find the eigenvalues and eigenvectors of A where A is a 2x2 matrix with entries a=1, b=1, c=-1, d=1

To find the eigenvalues and eigenvectors of matrix \( A \), where:

\[
A = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix},
\]

we will follow these steps:

### Step 1: Find the Eigenvalues
Eigenvalues \( \lambda \) satisfy the characteristic equation:

\[
\det(A - \lambda I) = 0,
\]
where \( I \) is the identity matrix. So, we calculate \( A - \lambda I \):

\[
A - \lambda I = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 - \lambda & 1 \\ -1 & 1 - \lambda \end{pmatrix}.
\]

Now, the determinant is:

\[
\det(A - \lambda I) = \det\begin{pmatrix} 1 - \lambda & 1 \\ -1 & 1 - \lambda \end{pmatrix} = (1 - \lambda)(1 - \lambda) - (-1)(1) = (1 - \lambda)^2 + 1.
\]

Simplifying:

\[
(1 - \lambda)^2 + 1 = 0 \quad \Rightarrow \quad (1 - \lambda)^2 = -1 \quad \Rightarrow \quad 1 - \lambda = \pm i,
\]
where \( i \) is the imaginary unit. Solving for \( \lambda \):

\[
\lambda = 1 \pm i.
\]

Thus, the eigenvalues are:

\[
\lambda_1 = 1 + i, \quad \lambda_2 = 1 - i.
\]

### Step 2: Find the Eigenvectors
To find the eigenvectors, we solve \( (A - \lambda I) \mathbf{v} = 0 \) for each eigenvalue.

#### Eigenvector for \( \lambda_1 = 1 + i \):
We solve:

\[
(A - (1 + i) I) \mathbf{v} = 0.
\]

Substitute \( \lambda_1 = 1 + i \) into \( A - \lambda I \):

\[
A - (1 + i) I = \begin{pmatrix} 1 - (1 + i) & 1 \\ -1 & 1 - (1 + i) \end{pmatrix} = \begin{pmatrix} -i & 1 \\ -1 & -i \end{pmatrix}.
\]

Now, solve \( \begin{pmatrix} -i & 1 \\ -1 & -i \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \).

From the first row:

\[
-i v_1 + v_2 = 0 \quad \Rightarrow \quad v_2 = i v_1.
\]

Thus, the eigenvector corresponding to \( \lambda_1 = 1 + i \) is:

\[
\mathbf{v_1} = \begin{pmatrix} 1 \\ i \end{pmatrix}.
\]

#### Eigenvector for \( \lambda_2 = 1 - i \):
Similarly, for \( \lambda_2 = 1 - i \), we solve:

\[
A - (1 - i) I = \begin{pmatrix} 1 - (1 - i) & 1 \\ -1 & 1 - (1 - i) \end{pmatrix} = \begin{pmatrix} i & 1 \\ -1 & i \end{pmatrix}.
\]

Solving \( \begin{pmatrix} i & 1 \\ -1 & i \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \), from the first row:

\[
i v_1 + v_2 = 0 \quad \Rightarrow \quad v_2 = -i v_1.
\]

Thus, the eigenvector corresponding to \( \lambda_2 = 1 - i \) is:

\[
\mathbf{v_2} = \begin{pmatrix} 1 \\ -i \end{pmatrix}.
\]

### Final Answer:
The eigenvalues and corresponding eigenvectors of \( A \) are:

1. \( \lambda_1 = 1 + i \), with eigenvector \( \mathbf{v_1} = \begin{pmatrix} 1 \\ i \end{pmatrix} \).
2. \( \lambda_2 = 1 - i \), with eigenvector \( \mathbf{v_2} = \begin{pmatrix} 1 \\ -i \end{pmatrix} \).

Would you like further details or have any questions?

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Here are 5 related questions:
1. How do you interpret complex eigenvalues geometrically?
2. How do eigenvectors behave when associated with real vs. complex eigenvalues?
3. How can the eigenvectors be normalized?
4. What happens if the matrix \( A \) is not invertible?
5. How would you apply this method to a 3x3 matrix?

**Tip:** Complex eigenvalues often indicate rotational behavior in the system associated with the matrix, especially in 2D transformations.

Discover how to find the eigenvalues and eigenvectors of a 2x2 matrix A = [[1, 1], [-1, 1]] with step-by-step calculations. Learn the method to solve the characteristic equation and calculate complex eigenvalues (λ = 1 ± i) and corresponding eigenvectors. Perfect for advanced linear algebra students.

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