MIT18.06 Linear Algebra (21)

lec21

eigen vector

find a function, in goes vector $x$ , out comes vector $Ax$ .

$Ax $parallel to$ x $i.e.$ Ax=\lambda x$

If $A$ is singular, $\lambda =0$ is an eigenvalue

projection matrix

for any $x$ in the plane, $Px=x$ i.e. $\lambda =1$

for any $x$ perpendicular to the plane, $Px = 0$ , i.e. $\lambda = 0$

All eigenvalues

$n\times n $matrix has$ n$ eigenvalues.

Sum of these eigenvalues = $\operatorname{tr}(A)$ (some if diagonal)

Product of eigenvalues = $\det(A)$

Solve $Ax=\lambda x$

$$(A-\lambda I)x=0$$

$$\Leftrightarrow  A-\lambda I \; \text{is singular} \Leftrightarrow  \det(A-\lambda I)=0$$

E.g.

$$A = \begin{bmatrix} 3 & 1 \\ 1 & 3 \\ \end{bmatrix}$$

$$\det(A-\lambda I)=\begin{vmatrix} 3-\lambda & 1\\ 1 & 3-\lambda \end{vmatrix} =(3-\lambda )^2-1 = \lambda ^2-{\color{Red} 6} \lambda +{\color{Red} 8} \\ (\text{6:trace, 8:det})$$

Example for rotation matrix

$$Q = \begin{bmatrix} 0 & -1\\ 1 & 0\end{bmatrix}$$

$$\mathrm{tr}(A)=0=\lambda_1 + \lambda_2 \\\det(A) = 1 = \lambda_1\lambda_2$$

$$\lambda_1 = i,\lambda_2=-i \text{(conjugated)}$$

if $A$ is symmetric, $\lambda$ will have no imaginary parts.

if $A=-A^\top$ , $\lambda$ will have pure imaginary parts.

Example for triangular matrix

for

$$\begin{bmatrix} 3 & 1\\ 0 & 3\end{bmatrix}$$

$$x_1=\begin{bmatrix}1 \\ 0 \end{bmatrix}$$

no independent $x_2$ .