can be found by solving the characteristic equation:

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

where \(A\) is the matrix and \(I\) is the identity matrix.

For our matrix:

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

The identity matrix \(I\) in \(\mathbb{C}^2\) is:

\[ 
I = \begin{pmatrix} 
1 & 0 \\ 
0 & 1 
\end{pmatrix} 
\]

The matrix \(A - \lambda I\) is:

\[ 
A - \lambda I = \begin{pmatrix} 
3 - \lambda & -2 \\ 
2 & 3 - \lambda 
\end{pmatrix} 
\]

The characteristic polynomial is given by the determinant:

\[ 
\text{det}(A - \lambda I) = \begin{vmatrix} 
3 - \lambda & -2 \\ 
2 & 3 - \lambda 
\end{vmatrix} 
\]

\[ 
= (3 - \lambda)(3 - \lambda) - (-2)(2) 
\]

\[ 
= (3 - \lambda)^2 - 4 
\]

\[ 
= 9 - 6\lambda + \lambda^2 - 4 
\]

\[ 
= \lambda^2 - 6\lambda + 5 
\]

Set the characteristic polynomial to zero and solve for \(\lambda\):

\[ 
\lambda^2 - 6 \lambda + 5 = 0 
\]

Factoring the quadratic equation, we get:

\[ 
(\lambda - 1)(\lambda - 5) = 0 
\]

Therefore, the eigenvalues are:

\[
Transcribed Image Text:### Problem 4: Finding Eigenvalues and Eigenvectors **Problem Statement:** Given the matrix \[ \begin{pmatrix} 3 & -2 \\ 2 & 3 \end{pmatrix} \] find the eigenvalues and a basis for each eigenspace in \(\mathbb{C}^2\). Show all of the work (no calculator). **Step-by-Step Solution:** **Step 1: Find the Eigenvalues** The eigenvalues \(\lambda\) of a matrix can be found by solving the characteristic equation: \[ \text{det}(A - \lambda I) = 0 \] where \(A\) is the matrix and \(I\) is the identity matrix. For our matrix: \[ A = \begin{pmatrix} 3 & -2 \\ 2 & 3 \end{pmatrix} \] The identity matrix \(I\) in \(\mathbb{C}^2\) is: \[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \] The matrix \(A - \lambda I\) is: \[ A - \lambda I = \begin{pmatrix} 3 - \lambda & -2 \\ 2 & 3 - \lambda \end{pmatrix} \] The characteristic polynomial is given by the determinant: \[ \text{det}(A - \lambda I) = \begin{vmatrix} 3 - \lambda & -2 \\ 2 & 3 - \lambda \end{vmatrix} \] \[ = (3 - \lambda)(3 - \lambda) - (-2)(2) \] \[ = (3 - \lambda)^2 - 4 \] \[ = 9 - 6\lambda + \lambda^2 - 4 \] \[ = \lambda^2 - 6\lambda + 5 \] Set the characteristic polynomial to zero and solve for \(\lambda\): \[ \lambda^2 - 6 \lambda + 5 = 0 \] Factoring the quadratic equation, we get: \[ (\lambda - 1)(\lambda - 5) = 0 \] Therefore, the eigenvalues are: \[
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