characteristic values and vectors (of a matrix)

Over the spectrum $\sigma (A)$ of a matrix $A$, its eigenvalues ${\lambda }_1,{\lambda }_2,\mathrm{\dots },{\lambda }_s$ possess multiplicities $n_1,n_2,\mathrm{\dots },n_s$, respectively, with ${\sum }_{k=1}^s{n}_k=n$. Its associated characteristic polynomial is then factored as

$$\mathrm{\Delta }(\lambda )\equiv |\lambda I-A|={\mathrm{\Pi }}_{k=1}^s{(\lambda -{\lambda }_k)}^{n_k}.$$ (1)

Let us set $\mathrm{mult}({\lambda }_k)=n_k$ for multiplicity of ${\lambda }_k$($k=1,\mathrm{\dots },s$). We will now prove the following theorem.

Theorem 1.

If $\sigma \mathit{}\mathrm{(}A\mathrm{)}\mathrm{=}{\mathrm{\{}{\lambda }_k\mathrm{\}}}_{k\mathrm{=}\mathrm{1}}^s$, $\mathrm{mult}\mathit{}\mathrm{(}{\lambda }_k\mathrm{)}\mathrm{=}{n}_k$, and $g\mathit{}\mathrm{(}\mu \mathrm{)}$ is a scalar polynomial, then $\sigma \mathit{}\mathrm{(}g\mathit{}\mathrm{(}A\mathrm{)}\mathrm{)}\mathrm{=}{\mathrm{\{}g\mathit{}\mathrm{(}{\lambda }_k\mathrm{)}\mathrm{\}}}_{k\mathrm{=}\mathrm{1}}^s$, $\mathrm{mult}\mathit{}\mathrm{(}g\mathit{}\mathrm{(}{\lambda }_k\mathrm{)}\mathrm{)}\mathrm{=}{n}_k$.

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

$$g(\mu )=a_0{\mathrm{\Pi }}_{i=1}^t{(\mu -{\mu }_i)}^{l_i},a_0\ne 0,\sum _{i=1}^t{l}_i=l.$$ (2)

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

$$g(A)=a_0{\mathrm{\Pi }}_{i=1}^t{(A-{\mu }_{i}I)}^{l_i},$$ (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 $A$).

$|g(A)|$ $=a_0^n{\mathrm{\Pi }}_{i=1}^t{|(-1)({\mu }_{i}I-A)|}^{l_i}=a_0^n{\mathrm{\Pi }}_{i=1}^t{(-1)}^{n{l}_i}{|{\mu }_{i}I-A|}^{l_i}$

$=a_0^n{(-1)}^{n{\sum }_{i=1}^t{l}_i}{\mathrm{\Pi }}_{i=1}^t{|{\mu }_{i}I-A|}^{l_i}=a_0^n{(-1)}^{nl}{\mathrm{\Pi }}_{i=1}^t\mathrm{\Delta }{({\mu }_i)}^{l_i}$

$=a_0^n{(-1)}^{nl}{\mathrm{\Pi }}_{i=1}^t{[{\mathrm{\Pi }}_{k=1}^s{({\mu }_i-{\lambda }_k)}^{n_k}]}^{l_i},$

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 $a_0^n=a_0^{{\sum }_{k=1}^s{n}_k}={\mathrm{\Pi }}_{k=1}^s{a}_0^{n_k}$, so that

$$|g(A)|={\mathrm{\Pi }}_{k=1}^s{[a_0{\mathrm{\Pi }}_{i=1}^t{({\lambda }_k-{\mu }_i)}^{l_i}]}^{n_k},$$

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

$$|g(A)|={\mathrm{\Pi }}_{k=1}^{s}g{({\lambda }_k)}^{n_k}.$$ (4)

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

$$\mathrm{\Delta }(g(A))\equiv |\lambda I-g(A)|={\mathrm{\Pi }}_{k=1}^s{[\lambda -g({\lambda }_k)]}^{n_k}.$$ (5)

This proves the theorem. ∎

As an important particular case we have: $\sigma (A^m)={\{{\lambda }_k^m\}}_{k=1}^s$, ($m=0,1,\mathrm{\cdots }$), $\mathrm{mult}({\lambda }_k)=n_k$.

