prove that if r is a matrix in echelon form, then a basis for row(R) consists of the nonzero rows of R

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

Round 14.6837 to the nearest HUNDREDTHS

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