Respuesta :
Let [tex]$\boldsymbol{r}_{1}, \ldots, \boldsymbol{r}_{k}$[/tex] be the nonzero rows of [tex]$R$[/tex] starting from the 1st row to the [tex]k[/tex]th row.
Step 1: We want to show that
R(R)= span [tex]$\boldsymbol{r}_{1}, \ldots, \boldsymbol{r}_{k}$[/tex]
For any vector [tex]v $\in \mathcal{R}(R)$[/tex], We may write
v= [tex]$c_{1} \boldsymbol{r}_{1}+c_{2} \boldsymbol{r}_{2}+\cdots+c_{k} \boldsymbol{r}_{k}+c_{k+1} \mathbf{0}+\cdots+c_{m} \mathbf{0}$[/tex]
Then v= [tex]$c_{1} r_{1}+c_{2} r_{2}+\cdots+c_{k} r_{k}$[/tex]
So [tex]v[/tex] belongs to span{ [tex]$\boldsymbol{r}_{1}, \ldots, \boldsymbol{r}_{k}$[/tex]}
Thus R(R)[tex]\leq[/tex] Span [tex]$\boldsymbol{r}_{1}, \ldots, \boldsymbol{r}_{k}$[/tex]
But trivially,
span [tex]$\boldsymbol{r}_{1}, \ldots, \boldsymbol{r}_{k}$[/tex]≤ span{[tex]$\boldsymbol{r}_{1}, \ldots, \boldsymbol{r}_{k}$[/tex], 0, . . . , 0} = R(R),
span [tex]$\boldsymbol{r}_{1}, \ldots, \boldsymbol{r}_{k}$[/tex]=R(R)
Step 2: We want to show that [tex]$\boldsymbol{r}_{1}, \ldots, \boldsymbol{r}_{k}$[/tex]are linearly independent. Suppose otherwise,
then we can write
[tex]$c_{1} \boldsymbol{r}_{i_{1}}+\cdots+c_{\ell} \boldsymbol{r}_{i_{\ell}}=\mathbf{0}$[/tex]
From Steps 1 and 2, we have proven that the nonzero rows of R form a basis for the row space.
#SPJ4
