A [linear map](/page/Linear%20Map) is often handed to us as a matrix, but the entries of that matrix are coordinates rather than permanent features of the map. If we change the basis, the same transformation may have a very different-looking matrix. The characteristic polynomial answers the need for a coordinate-independent algebraic fingerprint: it packages the values where $tI-A$ loses invertibility, and from that single polynomial we recover eigenvalues, determinant, trace, and relations among powers of the operator.

The first reason to introduce this polynomial is the eigenvalue problem. The equation $Av=\lambda v$ asks for a scalar $\lambda$ and a nonzero vector $v$. Rewriting it as $(\lambda I-A)v=0$ shows that such a vector exists exactly when $\lambda I-A$ is singular. Determinants turn singularity into one equation in the scalar parameter.

Here $M_n(k)$ denotes the set of $n\times n$ matrices with entries in a field $k$, $I_n$ denotes the $n\times n$ identity matrix, and $k[t]$ denotes the [polynomial ring](/page/Polynomial%20Ring) in the formal variable $t$ with coefficients in $k$. Thus $M_2(\mathbb R)$ is the space of $2\times2$ real matrices.

[example: Searching for Eigenvalues by a Determinant] Let $A=\begin{pmatrix}2&1\cr1&2\end{pmatrix}\in M_2(\mathbb R)$ and let $v=\begin{pmatrix}v_1\cr v_2\end{pmatrix}$. The equation $Av=\lambda v$ has a nonzero solution exactly when the homogeneous system $(\lambda I_2-A)v=0$ has a nonzero solution, which for a $2\times 2$ matrix is equivalent to $\det(\lambda I_2-A)=0$. Here \begin{align*} \lambda I_2-A=\begin{pmatrix}\lambda&0\cr0&\lambda\end{pmatrix}-\begin{pmatrix}2&1\cr1&2\end{pmatrix}=\begin{pmatrix}\lambda-2&-1\cr-1&\lambda-2\end{pmatrix}. \end{align*} Using $\det\begin{pmatrix}x&y\cr z&w\end{pmatrix}=xw-yz$, we get \begin{align*} \det(\lambda I_2-A)=(\lambda-2)(\lambda-2)-(-1)(-1). \end{align*} Thus \begin{align*} \det(\lambda I_2-A)=(\lambda-2)^2-1. \end{align*} Expanding and factoring, \begin{align*} (\lambda-2)^2-1=\lambda^2-4\lambda+4-1. \end{align*} \begin{align*} \lambda^2-4\lambda+4-1=\lambda^2-4\lambda+3. \end{align*} \begin{align*} \lambda^2-4\lambda+3=(\lambda-1)(\lambda-3). \end{align*} Therefore $\det(\lambda I_2-A)=0$ exactly for $\lambda=1$ or $\lambda=3$. For $\lambda=1$, \begin{align*} A-I_2=\begin{pmatrix}1&1\cr1&1\end{pmatrix}, \end{align*} so $(A-I_2)v=0$ means \begin{align*} v_1+v_2=0. \end{align*} The corresponding eigenvectors are the nonzero multiples of $\begin{pmatrix}1\cr-1\end{pmatrix}$. For $\lambda=3$, \begin{align*} A-3I_2=\begin{pmatrix}-1&1\cr1&-1\end{pmatrix}, \end{align*} so $(A-3I_2)v=0$ means \begin{align*} -v_1+v_2=0, \end{align*} equivalently $v_1-v_2=0$. The corresponding eigenvectors are the nonzero multiples of $\begin{pmatrix}1\cr1\end{pmatrix}$. The determinant polynomial has therefore found the two lines on which $A$ acts by scalar multiplication. [/example]

This example also fixes a convention. Some books use $\det(A-tI_n)$, but this page uses $\det(tI_n-A)$ because it gives a monic polynomial in $t$.

## Definition

The determinant condition becomes useful only after we stop substituting a particular scalar and leave the parameter formal. Otherwise we only test one possible eigenvalue at a time, with no single object that records all the values where invertibility fails. The formal determinant below packages that whole obstruction into one polynomial: its roots are candidates for eigenvalues, and its coefficients encode familiar invariants.

[definition: Characteristic Polynomial of a Matrix] Let $k$ be a field, let $n\in\mathbb N$, and let $A\in M_n(k)$. The characteristic polynomial of $A$ is the polynomial $\chi_A(t)\in k[t]$ defined by \begin{align*} \chi_A(t)=\det(tI_n-A). \end{align*} [/definition]

The variable $t$ is formal, so $tI_n-A$ is a matrix with entries in $k[t]$. This guarantees that the determinant is a genuine polynomial rather than a collection of separate scalar tests. Its general coefficient pattern will be recorded later when trace and determinant are compared systematically.

