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## Jan 5 The Characteristic Equation

Recall to the previous post (and the TFAE), in order for a matrix to have nontrivial solutions to Ax = 0, the matrix cannot be invertible, meaning its determinant is 0. We can use this same principle in calculating eigenvalues and eigenvectors.

# Characteristic Equation

MathJax TeX Test Page $$(A - \lambda{}I)x = 0$$ We want the matrix to not be invertible, so we get this equation: $$\boxed{det(A - \lambda{}I) = 0}$$

# Characteristic Polynomial

This is polynomial formed after calculating the determinant.

MathJax TeX Test Page The characteristic polynomial for the matrix $\begin{bmatrix} 1 & 2 \\ -1 & 4\end{bmatrix}$ is $\begin{vmatrix} 1 - \lambda{} & 2 \\ -1 & 4- \lambda{} \end{vmatrix}$ = $\lambda{}^2 - 5 \lambda{} + 6$ = $(\lambda{} - 2)(\lambda{} - 3)$. So, the eigenvalues are 2 and 3.

## Theorem: Similarity

If two matrices are similar, then they have the same characteristic polynomial, therefore the same eigenvalues (with the same multiplicities). We'll get to what that means in the next section.