Over the spectrum $\sigma(A)$ of a matrix $A$ , its eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_s$ possess multiplicities $n_1, n_2, \ldots, n_s$ , respectively, with $\sum_{k=1}^s n_k=n$ . Its associated characteristic polynomial is then factored as \begin{equation} \Delta(\lambda)\equiv |\lambda I-A|=\Pi_{k=1}^s (\lambda-\lambda_k)^{n_k}. \end{equation} Let us set $\mathrm{mult}(\lambda_k)=n_k$ for multiplicity of $\lambda_k$ ($k=1,\ldots,s$ ). We will now prove the following theorem.

Proof. Let $g(\mu)$ be an arbitrary scalar polynomial. We want to find the characteristic values of $g(A)$ . For this purpose we split $g(\mu)$ into linear factors \begin{equation} g(\mu)=a_0\Pi_{i=1}^t(\mu-\mu_i)^{l_i}, \qquad a_0\neq 0, \qquad \sum_{i=1}^t l_i=l. \end{equation}On substitution $\mu \mapsto A$ , we have \begin{equation} g(A)=a_0\Pi_{i=1}^t(A-\mu_i I)^{l_i}, \end{equation}being $I$ the identity matrix. Let us compute the determinant of $g(A)$ . (Coefficient $a_0$ will be powered to $n$ , the order of the square matrix $A$ ).

because on substitution $\lambda \mapsto \mu_i$ in (1). Next we commute the binomial by introducing $(-1)^{nl}$ into the product signs and also we note that $a_0^n=a_0^{\sum_{k=1}^s n_k}=\Pi_{k=1}^s a_0^{n_k}$ , so that \begin{equation*} |g(A)|=\Pi_{k=1}^s[a_0\Pi_{i=1}^t(\lambda_k-\mu_i)^{l_i}]^{n_k}, \end{equation*}and we may use (2) for $\mu=\lambda_k$ to obtain \begin{equation} |g(A)|=\Pi_{k=1}^s g(\lambda_k)^{n_k}. \end{equation}Finally we substitute the polynomial $g(\mu)$ by $\lambda-g(\mu)$ , where $\lambda$ is an arbitrary parameter, getting for (4) \begin{equation} \Delta(g(A))\equiv|\lambda I-g(A)|=\Pi_{k=1}^s[\lambda-g(\lambda_k)]^{n_k}. \end{equation}This proves the theorem.

Connection between the characteristic polynomial $\Delta(\lambda)$ and the adjugate matrix $B(\lambda)$ of $A$ .
As it is well known, the adjugate matrix $B$ of a matrix $A$ there corresponds to the algebraic complement or cofactor matrix of the transpose of $A$ . From this definition we have \begin{equation} B(\lambda)(\lambda I-A)=\Delta(\lambda)I\qquad\mathrm{and}\qquad (\lambda I-A)B(\lambda)=\Delta(\lambda)I. \end{equation}Let us suppose $\Delta(\lambda)$ is given by \begin{equation} \Delta(\lambda)=\lambda^n-\sum_{k=1}^n c_k\lambda^{n-k}. \end{equation}It is clear that the difference $\Delta(\lambda)-\Delta(\mu)$ is divisible by $\lambda-\mu$ without remainder, hence \begin{equation} \delta(\lambda,\mu)\equiv \frac{\Delta(\lambda)-\Delta(\mu)}{\lambda-\mu}=\lambda^{n-1}+(\mu-c_1)\lambda^{n-2}+(\mu^2-c_1\mu-c_2)\lambda^{n-3}+\cdots \end{equation}is a polynomial in $\lambda, \mu$ . If we replace in (8) $(\lambda, \mu)$ by the permutable matrices $(\lambda I, A)$ and recalling that from Cayley-Hamilton theorem $\Delta(A)=0$ , then \begin{equation} \delta(\lambda I,A)(\lambda I-A)= \Delta(\lambda)I, \end{equation}which by comparing it with (6)${}_1$ we conclude that \begin{equation} B(\lambda)=\delta(\lambda I, A) \end{equation}is the desired formula by virtue of the uniqueness of the quotient. Therefore (10) and (8) let to write the adjugate $B(\lambda)$ as the matrix polynomial \begin{equation} B(\lambda)=I\lambda^{n-1}+\sum_{k=1}^{n-1}B_k\lambda^{n-k-1}, \end{equation}where ($\mu\mapsto A$ in (8)) \begin{equation} B_k=A^k-\sum_{i=1}^k c_iA^{k-i}, \qquad (k=1,\ldots, n-1), \end{equation}which can also be obtained from the recurrence equation \begin{equation} B_k=AB_{k-1}-c_kI, \qquad (k=1, \ldots, n-1; \quad B_0=I). \end{equation}What is more, \begin{equation} AB_{n-1}-c_nI=0\equiv B_n. \end{equation}(13) as well as (14) follow inmediately from (6)${}_2$ if we equate the coefficients of equal powers of $\lambda$ on both sides. Also, if we substitute $B_{n-1}$ from (12), into (14), we get $\Delta(A)=0$ (Cayley-Hamilton), an implicit consequence of generalized Bézout theorem. On the other hand, by setting $\lambda=0$ in (7) we obtain $c_n=\Delta(0)/(-1)=|-A|/(-1)=(-1)^{n-1}|A|\neq 0$ , whenever $A$ be non- singular. From this and from (14) follow that \begin{equation} A^{-1}=\frac{1}{c_n}B_{n-1}. \end{equation}Let now $\lambda_c$ be a characteristic value of $A$ , then $\Delta(\lambda_c)=0$ and (6)${}_2$ becomes \begin{equation} (\lambda_c I-A)B(\lambda_c)=0. \end{equation}Let us assume that $B(\lambda_c)\neq 0$ and denote by $\mathbf{b}$ an arbitrary non-zero column of this matrix. From (16) we have $(\lambda_cI-A)\mathbf{b}=\mathbf{0}$ . That is, \begin{equation} A\mathbf{b}=\lambda_c\mathbf{b}. \end{equation}Therefore every non-zero column of $B(\lambda_c)$ determines a characteristic vector corresponding to the characteristic value $\lambda_c$ . Moreover, if to the characteristic value $\lambda_c$ there correspond $l$ linearly independent characteristic vectors, $n-l$ will be the rank of $\lambda_cI-A$ and so the rank of $B(\lambda_c)$ does not exceed $l$ . In particular, if only one characteristic vector there corresponds to $\lambda_c$ , then in $B(\lambda_c)$ the elements of any two columns will be proportional (In such a case $l=1$ , hence the rank of $\lambda_cI-A$ will be $n-1$ ).
In conclusion: If the coefficients of the characteristic polynomial are known, then the adjugate matrix may be found by (10). In addition, if the given matrix $A$ is non-singular, then the inverse matrix $A^{-1}$ can be found from (15). Also if $\lambda_c$ is a characteristic value of $A$ , the non-zero columns of $B(\lambda_c)$ are characteristc vectors of A for $\lambda=\lambda_c$ .

