91影视

Let A be an $n脳m$matrix such that the rank of A is r. Let$v_1,v_2,鈥�,v_m$be the column vector of A.

So, we can write$A=\left(v_1,v_2,鈥�,v_m\right)$

Since, the rank of A is r, therefore the dimension of the vector space$span\left\{v_1,v_2,鈥�,v_m\right\}$ is also r.

Hence, there exists $u_1,u_2,鈥�,u_r$in ${\mathrm{鈩�}}^n$such that $\left\{u_1,u_2,鈥�,u_r\right\}$forms a basis of$span\left\{v_1,v_2,鈥�,v_m\right\}$.

Thus, there exists$b_{i1},b_{i2},鈥�,b_{ir}$ in $\mathrm{鈩�}$such that

$v_i=b_{i1}{u}_1+b_{i2}{u}_2+鈥�+b_{ir}{u}_r鈭赌i=1,2,鈥�,n$

Using this we can write

$\begin{array}{rcl}A& =& \left(b_{11}{u}_1+b_{12}{u}_2+鈥�+b_{1r}{u}_r,鈥�,b_{n1}{u}_1+b_{n2}{u}_2+鈥�+b_{nr}{u}_r\right)\\ & =& \underset{k=1}{\overset{r}{鈭�}}\left(b_{1k}{u}_k,b_{2k}{u}_k,L,b_{nk}{u}_k\right)\\ & =& \underset{k=1}{\overset{r}{鈭�}}{C}_k\\ & & \end{array}$

Where, $C_k=\left(b_{1k}{u}_k,b_{2k}{u}_k,鈥�,b_{nk}{u}_k\right).$Thus every column of the matrix$C_k$ is a scalar multiple of the vector $u_k.$Since $u_k$a member of a basis, therefore it is a nonzero vector. Hence the rank of the matrix$C_{k}is1$. This holds for all$k=1,2,鈥�,r.$

Thus,$A=\underset{k=1}{\overset{r}{鈭�}}{C}_k$ where each $C_k$is a matrix of rank 1.

Hence, any matrix of rank rank be written as the sum of r matrices of rank 1.