Direct proof of matrix similarity standard type

On the interpretation of similar standard type, the usual advanced algebra textbook introduces the concept of $\lambda$-matrix, transforms the similarity problem of the digital matrix $A $ into the offset problem of the characteristic matrix $\lambda i-a$, and then find the French, invariant factor group and the initial The rational standard type of matrix and the Jordan standard type can be obtained by the same factor group. This teaching method is very general and suitable for beginners, because it is based on proving the existence and uniqueness of standard type, and gives the concrete calculation method of the standard type. The only disadvantage is that the digital matrix $A $ of the similar problem into the characteristics of the matrix $\lambda i-a$, the transition is more blunt, lack of further explanation, so that beginners do not understand. In fact, to really understand this transition requires the model theory of the main ideal region, but the acceptance of this theory is too difficult for freshman students. Therefore, without using the knowledge of the $\lambda$-matrix, the existence proof of the standard type is given directly, which is also a compromise method.

We first give a direct proof of the rational standard type.

The theorem is set $A $ is the $n $-square matrix on a number of domain $\mathbb{k}$, then there is a non-heterogeneous array on the $\mathbb{k}$ $P $, which makes $ $P ^{-1}ap=\mathrm{diag}\{f (D_k (x)), F (d_{k- 1} (x)), \cdots,f (D_1 (x)) \},$$ where $d _i (x) $ is the first polynomial of the very number, $d _i (x) \mid d_{i+1} (x) \, (1\leq I\leq k-1) $, $F (d_i (x)) $ is accompanied by $d _i (x) $ of the friend array.

prove  The order of $A $ $n $ .  when $n =1$, the conclusion is clearly established. If the order is less than $n $ when the conclusion is established, it is proved that the $n $ order matrix $A $ in the case. The minimum polynomial set $A $ is $m (x) $, $\deg m (x) =m$. According to the nature of the Friends, the $F (D_i (x)) $ of the minimum polynomial is $d _i (x) $, and then by the nature of the minimum polynomial, if the conclusion of the theorem is true, then $ $m (x) =[d_k (x), d_{k-1} (x), \cdots,d_1 (x)]=d_k (x). $$ The conclusion from study Questions 9 (6) is that there is $\alpha\in\mathbb{k}^n$, which makes the $\alpha$ minimum polynomial equal to the minimum polynomial of $A $ $m (x) $, making $d _k (x) =m (x) $. Again by the conclusion of Study questions 9 (2), $\alpha,a\alpha,\cdots,a^{m-1}\alpha$ is a set of bases of the cyclic subspace $C (A,\alpha) $, and $A $ limit on $C (A,\alpha) $ on the representation matrix under this set of bases is $F (D_k (x)) $. A set of base $\{\alpha,a\alpha,\cdots,a^{m-1}\alpha,\beta_1,\ that expands $\alpha,a\alpha,\cdots,a^{m-1}\alpha$ to $\mathbb{k}^n$ cdots,\beta_{n-m}\}$, there is $ $A (\alpha,a\alpha,\cdots,a^{m-1}\alpha,\beta_1,\cdots,\beta_{n-m}) = (\alpha,A\alpha,\ CDOTS,A^{M-1}\ALPHA,\BETA_1,\CDOTS,\BETA_{N-M}) \begin{pmatrix} F (D_k (x)) & * \ \ 0 & b \ \end{pmatrix},\ Cdots\cdots (1) $$ where $B $ is the $n-m$ order matrix on the $\mathbb{k}$. By inductive hypothesis, there exists $\mathbb{k}$ on the $n-m$-order non-heterogeneous array $Q $, which makes $ $Q ^{-1}bq=\mathrm{diag}\{f (D_{k-1} (x)), \cdots,f (D_1 (x)) \},$$ where $d _i (x) $ is non- The first number of regularItem, $d _i (x) \mid d_{i+1} (x) \, (1\leq I\leq k-2) $. In particular, $d _{k-1} (x) $ is a minimum polynomial of $B $. By the (1) formula $A $ similar to $\begin{pmatrix} F (D_k (x)) & * \ \ 0 & b \ \end{pmatrix}$, thus by $m (A) =0$ can be $m (B) =0$, and then by a small number of The nature of the formula can be $d _{k-1} (x) \mid m (x) =d_k (x) $. Make $ $P = (\alpha,a\alpha,\cdots,a^{m-1}\alpha,\beta_1,\cdots,\beta_{n-m}) \begin{pmatrix} i_m & 0 \ 0 & Q \ \end{pmatrix}= (\alpha,a\alpha,\cdots,a^{m-1}\alpha,\gamma_1,\cdots,\gamma_{n-m}), $$ $ $AP =p\begin {Pmatrix} F (D_k (x)) & * & \cdots & * \\ & f (D_{k-1} (x)) & & \\  & & \ddots & \ & & & F (D_1 (x)) \ \ \end{pmatrix}.\cdots\cdots (2) $$ The conclusion of the comparison theorem, we just need to prove that the linear independent column vector substitution can be selected appropriately $\{\gamma_1,\ cdots,\gamma_{n-m}\}$, which removes the $*$ number in the first row of the right-hand matrix (2), finally obtains the desired block diagonal array $\mathrm{diag}\{f (D_k (x)), F (D_{k-1} (x)), \cdots,f (D_1 (x ))\}$. Since the discussion of each chunked $F (d_i (x)) $ is similar, we only deal with $F (d_{k-1} (x)) $.

