Eigenvalues/Eigenvector Equation

Eigenvectors are never the zero vector.

Definition: given a linear transform L: V → V, an eigenvector for/of L is a non-zero vector $\vec v\in V$such that $L(\vec v)$ is a multiple of $\vec v$. That is:

$$ L(\vec v)=\lambda\vec v $$

In matrix form:

$$ A\vec v=\lambda\vec v $$

<aside> 💡 Vectors in the direction of {eigenvector} get scaled by the determinant {eigenvalue}

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Computing the Eigenvalues of a Matrix

<aside> 💡 There are n eigenvalues for a n x n matrix.

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$$ Av=\lambda v\\\ce{isolate}\hspace{.1cm}v\\ (Av-\lambda v)=\vec 0\\ \ce{here, we can't just group!}\\ \therefore(A-I\lambda)v=\vec 0 $$

Knowns

$$ (matrix)\cdot v=\vec 0\\ \implies \vec v\in\ker(matrix)\\ \implies\dim\ker(B)>0\\ \implies B\hspace{.1cm}\ce{is not injective}\\ \therefore\ce{the matrix is not invertible}. $$

Eigenvalue condition

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

Then solve the eigenpolynomial

Eigenpolynomials of 3 x 3 Matrices

Cross out rows by finding the cofactor.

From Eigenvalues to Eigenvectors to Eigenspaces