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Consider vectors ${\stackrel{鈫�}{v}}_1.\stackrel{鈫�}{v_2}鈥�,{\stackrel{鈫�}{v}}_n鈭�{\mathrm{鈩�}}^n$.

a. We say that a vector ${\stackrel{鈫�}{v}}_i$in the list ${\stackrel{鈫�}{v}}_1.\stackrel{鈫�}{v_2}鈥�,{\stackrel{鈫�}{v}}_n鈭�{\mathrm{鈩�}}^n$is redundant if ${\stackrel{鈫�}{v}}_i$is a linear combination of the preceding vecto${\stackrel{鈫�}{v}}_1.\stackrel{鈫�}{v_2}鈥�,{\stackrel{鈫�}{v}}_1$

b. The vectors ${\stackrel{鈫�}{v}}_1.\stackrel{鈫�}{v_2}鈥�,{\stackrel{鈫�}{v}}_n鈭�{\mathrm{鈩�}}^n$are called linearly independent if none of them is

redundant. Otherwise, the vectors are called linearly dependent (meaning

that at least one of them is redundant).

Let us consider$A=\left[\begin{array}{cc}1& 5\\ 2& 6\\ 3& 7\\ 5& 8\end{array}\right]$

Applying,

$\begin{array}{l}{R}_2-2R_1-R_2\\ R_3鈫�3R_1-R_3\\ R_4鈫�5R_1-R_4\end{array}$

$鈬�A=\left[\begin{array}{cc}1& 5\\ 0& 4\\ 0& 8\\ 0& 17\end{array}\right]$

Apply$R_3鈫�2R_2-R_3$

$鈬�A=\left[\begin{array}{cc}1& 5\\ 0& 4\\ 0& 0\\ 0& 17\end{array}\right]$

Since both the vectors are linearly independent as there are two pivot elements, therefore there is no redundant vectors.

And basis of the image of a matrix is equal to the linearly independent vectors.

Hence, basis of image of A is $\left\{\left[\begin{array}{c}1\\ 2\\ 3\\ 5\end{array}\right],\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right]\right\}$.

Since there is no zero row in the reduced echelon form of the given matrix, therefore there is no redundant vectors.

Hence, basis of image of A is $\left\{\left[\begin{array}{c}1\\ 2\\ 3\\ 5\end{array}\right],\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right]\right\}$.