2.3 KiB
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#math/linear_algebra
definition
#card If a matrix that multiplies a vector results in the same vector that a scalar multiplying that vector would create, that vector is an eigenvector and that scalar is an eigenvalue. ^1653443848519
[; \frac{(0,1) \cdot (1,2)}{(1,2) \cdot (1,2)} \cdot (1,2) = \frac{2}{5} (1,2) = (2/5,4/5) ;]formula to get eigenvalues
#card/reverse
All solutions of the equation \operatorname{det}(A-\lambda I)=0.
example:
get eigenvalues and eigenvectors of $\left[\begin{array}{ll}3 & 1 \ 1 & 3\end{array}\right]$
$\operatorname{det}(A-\lambda I)=\operatorname{det}\left[\begin{array}{cc}3-\lambda & 1 \ 1 & 3-\lambda\end{array}\right]=(3-\lambda)(3-\lambda)-1=\lambda^{2}-6 \lambda+8$
Solve the quadratic equation \lambda^{2}-6 \lambda+8=0. it's \lambda=4 and \lambda=2. These are the eigenvalues of A.
To find the eigenvectors corresponding to the eigenvalue \lambda=4, we compute the null space of A-4 I. We find
[A-4 I \mid \mathbf{0}]=\left[\begin{array}{rr|r}
-1 & 1 & 0 \\
1 & -1 & 0
\end{array}\right] \rightarrow\left[\begin{array}{rr|r}
1 & -1 & 0 \\
0 & 0 & 0
\end{array}\right]
from which it follows that \mathbf{x}=\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right] is an eigenvector corresponding to \lambda=4 if and only if x_{1}-x_{2}=0 or x_{1}=x_{2}. Hence, the eigenspace E_{4}=\left\{\left[\begin{array}{l}x_{2} \\ x_{2}\end{array}\right]\right\}=\left\{x_{2}\left[\begin{array}{l}1 \\ 1\end{array}\right]\right\}= \operatorname{span}\left(\left[\begin{array}{l}1 \\ 1\end{array}\right]\right).
Similarly, for \lambda=2, we have
[A-2 I \mid 0]=\left[\begin{array}{ll|l}
1 & 1 & 0 \\
1 & 1 & 0
\end{array}\right] \rightarrow\left[\begin{array}{ll|l}
1 & 1 & 0 \\
0 & 0 & 0
\end{array}\right]
so \mathbf{y}=\left[\begin{array}{l}y_{1} \\ y_{2}\end{array}\right] is an eigenvector corresponding to \lambda=2 if and only if y_{1}+y_{2}=0 or y_{1}=-y_{2}. Thus, the eigenspace E_{2}=\left\{\left[\begin{array}{c}-y_{2} \\ y_{2}\end{array}\right]\right\}=\left\{y_{2}\left[\begin{array}{r}-1 \\ 1\end{array}\right]\right\}=\operatorname{span}\left(\left[\begin{array}{r}-1 \\ 1\end{array}\right]\right).
(this is from the book)
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