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characteristic values and vectors (of a matrix) (Topic)

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

$\displaystyle \Delta(\lambda)\equiv \vert\lambda I-A\vert=\Pi_{k=1}^s (\lambda-\lambda_k)^{n_k}.$ (1)

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.

Theorem   If $ \sigma(A)=\{\lambda_k\}_{k=1}^s$, $ \mathrm{mult}(\lambda_k)=n_k$, and $ g(\mu)$ is a scalar polynomial, then $ \sigma(g(A))=\{g(\lambda_k)\}_{k=1}^s$, $ \mathrm{mult}(g(\lambda_k))=n_k$.
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
$\displaystyle g(\mu)=a_0\Pi_{i=1}^t(\mu-\mu_i)^{l_i}, \qquad a_0\neq 0, \qquad \sum_{i=1}^t l_i=l.$ (2)

On substitution $ \mu \mapsto A$, we have
$\displaystyle g(A)=a_0\Pi_{i=1}^t(A-\mu_i I)^{l_i},$ (3)

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$).
$\displaystyle \vert g(A)\vert$ $\displaystyle = a_0^n\Pi_{i=1}^t\vert(-1)(\mu_iI-A)\vert^{l_i}=a_0^n\Pi_{i=1}^t(-1)^{nl_i}\vert\mu_iI-A\vert^{l_i}$    
  $\displaystyle = a_0^n(-1)^{n\sum_{i=1}^t l_i}\Pi_{i=1}^t\vert\mu_iI-A\vert^{l_i}=a_0^n(-1)^{nl}\Pi_{i=1}^t\Delta(\mu_i)^{l_i}$    
  $\displaystyle = a_0^n(-1)^{nl}\Pi_{i=1}^t[\Pi_{k=1}^s(\mu_i-\lambda_k)^{n_k}]^{l_i},$    

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
$\displaystyle \vert g(A)\vert=\Pi_{k=1}^s[a_0\Pi_{i=1}^t(\lambda_k-\mu_i)^{l_i}]^{n_k},$    

and we may use (2) for $ \mu=\lambda_k$ to obtain
$\displaystyle \vert g(A)\vert=\Pi_{k=1}^s g(\lambda_k)^{n_k}.$ (4)

Finally we substitute the polynomial $ g(\mu)$ by $ \lambda-g(\mu)$, where $ \lambda$ is an arbitrary parameter, getting for (4)
$\displaystyle \Delta(g(A))\equiv\vert\lambda I-g(A)\vert=\Pi_{k=1}^s[\lambda-g(\lambda_k)]^{n_k}.$ (5)

This proves the theorem. $ \qedsymbol$
As an important particular case we have: $ \sigma(A^m)=\{\lambda_k^m\}_{k=1}^s$, ( $ m=0, 1, \cdots $), $ \mathrm{mult}(\lambda_k)=n_k$.

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

$\displaystyle B(\lambda)(\lambda I-A)=\Delta(\lambda)I\qquad\mathrm{and}\qquad (\lambda I-A)B(\lambda)=\Delta(\lambda)I.$ (6)

Let us suppose $ \Delta(\lambda)$ is given by
$\displaystyle \Delta(\lambda)=\lambda^n-\sum_{k=1}^n c_k\lambda^{n-k}.$ (7)

It is clear that the difference $ \Delta(\lambda)-\Delta(\mu)$ is divisible by $ \lambda-\mu$ without remainder, hence
$\displaystyle \delta(\lambda,\mu)\equiv \frac{\Delta(\lambda)-\Delta(\mu)}{\lam... ...mu}=\lambda^{n-1}+(\mu-c_1)\lambda^{n-2}+(\mu^2-c_1\mu-c_2)\lambda^{n-3}+\cdots$ (8)

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
$\displaystyle \delta(\lambda I,A)(\lambda I-A)= \Delta(\lambda)I,$ (9)

which by comparing it with (6)$ {}_1$ we conclude that
$\displaystyle B(\lambda)=\delta(\lambda I, A)$ (10)

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
$\displaystyle B(\lambda)=I\lambda^{n-1}+\sum_{k=1}^{n-1}B_k\lambda^{n-k-1},$ (11)

where ( $ \mu\mapsto A$ in (8))
$\displaystyle B_k=A^k-\sum_{i=1}^k c_iA^{k-i}, \qquad (k=1,\ldots, n-1),$ (12)

which can also be obtained from the recurrence equation
$\displaystyle B_k=AB_{k-1}-c_kI, \qquad (k=1, \ldots, n-1; \quad B_0=I).$ (13)

What is more,
$\displaystyle AB_{n-1}-c_nI=0\equiv B_n.$ (14)

