Finding Eigenvalues and Eigenvectors

1. So we find $\lambda$s that make $\det (A-\lambda I)=0$
2. Then for each eigenvalue, we solve $(A-\lambda I)\vec{x}=\vec{0}$.

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• Finding them isn’t too bad.
• We start with the definition and then convert it into a homogenous equation:

$$(A-\lambda I)\vec{x}=\vec{0}$$

• The above equation basically represents x getting squished into the origin, this only happens when the det = 0.
• Also, by Invertible Matrix Theorem, we know that the non-trivial solutions exist only if the matrix is not invertible.
• So for non-trivial solutions to exist, $\det (A-\lambda I)=0$, since det is non-zero for invertible matrices.

To solve for this, we define the Characteristic Polynomial of $A$:

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

• We find the $\lambda$s that make $C_A(\lambda )=0$.
• This is just a polynomial, so we solve it like normal.