\[ \huge{T = \begin{bmatrix} 3 & 1\\ 0 & 2 \end{bmatrix}} \]

\[ u = \begin{bmatrix} -1\\ 1 \end{bmatrix} \]

\[ \begin{aligned} Tu &= \begin{bmatrix} -2\\ 2 \end{bmatrix} \end{aligned} \]

\[ u = \begin{bmatrix} -1\\ 1 \end{bmatrix} \]

\[ \begin{aligned} Tu &= \begin{bmatrix} -2\\ 2 \end{bmatrix}\\\\ &= 2u \end{aligned} \]

\[ u = \begin{bmatrix} 1\\ 1 \end{bmatrix} \]

\[ \begin{aligned} Tu &= \begin{bmatrix} 4\\ 2 \end{bmatrix} \end{aligned} \]

\[ u = \begin{bmatrix} 1\\ 1 \end{bmatrix} \]

\[ \begin{aligned} Tu &= \begin{bmatrix} 4\\ 2 \end{bmatrix}\\\\ &\neq \lambda u \end{aligned} \]

For a linear transformation \(T\), a non-zero vector \(v\) is called an eigenvector with eigenvalue \(\lambda\) if:

For a linear transformation \(T\), a non-zero vector \(v\) is called an eigenvector with eigenvalue \(\lambda\) if:

\[ \boxed{\huge{T v = \lambda v}} \]

For a linear transformation \(T\), a non-zero vector \(v\) is called an eigenvector with eigenvalue \(\lambda\) if:

\[ \boxed{\huge{T v = \lambda v}} \]

\((\lambda, v)\) is called an eigenpair of \(T\)

For a matrix \(T\), a non-zero vector \(v\) is called an eigenvector with eigenvalue \(\lambda\) if:

\[ \boxed{\huge{T v = \lambda v}} \]

\((\lambda, v)\) is called an eigenpair of \(T\)

\[ T = \begin{bmatrix} 3 & 1\\ 0 & 2 \end{bmatrix} \]

\[ \left (2, \begin{bmatrix}-1\\1\end{bmatrix} \right) \]
\[ \left (3, \begin{bmatrix}1\\0\end{bmatrix} \right) \]

Why should an eigenvector be non-zero?

Why should an eigenvector be non-zero?

\[ T0 = \lambda 0 \]

Why should an eigenvector be non-zero?

\[ T0 = \lambda 0 \]

How many values can \(\lambda\) take?

If \(u\) is an eigenvector of \(T\) with eigenvalue \(\lambda\), then what can you say about \(ku\)?

\[ \begin{aligned} T(ku) &= \end{aligned} \]

If \(u\) is an eigenvector of \(T\) with eigenvalue \(\lambda\), then what can you say about \(ku\)?

\[ \begin{aligned} T(ku) &= k \cdot Tu \end{aligned} \]

If \(u\) is an eigenvector of \(T\) with eigenvalue \(\lambda\), then what can you say about \(ku\)?

\[ \begin{aligned} T(ku) &= k \cdot Tu\\\\ &= k \cdot \lambda u \end{aligned} \]

If \(u\) is an eigenvector of \(T\) with eigenvalue \(\lambda\), then what can you say about \(ku\)?

\[ \begin{aligned} T(ku) &= k \cdot Tu\\\\ &= k \cdot \lambda u\\\\ &= \lambda \cdot (ku) \end{aligned} \]

If \(u\) is an eigenvector of \(T\) with eigenvalue \(\lambda\), then what can you say about \(ku\)?

\[ \begin{aligned} T(ku) &= k \cdot Tu\\\\ &= k \cdot \lambda u\\\\ &= \lambda \cdot (ku) \end{aligned} \]

\(ku\) is an eigenvector of \(T\) with eigenvalue \(\lambda\), where \(k \neq 0\)

If \(u\) and \(v\) are eigenvectors of \(T\) with eigenvalue \(\lambda\), then what can you say about \(u + v\)?

\[ \begin{aligned} T(u + v) &= \end{aligned} \]

If \(u\) and \(v\) are eigenvectors of \(T\) with eigenvalue \(\lambda\), then what can you say about \(u + v\)?

\[ \begin{aligned} T(u + v) &= Tu + Tv \end{aligned} \]

If \(u\) and \(v\) are eigenvectors of \(T\) with eigenvalue \(\lambda\), then what can you say about \(u + v\)?

\[ \begin{aligned} T(u + v) &= Tu + Tv\\\\ &= \lambda u + \lambda v \end{aligned} \]

If \(u\) and \(v\) are eigenvectors of \(T\) with eigenvalue \(\lambda\), then what can you say about \(u + v\)?

\[ \begin{aligned} T(u + v) &= Tu + Tv\\\\ &= \lambda u + \lambda v\\\\ &= \lambda(u + v) \end{aligned} \]

If \(u\) and \(v\) are eigenvectors of \(T\) with eigenvalue \(\lambda\), then what can you say about \(u + v\)?

\[ \begin{aligned} T(u + v) &= Tu + Tv\\\\ &= \lambda u + \lambda v\\\\ &= \lambda(u + v) \end{aligned} \]

\(u + v\) is an eigenvector of \(T\) with eigenvalue \(\lambda\)

\[ \huge{E = \{u\ \vert\ Tu = \lambda u, u \in \mathbb{R}^n \}} \]