고유값과 고유벡터

Eigenvectors

$Ax$ parallel to $x$

What does it mean to be parallel? ⇒ $Ax=\lambda x$

Example 1.1 $\lambda=0$

$\lambda=0$ 이면, Eigen vector들은 $A$의 Null Space가 된다. 즉 $A$가 non singular면 $\lambda=0$인 Eigen Value를 가진다.

Example 1.2 Projection

• For any $x$ in plane , $Px=x$.
  • $\lambda =1$ and the whole plane becomes eigen vector
• For any $x\perp plane$ , $Px=0$
  • eigen value becomes 0

⇒ Therefore, eigen value of projection matrix becomes 0 and 1.

Example 1.3 Permutation

$A=\begin{bmatrix}0&1\\1&0\end{bmatrix}$

For $x=\begin{bmatrix}1\\1\end{bmatrix}, Ax=x$ , $\lambda=1$

For $x=\begin{bmatrix}-1\\1\end{bmatrix}, Ax=-x,\lambda=-1$

Remark 1.4 (Sum of eigenvalues)

sum of diagonals of matrix = sum of eigenvalues

Remark 1.5 (Solving eigenvalues)

$Ax=\lambda x$

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

For $x$ to exist, $A-I\lambda$ should be singular matrix.

Which means, $det(A-\lambda I)=0$

⇒ Find $\lambda$

Example 1.6 (Symmetric Matrix)

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

$det(A-\lambda I)=(3-\lambda)^2-1=\lambda^2-6\lambda+8=0$

• Observe that 6 is the trace, 8 is the determinant of A

$\lambda = 2,4$

Example 1.7 ( Adding $I$)

Example 1.6 의 symmetric matrix는 Example 1.3의 matrix에 $3I$를 더한 것인데, Eigen Value도 3씩 증가한 것을 확인할 수 있다.

Then is it true that eigen values of $A$ and $B$, then $A+B$ has eigenvalues $\alpha+\beta$?

This is false because the corresponding eigen vecotrs may be different.

$Ax=\alpha x$, $By=\beta y$ but it still holds for $I$

Example 1.8 ( Rotation Matrix )

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

Trace = sum of eigenvalues

det = multiple of eigenvalues

This Matrix rotates vector 90 degrees. What vector stays still after 90 degrees rotation?

$det(Q-\lambda I)=\lambda^2+1=0$

$\lambda=i,=i$

Example 1.9 (Transpose)

Transpose Matrix has same eigen values.

$(A-\lambda I)^T=A^T-\lambda I$

Therefore,

$det(A-\lambda I)=det((A-\lambda I)^T)=det(A^T-\lambda I)$

Therefore, they have the same eigen values.