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

1. We say that a vector${\stackrel{鈫�}{v}}_i$in the listlocalid="1659359112566" ${\stackrel{鈫�}{v}}_1,\stackrel{鈫�}{v_2},鈥�,{\stackrel{鈫�}{v}}_n鈭�{\mathrm{鈩�}}^n$is redundant if${\stackrel{鈫�}{v}}_i$is a linear combination of the preceding vectors${\stackrel{鈫�}{v}}_1,\stackrel{鈫�}{v_2},鈥�,{\stackrel{鈫�}{v}}_{i-1}$
2. The vectorslocalid="1659359124724" ${\stackrel{鈫�}{v}}_1,\stackrel{鈫�}{v_2},鈥�,{\stackrel{鈫�}{v}}_n鈭�{\mathrm{鈩�}}^n$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

$A=\left[\begin{array}{cccccc}0& 1& 2& 0& 3& 0\\ 0& 0& 0& 1& 4& 0\\ 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0& 0\end{array}\right]$

Let${\stackrel{鈫�}{u}}_1=\left[\left.\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right],{\stackrel{鈫�}{u}}_2=\left[\left.\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],{\stackrel{鈫�}{u}}_3=\left[\begin{array}{c}2\\ 0\\ 0\\ 0\end{array}\right],{\stackrel{鈫�}{u}}_4=\left[\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right],{\stackrel{鈫�}{u}}_5=\left[\begin{array}{c}3\\ 4\\ 0\\ 0\end{array}\right.\right],{\stackrel{鈫�}{u}}_6=\left[\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right.\right]$

Here we have

localid="1659356787292" ${\stackrel{鈫�}{u}}_1=0.{\stackrel{鈫�}{u}}_2,{\stackrel{鈫�}{u}}_3=2.{\stackrel{鈫�}{u}}_2$and localid="1659356770031" ${\stackrel{鈫�}{u}}_5=3{\stackrel{鈫�}{u}}_2+4{\stackrel{鈫�}{u}}_4$

localid="1659356744165" $鈬�{\stackrel{鈫�}{u}}_1,{\stackrel{鈫�}{u}}_3,{\stackrel{鈫�}{u}}_5$are redundant vectors and localid="1659356733449" ${\stackrel{鈫�}{u}}_2,{\stackrel{鈫�}{u}}_4,{\stackrel{鈫�}{u}}_6$are linearly independent vectors.

Since the vectors${\stackrel{鈫�}{u}}_2,{\stackrel{鈫�}{u}}_4,{\stackrel{鈫�}{u}}_6$are linearly independent and the vectors${\stackrel{鈫�}{u}}_1,{\stackrel{鈫�}{u}}_3,{\stackrel{鈫�}{u}}_5$are redundant vectors.

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

Hence, basis of image of A is $\left\{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right]\right\}$.

Since the vectors ${\stackrel{鈫�}{u}}_2,{\stackrel{鈫�}{u}}_4,{\stackrel{鈫�}{u}}_6$are linearly independent and the vectors${\stackrel{鈫�}{u}}_1,{\stackrel{鈫�}{u}}_3,{\stackrel{鈫�}{u}}_5$ are redundant vectors

Hence, basis of image of A is $\left\{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right]\right\}$.