Question

Find a basis of the image of the matrix $\mathbf{\left[}\begin{array}{cccccc}\mathbf{0}& \mathbf{1}& \mathbf{2}& \mathbf{0}& \mathbf{0}& \mathbf{3}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{1}& \mathbf{0}& \mathbf{4}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{1}& \mathbf{5}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}\end{array}\mathbf{\right]}$.

Short answer

The basis of the image of the matrix $\left[\begin{array}{cccccc}0& 1& 2& 0& 0& 3\\ 0& 0& 0& 1& 0& 4\\ 0& 0& 0& 0& 1& 5\\ 0& 0& 0& 0& 0& 0\end{array}\right]$ is $\left\{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\end{array}\right]\right\}$.

Step-by-step solution

  Concept of redundant vectors; linear independence; basis

Consider vectors ${\overrightarrow{v}}_1,\overrightarrow{v_2},\dots ,{\overrightarrow{v}}_n\in {\mathrm{ℝ}}^n$.

1. We say that a vector ${\overrightarrow{v}}_i$in the list ${\overrightarrow{v}}_1,\overrightarrow{v_2},\dots ,{\overrightarrow{v}}_n\in {\mathrm{ℝ}}^n$is redundant if ${\overrightarrow{v}}_i$is a linear combination of the preceding vectors${\overrightarrow{v}}_1,\overrightarrow{v_2},\dots ,{\overrightarrow{v}}_{i-1}$
2. The vectors localid="1659362216543" ${\overrightarrow{v}}_1,\overrightarrow{v_2},\dots ,{\overrightarrow{v}}_n\in {\mathrm{ℝ}}^n$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).

1. We say that the vectors ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,\dots ,{\overrightarrow{v}}_n$in a subspace V of ${\mathrm{ℝ}}^n$form a basis of V if they span V and are linearly independent.

  Finding the redundant vectors of the given matrix, if any

Let us consider

Let${\overrightarrow{u}}_1=\left[\left.\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right],{\overrightarrow{u}}_2=\left[\left.\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],{\overrightarrow{u}}_3=\left[\begin{array}{c}2\\ 0\\ 0\\ 0\end{array}\right],{\overrightarrow{u}}_4=\left[\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right],{\overrightarrow{u}}_5=\left[\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right.\right],{\overrightarrow{u}}_6=\left[\begin{array}{c}3\\ 4\\ 5\\ 0\end{array}\right.\right]$

Here we have

${\overrightarrow{u}}_1=0.{\overrightarrow{u}}_2,{\overrightarrow{u}}_3=2.{\overrightarrow{u}}_2$ and ${\overrightarrow{u}}_6=3{\overrightarrow{u}}_2+4{\overrightarrow{u}}_4+5{\overrightarrow{u}}_5$

$\Rightarrow {\overrightarrow{u}}_1,{\overrightarrow{u}}_3,{\overrightarrow{u}}_6$are redundant vectors and ${\overrightarrow{u}}_2,{\overrightarrow{u}}_4,{\overrightarrow{u}}_5$are linearly independent vectors.

Finding basis of the image of the given matrix

Since the vectors ${\overrightarrow{u}}_2,{\overrightarrow{u}}_4,{\overrightarrow{u}}_5$are linearly independent and the vectors ${\overrightarrow{u}}_1,{\overrightarrow{u}}_3,{\overrightarrow{u}}_6$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$\left\{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right]\right\}$.

  Final Answer

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

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