Fudan University 2014--2015 first semester (level 14) Advanced Algebra I final exam eighth big question answer

Viii. (10 points) set \ (a,b\) are the \ (m\times n\) matrix, meet the \ (r (a+b) =r (A) +r (B) \), proving: existence \ (m\) Order non-heterogeneous array \ (p\), \ (n\) Order non-heterogeneous (q\), so that \[paq=\begin{pmatrix} I_r & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix},\,\,\,\,pbq=\begin{pmatr IX} 0 & 0 & 0 \ 0 & i_s & 0 \ 0 & 0 & 0 \end{pmatrix}.\]

Certificate Law One (algebraic evidence method)

Set \ (r (A) =r\),  \ (R (B) =s\), then  \ (R (a+b) =r+s\), and existence of \ (m\) Order non-heterogeneous array (s\),  \ (n\) Order non-heterogeneous (t\), so that \[sat=\begin{ Pmatrix} i_r & 0 \ 0 & 0 \end{pmatrix},\,\,\,\,sbt=\begin{pmatrix} b_{11} & B_{12} \ B_{21} & B_{22} \en D{pmatrix},\,\,\,\,s (a+b) T=\begin{pmatrix} i_r+b_{11} & B_{12} \ B_{21} & B_{22} \end{pmatrix}.\] because  \ (R ( A+B) =r+s\), delete the front \ (r\) line of \ (S (a+b) t\), the rank of the after \ (m-r\) line must be greater than or equal to \ (s\), i.e. \ (R (b_{21},b_{22}) \geq s\). On the other hand, we also have \ (R (b_{21},b_{22}) \leq R (b) =s\), so  \ (R (b_{21},b_{22}) =r (b) =s\), thus the Maximal independent group of \ ((b_{21},b_{22}) \) line vectors is also \ ( sbt\) The maximal independent group of line vector groups. Therefore, using the elementary line transformation of the sbt\ (m-r\) line can eliminate the front \ (r\) line of \ (sbt\), and the Elementary column transformation of the post \ ( \) column of sbt\ (n-r\) can eliminate the former \ (sbt\) column of \ (r\), that is, the existence  \ (m\) Order non-heterogeneous array \ (u\), \ (n\) Order non-heterogeneous array (v\), make \[usatv=\begin{pmatrix} i_r & 0 \ 0 & 0 \end{pmatrix},\,\,\,\,usbtv= \begin{pmatrix} 0 & 0 \ 0 & b_{22} \end{pmatrix}.\] There is a  \ (m-r\) Order non-heterogeneous array \ (c\), \ (n-r\) Order non-heterogeneous array \ (d\), making \ (cb_{22 }d=\begin{pmatrix} i_s & 0 \ 0 & 0\end{pmatrix}\). Make \[p=\begin{pmatrix} i_r & 0 \ 0 & C \end{pmatrix}us,\,\,\,\,q=tv\begin{pmatrix} i_r & 0 \ 0 & D \end{p Matrix},\] \ (p\) is  \ (m\) Order non-heterogeneous array, \ (q\) is  \ (n\) Order non-heterogeneous array and satisfies the problem conclusion.

Certificate Law II (geometrical proof method)

Translate the topic into the language of geometry: set \ (v=\mathbb{k}^n\) to \ (n\) Willi vector space, \ (u=\mathbb{k}^m\) for \ (m\) Willi vector space, \ (\varphi_a,\varphi_b:v\to u\) is the matrix \ (a,b\) The linear mapping induced by the left multiplier satisfies \ (r (\varphi_a+\varphi_b) =r (\varphi_a) +r (\varphi_b) \), proving that there is a set of bases of \ (v\), a set of bases of \ (u\), so that \ (\varphi_a , \varphi_b\) The representation matrices under these two sets of bases are \[\begin{pmatrix} I_r & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix}, \,\,\,\,\begin{pmatrix} 0 & 0 & 0 \ 0 & i_s & 0 \ 0 & 0 & 0 \end{pmatrix}.\]

Set \ (r (A) =r\), \ (R (B) =s\), then  \ (R (a+b) =r+s\) .  Note \[r (a+b) \leq R\begin{pmatrix} a \ b \end{pmatrix}\leq R (a) +r (b), \] so \ (R\begin{pmatrix} A \ B \end{pmatrix}=r+s\), thereby \ (\dim (\mathrm{ker\,}\varphi_a\cap\mathrm{ker\,}\varphi_ B) =n-(r+s) \). The dimension formula of child space can be obtained \[\dim (\mathrm{ker\,}\varphi_a+\mathrm{ker\,}\varphi_b) = (n-r) + (n-s)-(n-r-s) =n,\] so there \ (V=\mathrm{ker\, }\varphi_a+\mathrm{ker\,}\varphi_b\). On the other hand, note \[r (a+b) =\dim\mathrm{im\,} (\varphi_a+\varphi_b) \leq \dim (\mathrm{im\,}\varphi_a+\mathrm{im\,}\varphi_b) \leq \dim\mathrm{im\,}\varphi_a+\dim\mathrm{im\,}\varphi_a=r (A) +r (B), \] So \[\mathrm{im\,} (\varphi_a+\varphi_b) =\ Mathrm{im\,}\varphi_a+\mathrm{im\,}\varphi_b=\mathrm{im\,}\varphi_a\oplus\mathrm{im\,}\varphi_b.\cdots (1) \]

(\mathrm{ker\,}\varphi_a\cap\mathrm{ker\,}\varphi_b\) A set of base \ (\{e_{r+s+1},\cdots,e_n\}\) to expand it to \ (\mathrm{Ker\,}\ Varphi_a\) a set of base \ (\{e_{r+1},\cdots,e_n\}\), and then expand it to \ (\mathrm{ker\,}\varphi_b\) a set of base \ (\{e_1,\cdots,e_r,e_{r+s+1},\ cdots,e_n\}\). According to the Gaodai of the 160th page subspace dimension formula of Fudan University, we know that \ (\{e_1,\cdots,e_n\}\) is exactly a group of \ (v=\mathrm{ker\,}\varphi_a+\mathrm{ker\,}\varphi_b\). According to another direct proof of the linear mapping dimension formula (as I said in the fourth chapter): \ (\{ae_1,\cdots,ae_r\}\) is a group of \ (\mathrm{im\,}\varphi_a\), \ (\{be_{r+1},\cdots, be_{r+s}\}\) is a set of bases for \ (\mathrm{im\,}\varphi_b\). Also by (1) know \ (\{ae_1,\cdots,ae_r,be_{r+1},\cdots,be_{r+s}\}\) linear Independent, it can be expanded to \ (u\) a group of base \ (\{ae_1,\cdots,ae_r,be_{r+1},\ cdots,be_{r+s},f_{r+s+1},\cdots,f_m\}\).

Finally easy to verify: \ (\varphi_a,\varphi_b\) in \ (v\) a set of base \ (\{e_1,\cdots,e_n\}\) and \ (u\) a set of base \ (\{ae_1,\cdots,ae_r,be_{r+1},\cdots,  be_{r+s},f_{r+s+1},\cdots,f_m\}\) The representation matrix is the required matrix. \ (\box\)

Fudan University 2014--2015 first semester (level 14) Advanced Algebra I final exam eighth big question answer