\[
A = \begin{bmatrix}
3 & 1\\
0 & 2
\end{bmatrix}
\]
\(\lambda = 2\) is an eigenvalue. What are the corresponding eigenvectors?
\[
A = \begin{bmatrix}
3 & 1\\
0 & 2
\end{bmatrix}
\]
\(\lambda = 2\) is an eigenvalue. What are the corresponding eigenvectors?
\[
\begin{aligned}
(A - 2I) x &= 0\\\\
\end{aligned}
\]
\[
A = \begin{bmatrix}
3 & 1\\
0 & 2
\end{bmatrix}
\]
\(\lambda = 2\) is an eigenvalue. What are the corresponding eigenvectors?
\[
\begin{aligned}
(A - 2I) x &= 0\\\\
\begin{bmatrix}
1 & 1\\
0 & 0
\end{bmatrix} \begin{bmatrix}
x_1\\
x_2
\end{bmatrix} &= \begin{bmatrix}
0\\
0
\end{bmatrix}
\end{aligned}
\]
\[
A = \begin{bmatrix}
3 & 1\\
0 & 2
\end{bmatrix}
\]
\(\lambda = 2\) is an eigenvalue. What are the corresponding eigenvectors?
\[
\begin{aligned}
(A - 2I) x &= 0\\\\
\begin{bmatrix}
1 & 1\\
0 & 0
\end{bmatrix} \begin{bmatrix}
x_1\\
x_2
\end{bmatrix} &= \begin{bmatrix}
0\\
0
\end{bmatrix}\\\\
x_1 + x_2 &= 0
\end{aligned}
\]
\[
A = \begin{bmatrix}
3 & 1\\
0 & 2
\end{bmatrix}
\]
\(\lambda = 2\) is an eigenvalue. What are the corresponding eigenvectors?
\[
\begin{aligned}
(A - 2I) x &= 0\\\\
\begin{bmatrix}
1 & 1\\
0 & 0
\end{bmatrix} \begin{bmatrix}
x_1\\
x_2
\end{bmatrix} &= \begin{bmatrix}
0\\
0
\end{bmatrix}\\\\
x_1 + x_2 &= 0
\end{aligned}
\] \(\begin{bmatrix}-1\\1\end{bmatrix}\) is an eigenvector of \(A\) with eigenvalue \(2\)
\[
A = \begin{bmatrix}
3 & 1\\
0 & 2
\end{bmatrix}
\]
\(\lambda = 2\) is an eigenvalue. What are the corresponding eigenvectors?
\[
\begin{aligned}
(A - 2I) x &= 0\\\\
\begin{bmatrix}
1 & 1\\
0 & 0
\end{bmatrix} \begin{bmatrix}
x_1\\
x_2
\end{bmatrix} &= \begin{bmatrix}
0\\
0
\end{bmatrix}\\\\
x_1 + x_2 &= 0
\end{aligned}
\] \(\begin{bmatrix}-1\\1\end{bmatrix}\) is an eigenvector of \(A\) with eigenvalue \(2\)
\(\text{span} \left ( \begin{bmatrix}-1\\1\end{bmatrix} \right)\) is the the eigenspace corresponding to \(\lambda = 2\)
If \((\lambda, v)\) is an eigenpair, then \((A - \lambda I) v = 0\)
Then, \(v\) belongs to the nullspace of \(A - \lambda I\)
Every non-zero vector in \(N(A - \lambda I)\) is an eigenvector for \(\lambda\)
Find a basis \(B\) for the nullspace of \(A - \lambda I\)
\(\text{span}(B)\) is nothing but the eigenspace corresponding to \(\lambda\)