Matrix analysis notes (II)

$\lambda$ matrix

set up $a_{ij}(\lambda )(i=1...m,j=1...n)$ It's a number field $F$ Upper polynomial , said
$$A(\lambda )=\left[\begin{array}{cccc}{a}_{11}(\lambda )& a_{12}(\lambda )& \cdots & a_{1n}(\lambda )\\ a_{21}(\lambda )& a_{22}(\lambda )& \cdots & a_{2n}(\lambda )\\ ⋮& ⋮& \ddots & ⋮\\ a_{m1}(\lambda )& a_{m2}(\lambda )& \cdots & a_{mn}(\lambda )\end{array}\right]$$ Is a polynomial matrix or $\lambda$ matrix
among $a_{ij}(\lambda )(i=1...m,j=1...n)$ The highest number of times is $A(\lambda )$ frequency

For example, digital matrix , Characteristic matrix $\lambda E-A$

Rank

if $\lambda$ matrix $A(\lambda )$ There is $r(r\ge 1)$ The order is not zero , And all $r+1$ Order subformula （ If any ） It's all zero , said $A(\lambda )$ The rank of is $r$, Write it down as $rankA(\lambda )=r$, The rank of the zero matrix is $0$

Elementary transformation

• That's ok （ Column ） transposition
• Multiply lines by numbers （ Column ）
• That's ok （ Column ） Doubling

Where the multiple is $\varphi (\lambda )$, by $\lambda$ A polynomial of

That's ok （ Column ） Transformation is equivalent to left （ Right ） Multiply the corresponding elementary matrix

Equivalent

if $A(\lambda )$ After a finite number of elementary transformations, it becomes $B(\lambda )$ said $A(\lambda )$ And $B(\lambda )$ Equivalent , Write it down as $A(\lambda )\simeq B(\lambda )$
Equivalence relation satisfies

• reflexivity
• symmetry
• Transitivity

Smith Standard shape

For any nonzero $\lambda$ matrix $A(\lambda )$ Are equivalent to one “ Diagonal matrix ”
$$A(\lambda )\simeq \left[\begin{array}{cccccc}{d}_1(\lambda )& & & & & \\ & \ddots & & & & \\ & & d_r(\lambda )& & & \\ & & & 0& & \\ & & & & \ddots & \\ & & & & & 0\end{array}\right]$$
among $r\ge 1$,$d_i(\lambda )$ The coefficient of the first term is $1$, And $d_i(\lambda )|d_{i+1}(\lambda )$
In this form $\lambda$ Matrix scale Smith Standard shape
$d_1(\lambda ),...,d_r(\lambda )$ call $A(\lambda )$ Invariant factors

Uniqueness

$A(\lambda )$ by $\lambda$ Matrix and $rank(A(\lambda ))=r$ For any positive integer $k,1\le k\le r$,$A(\lambda )$ There must be non-zero $k$ Order subformula ,$A(\lambda )$ All $k$ The coefficient of the first term of the order formula is $1$ The greatest common factor of $D_k(\lambda )$ be called $A(\lambda )$ Of $k$ Determinant factor of order

obviously , if $rank(A(\lambda ))=r$, Then determinant factors share $r$ individual

equivalent $\lambda$ A matrix has the same determinant factor of each order , So they have the same rank

$\lambda$ matrix $A(\lambda )$ Of Smith The canonical form is the only

The determinant factor of each order of the canonical form is
$$\begin{gathered} D_1(\lambda )=d_1(\lambda ) \\ {D}_2(\lambda )=d_1(\lambda )d_2(\lambda ) \\ ⋮ \\ {D}_r(\lambda )=d_1(\lambda )d_2(\lambda )\cdots d_r(\lambda ) \end{gathered}$$
Thus there are
$$\begin{gathered} d_1(\lambda )=D_1(\lambda ) \\ {d}_2(\lambda )=\frac{D_2(\lambda )}{D_1(\lambda )} \\ ⋮ \\ {d}_r(\lambda )=\frac{D_r(\lambda )}{D_{r-1}(\lambda )} \end{gathered}$$
because $A(\lambda )$ With the above Smith The canonical form has the same determinant factor of each order
therefore $A(\lambda )$ The determinant factor of each order of is $D_1(\lambda ),...,D_r(\lambda )$

