# characteristic values and vectors (of a matrix)

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

 $\Delta(\lambda)\equiv|\lambda I-A|=\Pi_{k=1}^{s}(\lambda-\lambda_{k})^{n_{k}}.$ (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 1.

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

 $g(\mu)=a_{0}\Pi_{i=1}^{t}(\mu-\mu_{i})^{l_{i}},\qquad a_{0}\neq 0,\qquad\sum_{% i=1}^{t}l_{i}=l.$ (2)

On substitution $\mu\mapsto A$, we have

 $g(A)=a_{0}\Pi_{i=1}^{t}(A-\mu_{i}I)^{l_{i}},$ (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  $A$).

 $\displaystyle|g(A)|$ $\displaystyle=a_{0}^{n}\Pi_{i=1}^{t}|(-1)(\mu_{i}I-A)|^{l_{i}}=a_{0}^{n}\Pi_{i% =1}^{t}(-1)^{nl_{i}}|\mu_{i}I-A|^{l_{i}}$ $\displaystyle=a_{0}^{n}(-1)^{n\sum_{i=1}^{t}l_{i}}\Pi_{i=1}^{t}|\mu_{i}I-A|^{l% _{i}}=a_{0}^{n}(-1)^{nl}\Pi_{i=1}^{t}\Delta(\mu_{i})^{l_{i}}$ $\displaystyle=a_{0}^{n}(-1)^{nl}\Pi_{i=1}^{t}[\Pi_{k=1}^{s}(\mu_{i}-\lambda_{k% })^{n_{k}}]^{l_{i}},$

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

 $|g(A)|=\Pi_{k=1}^{s}[a_{0}\Pi_{i=1}^{t}(\lambda_{k}-\mu_{i})^{l_{i}}]^{n_{k}},$

and we may use (2) for $\mu=\lambda_{k}$ to obtain

 $|g(A)|=\Pi_{k=1}^{s}g(\lambda_{k})^{n_{k}}.$ (4)

Finally we substitute the polynomial $g(\mu)$ by $\lambda-g(\mu)$, where $\lambda$ is an arbitrary parameter, getting for (4)

 $\Delta(g(A))\equiv|\lambda I-g(A)|=\Pi_{k=1}^{s}[\lambda-g(\lambda_{k})]^{n_{k% }}.$ (5)

This proves the theorem. ∎

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

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

Let us suppose $\Delta(\lambda)$ is given by

 $\Delta(\lambda)=\lambda^{n}-\sum_{k=1}^{n}c_{k}\lambda^{n-k}.$ (7)

It is clear that the difference $\Delta(\lambda)-\Delta(\mu)$ is divisible by $\lambda-\mu$ without remainder, hence

 $\delta(\lambda,\mu)\equiv\frac{\Delta(\lambda)-\Delta(\mu)}{\lambda-\mu}=% \lambda^{n-1}+(\mu-c_{1})\lambda^{n-2}+(\mu^{2}-c_{1}\mu-c_{2})\lambda^{n-3}+\cdots$ (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

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

which by comparing it with (6)${}_{1}$ we conclude that

 $B(\lambda)=\delta(\lambda I,A)$ (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

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

where ($\mu\mapsto A$ in (8))

 $B_{k}=A^{k}-\sum_{i=1}^{k}c_{i}A^{k-i},\qquad(k=1,\ldots,n-1),$ (12)

which can also be obtained from the recurrence equation

 $B_{k}=AB_{k-1}-c_{k}I,\qquad(k=1,\ldots,n-1;\quad B_{0}=I).$ (13)

What is more,

 $AB_{n-1}-c_{n}I=0\equiv B_{n}.$ (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 $c_{n}=\Delta(0)/(-1)=|-A|/(-1)=(-1)^{n-1}|A|\neq 0$, whenever $A$ be non- singular. From this and from (14) follow that

 $A^{-1}=\frac{1}{c_{n}}B_{n-1}.$ (15)

Let now $\lambda_{c}$ be a characteristic value of $A$, then $\Delta(\lambda_{c})=0$ and (6)${}_{2}$ becomes

 $(\lambda_{c}I-A)B(\lambda_{c})=0.$ (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 $(\lambda_{c}I-A)\mathbf{b}=\mathbf{0}$. That is,

 $A\mathbf{b}=\lambda_{c}\mathbf{b}.$ (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 $\lambda_{c}I-A$ 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 $\lambda_{c}I-A$ 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

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

From (1),

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

Comparing with (7), we have

 $c_{1}=8,\qquad c_{2}=-20,\qquad c_{3}=16.$

Next we use (8),

 $\delta(\lambda,\mu)=\frac{\Delta(\lambda)-\Delta(\mu)}{\lambda-\mu}=\lambda^{2% }+(\mu-8)\lambda+\mu^{2}-8\mu+20,$

so that from (11)

 $B(\lambda)=\delta(\lambda I,A)=\lambda^{2}I+(\underbrace{A-8I}_{B_{1}})\lambda% +\underbrace{A^{2}-8A+20I}_{B_{2}}.$

We will now evaluate $B_{1}$ and $B_{2}$ by using (12) and (13), respectively.

 $B_{1}=A-8I=\begin{bmatrix}-5&-3&2\\ -1&-3&-2\\ -1&3&-8\end{bmatrix},\qquad B_{2}=AB_{1}+20I=\begin{bmatrix}6&6&-4\\ 2&2&4\\ 2&-6&12\end{bmatrix},$

thus $B(\lambda)$ is

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

Also $|A|=16$ and $A^{-1}$ is obtained from (15), i.e.

 $A^{-1}=\frac{1}{16}B_{2}=\frac{1}{8}\begin{bmatrix}3&3&-2\\ 1&1&2\\ 1&-3&6\end{bmatrix}.$

Furthermore,

 $\Delta(\lambda)=(\lambda-2)^{2}(\lambda-4).$

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 $\lambda-2$ ($|B(2)|=0$, i.e. $\lambda=2$ annihilates it), therefore it can be reduced which makes instructive this problem. Thus,

 $C(\lambda)=\begin{bmatrix}\lambda-3&-3&2\\ -1&\lambda-1&-2\\ -1&3&\lambda-6\end{bmatrix},$

which for $\lambda=2$ it becomes

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

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 $\lambda=4$

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

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

Title characteristic values and vectors (of a matrix) CharacteristicValuesAndVectorsofAMatrix 2013-03-22 17:43:58 2013-03-22 17:43:58 perucho (2192) perucho (2192) 6 perucho (2192) Topic msc 15A18 eigenvalues eigenvectors