### Characteristic polynomial

If $A$ is a matrix in $\mathbb{F}^{n\times n}$ for some field $\mathbb{F}$, then the **characteristic polynomial** $p_A(t)$ is defined as the polynomial obtained by expanding the determinant of $A-t I$ in terms of $t$:

This polynomial offers a straightforward technique of calculating the eigenvalues of a matrix: simply calculate the roots of its characteristic polynomial. Why must the roots of $p_A$ be eigenvalues of $A$? Well, if $A$ has an eigenvector $v$ with corresponding eigenvalue $\lambda$, then $Av=\lambda v$, or $Av=\lambda I v$, meaning that $(A-\lambda I)v=0$, making $A-\lambda I$ a singular matrix. This means that $\det(A-\lambda I)=p_A(\lambda)=0$ for eigenvalues $\lambda$ of $A$.

If $\mathbb{F}$ is algebraically closed, then $p_A(t)$ can be factored into linear factors corresponding to the eigenvalues of $A$. The multiplicities of these linear factors are called the **algebraic multiplicities** of the eigenvalues of $A$. The multiplicities of these factors do not necessarily equal the dimensions of the corresponding eigenspaces, as one might think. For instance, the multiplicity with which $t$ is a factor of $p_A(t)$ is not equal to the dimension of the null space of $A$, but rather the generalized null space $\mathcal{N}(A^n)$.

There is a nice formula for the characteristic polynomial of a rank-one matrix:

**Proposition 1.** If $A=uv^\ast$, then

# linear-algebra

# matrices

# polynomials

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