Set $d _{k-1} (x) =x^t+c_{t-1}x^{t-1}+\cdots+c_1t+c_0$, $ $F (D_{k-1} (x)) =\begin{pmatrix} & & &-c_0 \ 1 & &A mp &-c_1 \ & \ddots & & \vdots \ & & 1 &-c_{t-1} \end{pmatrix},$$ so the (2) type can get $ $A \gamma_i=g_i (a) \alpha+\gamma_{i+1}\, (1\leq i\leq t-1), \,\,\,\,a\gamma_t=g_t (a) \alpha-c_0\gamma_1-c_1\gamma_2-\cdots-c_{t-1}\ gamma_t,$$ where $g _i (x) \, (1\leq I\leq t) $ is a polynomial that is determined by the $*$ number less than $m $. The above equation is constantly iterative and can be $$\gamma_i=a^{i-1}\gamma_1+h_i (a) \alpha\, (1\leq i\leq t), \cdots\cdots (3) $$ $ $d _{k-1} (a) \gamma_1=h (a) \ Alpha, \cdots\cdots (4) $$ where $h _i (x), H (x) $ is a polynomial. Set $m (x) =d_k (x) =u (x) D_{k-1} (x) $, because $m (a) =0$, it is on both sides of (4) the same time left multiply $u (a) $ available $$0=m (a) \gamma_1=u (a) d_{k-1} (a) \gamma_1=u (a) h (a) \ alpha.$$ due to the minimal polynomial of the $\alpha$ is $m (x) $, so the study questions 9 (1) can be $m (x) \mid U (x) h (x) $, thereby $d _{k-1} (x) \mid h (x) $. Set $h (x) =d_{k-1} (x) v (x) $, the   (4) type is available   $d _{k-1} (a) (\gamma_1-v (a) \alpha) =0$.  Order $\delta=\gamma_1-v (a) \ alpha$, we assert that $\{\alpha,a\alpha,\cdots,a^{m-1}\alpha,\delta,a\delta,\cdots,a^{t-1}\delta\}$ is linearly independent, otherwise the (3) type is easy to verify that the $\{\alpha,a\alpha,\cdots,a^{m-1}\alpha,\gamma_1,\cdots,\gamma_t\}$ will be linearly correlated, which contradicts the hypothesis. Linearly independent of the aforementioned vector groups can also be introduced $\{\delta,a\delta,\cdots,a^{t-1}\delta\}$ linearly independent, thus $d _{k-1} (x) $ is a minimal polynomial of $\delta$. The same discussion can be done on all the chunking $F (d_i (x)) $, so we find a new set of $\mathbb{k}^n$, which are spelled out as non-heterogeneous $P $, making the conclusion of the theorem .  $\box$

The direct proof of the Jordan standard has already been given in the new white paper's $\s$ 7.2.10, which shows the proof of another inductive approach to example 7.63 of the new white Paper, which proves to be less skillful but more understandable.

Direct proof of matrix similarity standard type