Example. We find out the characteristic values and vectors from the matrix \begin{equation*} A= \begin{bmatrix} 3 & -3 & 2 \\ -1 & 5 & -2 \\ -1 & 3 & 0 \end{bmatrix} . \end{equation*}From (1), \begin{equation*} \Delta(\lambda)=|\lambda I-A|= \begin{vmatrix} \lambda-3 & 3 & -2 \\ 1 & \lambda-5 & 2 \\ 1 & -3 & \lambda \end{vmatrix} =\lambda^3-8\lambda^2+20\lambda-16. \end{equation*}Comparing with (7), we have \begin{equation*} c_1=8, \qquad c_2=-20, \qquad c_3=16. \end{equation*}Next we use (8), \begin{equation*} \delta(\lambda,\mu)=\frac{\Delta(\lambda)-\Delta(\mu)}{\lambda-\mu}=\lambda^2+(\mu-8)\lambda+\mu^2-8\mu+20, \end{equation*}so that from (11) \begin{equation*} B(\lambda)=\delta(\lambda I,A)=\lambda^2 I+(\underbrace{A-8I}_{B_1})\lambda+\underbrace{A^2-8A+20I}_{B_2}. \end{equation*}We will now evaluate $B_1$ and $B_2$ by using (12) and (13), respectively. \begin{equation*} B_1=A-8I= \begin{bmatrix} -5 & -3 & 2 \\ -1 & -3 & -2 \\ -1 & 3 & -8 \end{bmatrix} , \qquad B_2=AB_1+20I= \begin{bmatrix} 6 & 6 & -4 \\ 2 & 2 & 4 \\ 2 & -6 & 12 \end{bmatrix} , \end{equation*}thus $B(\lambda)$ is \begin{equation*} B(\lambda)= \begin{bmatrix} \lambda^2-5\lambda+6 & -3\lambda+6 & 2\lambda-4 \\ -\lambda+2 & \lambda^2-3\lambda+2 & -2\lambda+4 \\ -\lambda+2 & 3\lambda-6 & \lambda^2-8\lambda+12 \end{bmatrix} . \end{equation*}Also $|A|=16$ and $A^{-1}$ is obtained from (15), i.e. \begin{equation*} A^{-1}=\frac{1}{16}B_2=\frac{1}{8} \begin{bmatrix} 3 & 3 & -2 \\ 1 &1 & 2 \\ 1 & -3 & 6 \end{bmatrix} . \end{equation*}Furthermore, \begin{equation*} \Delta(\lambda)=(\lambda-2)^2(\lambda-4). \end{equation*}We notice the eigenvalue $\lambda=2$ possesses multiplicity $2$ and also that all the entries of the adjugate $B(\lambda)$ are divisible by the binomial $\lambda-2$ ($|B(2)|=0$ , i.e. $\lambda=2$ annihilates it), therefore it can be reduced which makes instructive this problem. Thus, \begin{equation*} C(\lambda)= \begin{bmatrix} \lambda-3 & -3 & 2 \\ -1 & \lambda-1 & -2 \\ -1 & 3 & \lambda-6 \end{bmatrix} , \end{equation*}which for $\lambda=2$ it becomes \begin{equation*} C(2)= \begin{bmatrix} -1 & -3 & 2 \\ -1 & 1 & -2 \\ -1 & 3 & -4 \end{bmatrix} . \end{equation*}From this we get the charactreristic vectors $(1,1,1)$ by multiplying the first colum by $-1$ , and also $(-3,1,3)$ , both correponding to $\lambda=2$ . Third column is a linear combination of the first two (subtract it). Likewise we find for the another characteristic value $\lambda=4$ \begin{equation*} C(4)= \begin{bmatrix} 1 & -3 & 2 \\ -1 & 3 & -2 \\ -1 & 3 & -2 \end{bmatrix} , \end{equation*}whence we get the eigenvector $(1,-1,-1)$ , being the remaining two columns clearly proportional to the first one.