Connection between the characteristic polynomial $\mathrm{\Delta }\mathbf{}\mathrm{(}\lambda \mathrm{)}$ and the adjugate matrix $B\mathbf{}\mathrm{(}\lambda \mathrm{)}$ 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

$$B(\lambda )(\lambda I-A)=\mathrm{\Delta }(\lambda )I\mathit{\hspace{1em}\hspace{1em}}\mathrm{and}\mathit{\hspace{1em}\hspace{1em}}(\lambda I-A)B(\lambda )=\mathrm{\Delta }(\lambda )I.$$ (6)

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

$$\mathrm{\Delta }(\lambda )={\lambda }^n-\sum _{k=1}^n{c}_k{\lambda }^{n-k}.$$ (7)

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

$$\delta (\lambda ,\mu )\equiv \frac{\mathrm{\Delta }(\lambda )-\mathrm{\Delta }(\mu )}{\lambda -\mu }={\lambda }^{n-1}+(\mu -c_1){\lambda }^{n-2}+({\mu }^2-c_1\mu -c_2){\lambda }^{n-3}+\mathrm{\cdots }$$ (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 $\mathrm{\Delta }(A)=0$, then

$$\delta (\lambda I,A)(\lambda I-A)=\mathrm{\Delta }(\lambda )I,$$ (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))

$$B_k=A^k-\sum _{i=1}^k{c}_i{A}^{k-i},(k=1,\mathrm{\dots },n-1),$$ (12)

which can also be obtained from the recurrence equation

$$B_k=A{B}_{k-1}-c_{k}I,(k=1,\mathrm{\dots },n-1;B_0=I).$$ (13)

What is more,

$$A{B}_{n-1}-c_{n}I=0\equiv B_n.$$ (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 $\mathrm{\Delta }(A)=0$ (Cayley-Hamilton), an implicit consequence of generalized Bézout theorem. On the other hand, by setting $\lambda =0$ in (7) we obtain $c_n=\mathrm{\Delta }(0)/(-1)=|-A|/(-1)={(-1)}^{n-1}|A|\ne 0$, 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 $\mathrm{\Delta }({\lambda }_c)=0$ and (6)${}_2$ becomes

$$({\lambda }_{c}I-A)B({\lambda }_c)=0.$$ (16)

Let us assume that $B({\lambda }_c)\ne 0$ and denote by $𝐛$ an arbitrary non-zero column of this matrix. From (16) we have $({\lambda }_{c}I-A)𝐛=\mathrm{𝟎}$. That is,

$$A𝐛={\lambda }_c𝐛.$$ (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 $A^{\mathrm{-}\mathrm{1}}$ can be found from (15). Also if ${\lambda }_c$ is a characteristic value of $A$, the non-zero columns of $B\mathit{}\mathrm{(}{\lambda }_c\mathrm{)}$ are characteristc vectors of A for $\lambda \mathrm{=}{\lambda }_c$.

Example.  We find out the characteristic values and vectors from the matrix

From (1),

Comparing with (7), we have

$$c_1=8,c_2=-20,c_3=16.$$

Next we use (8),

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

so that from (11)

$$B(\lambda )=\delta (\lambda I,A)={\lambda }^{2}I+(\underset{B_1}{\underbrace{A-8I}})\lambda +\underset{B_2}{\underbrace{A^2-8A+20I}}.$$

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

thus $B\mathit{}\mathrm{(}\lambda \mathrm{)}$ is

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

Furthermore,

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

We notice the eigenvalue $\lambda \mathrm{=}\mathrm{2}$ possesses multiplicity $\mathrm{2}$ and also that all the entries of the adjugate $B\mathit{}\mathrm{(}\lambda \mathrm{)}$ are divisible by the binomial $\lambda \mathrm{-}\mathrm{2}$ ($\mathrm{|}B\mathit{}\mathrm{(}\mathrm{2}\mathrm{)}\mathrm{|}\mathrm{=}\mathrm{0}$, i.e. $\lambda \mathrm{=}\mathrm{2}$ annihilates it), therefore it can be reduced which makes instructive this problem. Thus,

which for $\lambda \mathrm{=}\mathrm{2}$ it becomes

From this we get the charactreristic vectors $\mathrm{(}\mathrm{1}\mathrm{,}\mathrm{1}\mathrm{,}\mathrm{1}\mathrm{)}$ by multiplying the first colum by $\mathrm{-}\mathrm{1}$, and also $\mathrm{(}\mathrm{-}\mathrm{3}\mathrm{,}\mathrm{1}\mathrm{,}\mathrm{3}\mathrm{)}$, both correponding to $\lambda \mathrm{=}\mathrm{2}$. Third column is a linear combination of the first two (subtract it). Likewise we find for the another characteristic value $\lambda \mathrm{=}\mathrm{4}$

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

Title	characteristic values and vectors (of a matrix)
Canonical name	CharacteristicValuesAndVectorsofAMatrix
Date of creation	2013-03-22 17:43:58
Last modified on	2013-03-22 17:43:58
Owner	perucho (2192)
Last modified by	perucho (2192)
Numerical id	6
Author	perucho (2192)
Entry type	Topic
Classification	msc 15A18
Synonym	eigenvalues
Synonym	eigenvectors