(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)=\vert-A\vert/(-1)=(-1)^{n-1}\vert A\vert\neq 0$, whenever $ A$ be non- singular. From this and from (14) follow that
$\displaystyle A^{-1}=\frac{1}{c_n}B_{n-1}.$ (15)

Let now $ \lambda_c$ be a characteristic value of $ A$, then $ \Delta(\lambda_c)=0$ and (6)$ {}_2$ becomes
$\displaystyle (\lambda_c I-A)B(\lambda_c)=0.$ (16)

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,
$\displaystyle A\mathbf{b}=\lambda_c\mathbf{b}.$ (17)

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

$\displaystyle A= \begin{bmatrix}3 & -3 & 2 \\ -1 & 5 & -2 \\ -1 & 3 & 0 \end{bmatrix} .$    

From (1),
$\displaystyle \Delta(\lambda)=\vert\lambda I-A\vert= \begin{vmatrix}\lambda-3 &... ...bda-5 & 2 \\ 1 & -3 & \lambda \end{vmatrix} =\lambda^3-8\lambda^2+20\lambda-16.$    

Comparing with (7), we have
$\displaystyle c_1=8, \qquad c_2=-20, \qquad c_3=16.$    

Next we use (8),
$\displaystyle \delta(\lambda,\mu)=\frac{\Delta(\lambda)-\Delta(\mu)}{\lambda-\mu}=\lambda^2+(\mu-8)\lambda+\mu^2-8\mu+20,$    

so that from (11)
$\displaystyle B(\lambda)=\delta(\lambda I,A)=\lambda^2 I+(\underbrace{A-8I}_{B_1})\lambda+\underbrace{A^2-8A+20I}_{B_2}.$    

We will now evaluate $ B_1$ and $ B_2$ by using (12) and (13), respectively.
$\displaystyle B_1=A-8I= \begin{bmatrix}-5 & -3 & 2 \\ -1 & -3 & -2 \\ -1 & 3 & ... ...AB_1+20I= \begin{bmatrix}6 & 6 & -4 \\ 2 & 2 & 4 \\ 2 & -6 & 12 \end{bmatrix} ,$    

thus $ B(\lambda)$ is
$\displaystyle B(\lambda)= \begin{bmatrix}\lambda^2-5\lambda+6 & -3\lambda+6 & 2... ... -2\lambda+4 \\ -\lambda+2 & 3\lambda-6 & \lambda^2-8\lambda+12 \end{bmatrix} .$    

Also $ \vert A\vert=16$ and $ A^{-1}$ is obtained from (15), i.e.
$\displaystyle A^{-1}=\frac{1}{16}B_2=\frac{1}{8} \begin{bmatrix}3 & 3 & -2 \\ 1 &1 & 2 \\ 1 & -3 & 6 \end{bmatrix} .$    

Furthermore,
$\displaystyle \Delta(\lambda)=(\lambda-2)^2(\lambda-4).$    

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$ ($ \vert B(2)\vert=0$, i.e. $ \lambda=2$ annihilates it), therefore it can be reduced which makes instructive this problem. Thus,
$\displaystyle C(\lambda)= \begin{bmatrix}\lambda-3 & -3 & 2 \\ -1 & \lambda-1 & -2 \\ -1 & 3 & \lambda-6 \end{bmatrix} ,$    

which for $ \lambda=2$ it becomes
$\displaystyle C(2)= \begin{bmatrix}-1 & -3 & 2 \\ -1 & 1 & -2 \\ -1 & 3 & -4 \end{bmatrix} .$    

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$
$\displaystyle C(4)= \begin{bmatrix}1 & -3 & 2 \\ -1 & 3 & -2 \\ -1 & 3 & -2 \end{bmatrix} ,$    

whence we get the eigenvector $ (1,-1,-1)$, being the remaining two columns clearly proportional to the first one.



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Other names:  eigenvalues, eigenvectors
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Cross-references: linear combination, reduced, eigenvalue, inverse, non-singular, addition, conclusion, rank, linearly independent, vector, column, singular, consequence, sides, powers, equate, equation, quotient, Cayley-Hamilton theorem, permutable, remainder, divisible, difference, clear, transpose, cofactor, algebraic complement, adjugate, connection, parameter, product, binomial, square matrix, order, coefficient, determinant, identity matrix, factors, characteristic, polynomial, scalar, characteristic polynomial, multiplicities, matrix, spectrum
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This is version 3 of characteristic values and vectors (of a matrix), born on 2008-01-10, modified 2008-01-22.
Object id is 10182, canonical name is CharacteristicValuesAndVectorsOfAMatrix.
Accessed 1464 times total.

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AMS MSC: 15A18 (Linear and multilinear algebra; matrix theory :: Eigenvalues, singular values, and eigenvectors)
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