Eigenstuff

Eigenvector

In Linear Algebra, an Eigenvector is a Coordinate Vector for which, applying the given Linear Transformation has the same effect as multiplying that vector by a scalar. In other words, vectors for which the Output Space of the given linear transformation is proportional to the Input Space.

• Let $A$ be a Square Matrix that represents a linear transformation.
• Let $\vec{v}$ be some Coordinate Vector that we need to solve for.
• Let $\lambda$ be some Scalar that we need to solve for.
• By solving the following equation, we find eigenstuff

$$A\vec{v}=\lambda \vec{v}$$

• Eigenvalues are nonzero scalar values $\lambda$ that solve the equation
• Eigenvectors are coordinate vectors $\vec{v}$ that solve the equation
• Eigenspaces are the spaces that extend from the eigenvectors (the spans of the eigenvectors)
• Each eigenvector has a corresponding eigenvalue, i.e. each eigenvector has a specific amount that the linear transformation scales it by
• The following GIF shows a linear transformation scaling 2D space. The red lines are the eigenspaces of the transformation. Any coordinate vector part of those spaces is an eigenvector.

Eigenvalue

$\lambda$ is an eigenvalue for the transformation

• $A\vec{v}=\lambda \vec{v}⟹\det (A-\lambda I)=0$
• Solve for $\lambda$ in $\det (A-\lambda I)=0$, which yields the Characteristic Equation for this system, e.g. in a 2D systems it is ${\lambda }^2-\text{tr}A\cdot \lambda +\det A=0$.

• $\lambda >0⟹A\vec{v},\vec{v}$ point same direction
• $\lambda <0⟹A\vec{v},\vec{v}$ point opposite direction
• $\lambda$ can be complex even if nothing else in the equation is
• Eigenvalues cannot be determined from the reduced version of a matrix ⭐
  • i.e. row reductions change the eigenvalues of a matrix
• The diagonal elements of a triangular matrix are its eigenvalues.
• A invertible iff 0 is not an eigenvalue of A.
• Stochastic matrices have an eigenvalue equal to 1.
• If ${\vec{v}}_1,{\vec{v}}_2,...,{\vec{v}}_k$ are eigenvectors that correspond to distinct eigenvalues, then ${\vec{v}}_1,{\vec{v}}_2,...,{\vec{v}}_k$ are linearly independent

Defective

An eigenvalue is defective if and only if it does not have a complete Set of Linearly Independent eigenvectors.
$\lambda =0⟹\text{ Defective}$ Due to $\mathrm{Im}\lambda$‘s contribution, $\mathrm{Re}\lambda =0/⟹\text{ Defective}$

Neutral Eigenvalue

$$\forall \lambda s.t.\mathrm{Re}\lambda =0$$

Eigenspace

• the span of the eigenvectors that correspond to a particular eigenvalue

• $Nul(A-\lambda I)$