Given $A(\lambda )$ The determinant factors for each row have been determined

So the invariant factor
$$d_1(\lambda ),...,d_r(\lambda )$$
Uniquely determined by determinant factor ,$A(\lambda )$ Of Smith The canonical form is the only

Equivalent

• $\lambda$ matrix $A(\lambda )$ And $B(\lambda )$ The necessary and sufficient condition for equivalence is that the determinant factors of each order are the same
• $\lambda$ matrix $A(\lambda )$ And $B(\lambda )$ The necessary and sufficient condition for equivalence is that they have the same invariant factor

Elementary factors

set up $\lambda$ matrix $A(\lambda )$ The invariant factor is $d_1(\lambda ),...,d_r(\lambda )$, In the complex number field, they are decomposed into the power product of the first-order factor
$$\begin{gathered} d_1(\lambda )=(\lambda -{\lambda }_1{)}^{e_{11}}(\lambda -{\lambda }_1{)}^{e_{12}}...(\lambda -{\lambda }_1{)}^{e_{1s}} \\ {d}_2(\lambda )=(\lambda -{\lambda }_1{)}^{e_{21}}(\lambda -{\lambda }_1{)}^{e_{22}}...(\lambda -{\lambda }_1{)}^{e_{2s}} \\ ⋮ \\ {d}_r(\lambda )=(\lambda -{\lambda }_1{)}^{e_{r1}}(\lambda -{\lambda }_1{)}^{e_{r2}}...(\lambda -{\lambda }_1{)}^{e_{rs}} \end{gathered}$$
among ${\lambda }_1,...,{\lambda }_s$ Are mutually exclusive plural numbers ,$e_{ij}$ Is a nonnegative integer , All factors with exponent greater than zero $(\lambda -{\lambda }_j{)}^{e_{ij}},e_{ij}>0,i=1...r,j=1...s$ call $A(\lambda )$ Elementary factors
And there are
$$\begin{gathered} 0\le e_{11}\le e_{21}\le ...\le e_{r1} \\ 0\le e_{12}\le e_{22}\le ...\le e_{r2} \\ ⋮ \\ 0\le e_{1s}\le e_{2s}\le ...\le e_{rs} \end{gathered}$$
$\lambda$ matrix $A(\lambda )$ and $B(\lambda )$ The necessary and sufficient conditions for equivalence are the same rank and elementary factor

if $\lambda$ matrix
$$A(\lambda )=\left[\begin{array}{cccc}{A}_1(\lambda )& & & \\ & A_1(\lambda )& & \\ & & \ddots & \\ & & & A_t(\lambda )\end{array}\right]$$
be $A_1(\lambda ),...,A_t(\lambda )$ All elementary factors are $A(\lambda )$ All elementary factors

$\lambda$ matrix
$$A(\lambda )=\left[\begin{array}{cccccc}{f}_1(\lambda )& & & & & \\ & \ddots & & & & \\ & & f_t(\lambda )& & & \\ & & & 0& & \\ & & & & \ddots & \\ & & & & & 0\end{array}\right]$$
be $f_1(\lambda ),...,f_t(\lambda )$ The powers of all first-order factors are all $A(\lambda )$ All elementary factors of

The number matrix is similar

For numeric matrices $A$, We call $\lambda I-A$ Determinant factor / The invariant factor is $A$ Determinant factor / Invariant factors , call $\lambda I-A$ The primary factor of is $A$ The elementary factor of

set up $A,B$ For two $n$ Order digital matrix , that $A$ And $B$ The necessary and sufficient condition for similarity is that their characteristic matrices $\lambda I-A$ And $\lambda I-B$ Equivalent

For any numerical matrix $A$,$|\lambda I-A|\ne 0$ be
$$rank(\lambda I-A)=n$$

Two square matrices of the same order $A,B$ The necessary and sufficient condition for similarity is that they have the same elementary factor
Two square matrices of the same order $A,B$ The necessary and sufficient condition for similarity is that they have the same determinant factor / Invariant factors