Before using that general coefficient result, it helps to compute the smallest non-scalar case. The next example is needed because the $2\times2$ formula is the model behind many quick characteristic polynomial calculations.

[example: The $2\times2$ Formula] Let $A=\begin{pmatrix}a&b\cr c&d\end{pmatrix}\in M_2(k)$. By the definition of the characteristic polynomial, \begin{align*} \chi_A(t)=\det(tI_2-A). \end{align*} Now \begin{align*} tI_2=\begin{pmatrix}t&0\cr0&t\end{pmatrix}. \end{align*} Subtracting entries gives \begin{align*} tI_2-A=\begin{pmatrix}t-a&0-b\cr0-c&t-d\end{pmatrix}. \end{align*} Hence \begin{align*} tI_2-A=\begin{pmatrix}t-a&-b\cr-c&t-d\end{pmatrix}. \end{align*} Using $\det\begin{pmatrix}x&y\cr z&w\end{pmatrix}=xw-yz$, with $x=t-a$, $y=-b$, $z=-c$, and $w=t-d$, we get \begin{align*} \chi_A(t)=(t-a)(t-d)-(-b)(-c). \end{align*} Since $(-b)(-c)=bc$, \begin{align*} \chi_A(t)=(t-a)(t-d)-bc. \end{align*} Expanding the first product, \begin{align*} (t-a)(t-d)=t^2-td-at+ad. \end{align*} Because $td=dt$ in the field $k$, \begin{align*} t^2-td-at+ad=t^2-(a+d)t+ad. \end{align*} Therefore \begin{align*} \chi_A(t)=t^2-(a+d)t+ad-bc. \end{align*} Here $\operatorname{tr}A=a+d$ and $\det A=ad-bc$, so \begin{align*} \chi_A(t)=t^2-(\operatorname{tr}A)t+\det A. \end{align*} The off-diagonal entries $b$ and $c$ enter only through the determinant term $ad-bc$, while the coefficient of $t$ is determined by the trace. [/example]

A diagonal matrix shows what the polynomial looks like when the coordinate axes are already eigenvector directions. The next example is the baseline for triangular and diagonalizable cases.

[example: Diagonal Matrices] Let $A=\operatorname{diag}(a_1,\dots,a_n)\in M_n(k)$. Since $I_n=\operatorname{diag}(1,\dots,1)$, multiplying by the formal variable $t$ gives \begin{align*} tI_n=\operatorname{diag}(t,\dots,t). \end{align*} Subtracting diagonal entries and subtracting $0$ from every off-diagonal entry gives \begin{align*} tI_n-A=\operatorname{diag}(t-a_1,\dots,t-a_n). \end{align*} To compute the determinant, use the permutation formula. The identity permutation contributes \begin{align*} (t-a_1)(t-a_2)\cdots(t-a_n). \end{align*} Every non-identity permutation $\sigma$ has some index $i$ with $\sigma(i)\ne i$, so the product $\prod_{i=1}^n (tI_n-A)_{i,\sigma(i)}$ contains an off-diagonal entry of the diagonal matrix $tI_n-A$, hence contains a factor $0$. Therefore all non-identity permutation terms vanish, and \begin{align*} \chi_A(t)=\det(tI_n-A)=\prod_{i=1}^n(t-a_i). \end{align*} Thus each diagonal entry $a_i$ appears as a linear factor $t-a_i$, so the roots of the characteristic polynomial are exactly the diagonal entries, counted with their repetitions in the diagonal list. [/example]

## Eigenvalues and Multiplicity

The characteristic polynomial would be much less useful if its roots were only formal artifacts. The main reason it matters is that its roots are exactly the eigenvalues, so the polynomial detects the scalar directions of the operator.

### Eigenvalues and Eigenspaces

To connect the polynomial to geometry, we first name the scalars that act on some nonzero direction by stretching. This definition is needed because the determinant equation will be interpreted through the existence of such directions.

[definition: Eigenvalue] Let $k$ be a field, let $V$ be a [vector space](/page/Vector%20Space) over $k$, and let $T:V\to V$ be a linear map. A scalar $\lambda\in k$ is an eigenvalue of $T$ if there exists a nonzero vector $v\in V$ such that \begin{align*} T(v)=\lambda v. \end{align*} [/definition]

For a fixed scalar $\lambda$, we need to collect all vectors that satisfy the eigenvalue equation. This collection is a subspace, and its dimension will later measure how many independent eigenvectors the polynomial root has produced.

[definition: Eigenspace] Let $k$ be a field, let $V$ be a vector space over $k$, let $T:V\to V$ be a linear map, and let $\lambda\in k$. The eigenspace of $T$ with eigenvalue $\lambda$ is \begin{align*} E_\lambda(T)=\ker(T-\lambda\operatorname{id}_V). \end{align*} [/definition]