MIT18.06 Linear Algebra (26)

lec26

Eigenvector of symmetric matrices

• A symmetric matrix has only real eigenvalues

• The eigenvectors can be chosen orthonormal

• proof

  Orthogonality of Eigenvectors of a Symmetric Matrix Corresponding to Distinct Eigenvalues

$$\begin{aligned}Au & = \lambda u\\Av & = \omega v\end{aligned}$$

$$\omega v^\top = v^\top A^\top = v^\top A \\ \omega v^\top u = v^\top Au=\lambda v^\top u \\ (\omega -\lambda )v^\top u = 0 \\ v^\top u = 0$$

We can choose orthonormal eigenvectors. Therefor, for $S=S^\top$ :

$$S=Q \Lambda Q^{-1}=Q \Lambda Q^\top$$

$$S^\top=Q \Lambda^\top Q^\top=Q\Lambda Q^\top=S$$

Why real eigenvalues?

$$Ax=\lambda x\Longrightarrow A\bar{x}=\bar{\lambda }\bar{x}\Longrightarrow\bar{x}^{\top} A^{\top}=\bar{x}^{\top} \bar{\lambda}$$

$$Ax=\lambda x \Longrightarrow \bar{x}^\top Ax=\lambda \bar{x}^{\top}x \\ \bar{x}^\top A^{\top}=\bar{x}^{\top} \bar{\lambda} \Longrightarrow\bar{x}^{\top}A x=\bar{\lambda}\bar{x}^{\top}x$$

$$⁍$$

$\lambda$ is real.

break down $A=Q\Lambda Q^\top$

$$A = \begin{bmatrix} \vdots & \vdots & \\ q_1 & q_1 & \cdots\\ \vdots & \vdots & \end{bmatrix} \begin{bmatrix} \lambda_1 & & \\ & \lambda_2 & \\ & & \ddots \end{bmatrix} \begin{bmatrix} \cdots & q_1^\top & \cdots \\ \cdots & q_2^\top & \cdots \\ & \vdots & \end{bmatrix}=\lambda_{1} q_{1} q_{1}^{\top}+\lambda_{2} q_{2} q_{2}^{\top}+\cdots$$

$q $is a unit vector, so$ q^\top q=1, qq^\top=\frac{qq^\top}{q^\top q}$, which is projection matrix.

every symmetricmatrix is a combination of mutually perpendicular projedtion matrices. (spectral theorem)

for symmetric matrices, the sign of pivots is the same as the sign of eigenvalues.

Intro to Positive definite symmetric matrix

• all eigenvalues are positive
• all pivots are positive

all subdeterminate are positive.