Square matrix ruoerdan standard form

call $n_i$ Order matrix
$$J_i=\left[\begin{array}{ccccc}{a}_i& 1& & & \\ & a_i& 1& & \\ & & \ddots & \ddots & \\ & & & \ddots & 1\\ & & & & a_i\end{array}\right]$$
by Jordan block
$(\lambda I-J_i)$ The determinant factor is
$$\begin{gathered} D_{n_i}(\lambda )=(\lambda -a_i{)}^{n_i} \\ {D}_{n_i-1}(\lambda )=...=D_1(\lambda )=1 \end{gathered}$$
therefore $J_i$ The primary factor is $(\lambda -a_i{)}^{n_i}$

if $J_1...J_s$ All for Jordan block , Then it is called quasi diagonal matrix
$$J=\left[\begin{array}{cccc}{J}_1& & & \\ & J_2& & \\ & & \ddots & \\ & & & J_s\end{array}\right]$$
by Jordan Standard
$J$ The primary factor of is $(\lambda -a_1{)}^{n_1},(\lambda -a_2{)}^{n_2},...,(\lambda -a_s{)}^{n_s}$

set up $A$ by $n$ Square matrix , The primary factor is
$$(\lambda -a_1{)}^{n_1},...,(\lambda -a_s{)}^{n_s}$$
be $A\sim J=diag(J_1,...,J_S)$

Jordan Block ordering is not important

$n$ Order matrix $A$ The necessary and sufficient condition for diagonalization is $A$ The elementary factors of are all primary factors

Similarity transformation matrix

set up $n$ Square matrix $A$ Of Jordan The standard form is $J$, Then there is an invertible matrix $P$ bring $P^{-1}AP=J$, call $P$ Is a similar transformation matrix

nature

• Every Jordan block $J_i$ Corresponding to ${\lambda }_i$ An eigenvector
• For a given ${\lambda }_i$ Corresponding Jordan The number of blocks is equal to ${\lambda }_i$ Geometric multiplicity of
• The eigenvalue ${\lambda }_i$ Corresponding to all Jordan The sum of the block orders is equal to ${\lambda }_i$ Algebraic repetition
• set up $n$ Square matrix $A$ is equivalent to Jordan Standard shape $J$, And $P^{-1}AP=J$, be
  $$\begin{gathered} rank({\lambda }_{i}E-A{)}^l=rank{P}^{-1}({\lambda }_{i}E-A{)}^{l}P=rank(P^{-1}({\lambda }_{i}E-A)P{)}^l=rank({\lambda }_{i}E-J{)}^l \\ l=1,2,... \end{gathered}$$
• about $n_i$ Step Jordan block $J_i$,$(J_i-{\lambda }_{i}E{)}^l$ The rank change of is as follows
  $$\begin{gathered} rank(J_i-{\lambda }_{i}E)=n_i-1, \\ rank(J_i-{\lambda }_{i}E{)}^2=n_i-2, \\ ⋮ \\ rank(J_i-{\lambda }_{i}E{)}^{n_i-1}=1, \\ rank(J_i-{\lambda }_{i}E{)}^h=0,(h\ge n_i) \end{gathered}$$
  if ${\lambda }_j\ne {\lambda }_i$, be
  $$rank(J_j-{\lambda }_{i}E{)}^l=n_j,(l=1,2,...)$$
  That is, with $l$ increase , influence $rank(J-{\lambda }_{i}E{)}^l$ Only ${\lambda }_i$ Corresponding Jordan block

Jordan Another solution of standard type

Yes $n$ Order matrix $A$, if
$$\begin{gathered} rank({\lambda }_{i}I-A)=s_1, \\ rank({\lambda }_{i}I-A{)}^2=s_2, \\ ⋮ \\ rank({\lambda }_{i}I-A{)}^l=s_l, \\ rank({\lambda }_{i}I-A{)}^{l+1}=s_l \end{gathered}$$
For $A$ The characteristic root of the $\lambda ={\lambda }_i$, share $n-s_1$ individual Jordan block , The highest order is $l$, Order $\ge 2$ Of Jordan There are $s_1-s_2$ individual , Order $\ge 3$ Of Jordan There are $s_2-s_3$ individual ,…,$l$ Have of order $s_{l-1}-s_l$ individual