Answer
Eigenvalues are $3+i$ and $3-i$ and the basis vectors for corresponding eigenspaces are $\begin{bmatrix} 2+i & 1 \end{bmatrix}$ and $\begin{bmatrix} 2-i & 1 \end{bmatrix}$ repectively.
Work Step by Step
Let the given matrix be $A$.
\[A = \begin{bmatrix}5 & -5 \\ 1 & 1 \end{bmatrix}\]
The characteristic equation for $A$ is $|A-\lambda I| = 0$. Where $\lambda$ is the eigenvalue. Substituting value of $A$ in the characteristic equation gives
\[\left| \begin{bmatrix}5 & -5 \\ 1 & 1 \end{bmatrix}- \lambda \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \right| = 0\]
\[\implies \left| \begin{bmatrix}5-\lambda & -5 \\ 1 & 1-\lambda \end{bmatrix} \right| = 0\]
\[\implies (5-\lambda)(1-\lambda)+5 = 0\]
\[\implies \lambda^2-6\lambda+10 = 0\]
Solving for $\lambda$ gives
\[\lambda = \frac{6\pm\sqrt{6^2-40}}{2}\]
\[\implies \lambda = 3\pm i\]
In what follows, we find the eigen vector for each eigenvalue of $A$.
Eigenvector corresponding to $3+i$:
The Eigenvector is the solution of the following equation.
\[(A-\lambda I)X = 0\]
where $\lambda = 3+i$ and $X = \begin{bmatrix}x_{1} & x_{2}\end{bmatrix}$.
Substituting for $\lambda$ and $A$ gives
\[\begin{bmatrix} 2-i & -5 \\ 1 & -2-i\end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2}\end{bmatrix} = 0\]
Which implies
\[ (2-i)x_{1}-5x_{2} = 0\]
and
\[x_{1}-(2+i)x_{2} = 0\]
\[\implies x_{1} = (2+i)x_{2}\]
Therefore, we have
\[\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 2+i \\ 1 \end{bmatrix} x_{2} \]
Therefore, $\begin{bmatrix} 2+i & 1 \end{bmatrix}$ forms the basis for eigenspace.
Eigenvector corresponding to $3-i$:
The Eigenvector is the solution of the following equation.
\[(A-\lambda I)X = 0\]
where $\lambda = 3-i$ and $X = \begin{bmatrix}x_{1} & x_{2}\end{bmatrix}$.
Substituting for $\lambda$ and $A$ gives
\[\begin{bmatrix} 2+i & -5 \\ 1 & -2+i\end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2}\end{bmatrix} = 0\]
Which implies
\[ (2+i)x_{1}-5x_{2} = 0\]
and
\[x_{1}+(-2+i)x_{2} = 0\]
\[\implies x_{1} = (2-i)x_{2}\]
Therefore, we have
\[\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 2-i \\ 1 \end{bmatrix} x_{2} \]
Therefore, $\begin{bmatrix} 2-i & 1 \end{bmatrix}$ forms the basis for eigenspace.
