Fudan University 2014--2015 second semester (level 14) Advanced Algebra II Final exam seventh big question answer

Seven, (10 points) set $A, b$ for $n $ order phalanx, meet $AB =ba=0$, $r (A) =r (a^2) $, verify: $ $r (a+b) =r (a) +r (B). $$

Analysis This is a problem, appeared in a variety of high-generation teaching materials or examination of the exam. This topic has at least three kinds of evidence method, the first method uses the block elementary transformation, this needs to be familiar with the proof technique of the matrix rank to think; The second method uses the theory of linear transformation, as long as the geometry concept and related skills mastery, and is not unattainable proof; The third method uses the Jordan standard form theory, which is the simplest and quickest method, and is also one of the important examples of the application of Jordan's standard form theory (unfortunately, forgotten in the new white paper).

Certificate Law One(Algebraic method--using the elementary transformation of blocks)  Similar to the solution theory of linear equations, it can be proved that the necessary and sufficient condition for the solution of matrix equation $AX =b$ is $r (A\mid B) =r (A) $. Considering the matrix equation $A ^2x=a$, we have $ $r (a^2) \leq R (A^2\mid a) =r\big (A (a\mid i_n) \big) \leq R (a), $$ again by $r (a) =r (a^2) $ knowable $r (a^2\mid A) =r (a^2 ) $, thus the matrix equation $A ^2x=a$ has a solution, it may be set as $X =x_0$. Similarly to $A ' $ Repeat the above discussion can be obtained, the matrix equation $YA ^2=a$ have a solution, may be set to $Y =y_0$. The following sub-block matrix is implemented as the following block elementary transformation, the first step is the first block column right multiply $X _0$ added to the second block column; The second step is to add the first block line to the second block line; The third step is the second block column right multiply $-a$ to the first block column, and use the $BA =0$ to simplify; The fourth step is to add the second block row to the left by $-y_0a$ to the first block line, and use the $AB =0$ to simplify: $$\begin{pmatrix} a^2 & 0 \ 0 & B \ \end{pmatrix}\to \begin{pmatr IX} a^2 & a \ 0 & B \ \end{pmatrix}\to \begin{pmatrix} a^2 & a \\ a^2 & a+b \ \end{pmatr ix}\to \begin{pmatrix} 0 & a \ 0 & a+b \ \end{pmatrix}\to \begin{pmatrix} 0 & 0 \ 0 & A + B \ \ \end{pmatrix}.$$  Because the rank of the matrix does not change under the elementary transformation of the block, the basic formula of rank can be obtained $ $r (A) +r (b) =r (a^2) +r (b) =r\begin{pmatrix} a^2 & 0 \ 0 & B \ \ \end{pmatrix}=r\begin{pmatrix} 0 & 0 \ 0 & a+b \ \end{pmatrix}=r (a+b). $$

Certificate Act Two (Geometric method-using linear transformation theory) refer to [question 2014a12] and its solution.

Certificate Law III (Algebraic method--using the Jordan standard form theory) to set $P $ for $n $ order non-heterogeneous array, so that the $P ^{-1}ap$ for the Jordan standard shape. On both sides of the equation $AB =ba=0$ $P ^{-1}$, right multiply by $P $ $$ (p^{-1}ap) (P^{-1}BP) = (P^{-1}BP) (P^{-1}AP) =0.$$ the same to $A, b$ At the same time do similar transformations do not change the rank of the conditions and conclusions, so may wish to start with the assumption that $A $ is the Jordan standard form. by $r (A) =r (a^2) $ $A $ of the Jordan block on the zero eigenvalue is 1 order, it can be set $A =\begin{pmatrix} a_{11} & 0 \ 0 & 0 \end{pmatrix}$, of which $A _{ 11}$ is a non-heterogeneous array. Set $B =\begin{pmatrix} b_{11} & B_{12} \ B_{21} & B_{22} \end{pmatrix}$ for the corresponding chunking, substituting $AB =ba=0$ can be $B _{11}$, $B _{12}$,  $B _{21}$ are all 0 matrices, thus $B =\begin{pmatrix} 0 & 0 \ 0 & b_{22} \end{pmatrix}$, which is the conclusion by the basic formula of rank. $\box$

Fudan University 2014--2015 second semester (level 14) Advanced Algebra II Final exam seventh big question answer