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Expert-verified Found in: Page 131 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Find a basis of the image of the matrix ${\mathbf{\left[}}\begin{array}{cccccc}\mathbf{0}& \mathbf{1}& \mathbf{2}& \mathbf{0}& \mathbf{0}& \mathbf{3}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{1}& \mathbf{0}& \mathbf{4}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{1}& \mathbf{5}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}\end{array}{\mathbf{\right]}}$.

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 $\left\{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\end{array}\right]\right\}$.

See the step by step solution

## Step 1:   Concept of redundant vectors; linear independence; basis

Consider vectors ${\stackrel{\to }{v}}_{1},\stackrel{\to }{{v}_{2}},\dots ,{\stackrel{\to }{v}}_{n}\in {\mathrm{ℝ}}^{n}$.

1. We say that a vector ${\stackrel{\to }{v}}_{i}$ in the list ${\stackrel{\to }{v}}_{1},\stackrel{\to }{{v}_{2}},\dots ,{\stackrel{\to }{v}}_{n}\in {\mathrm{ℝ}}^{n}$ is redundant if ${\stackrel{\to }{v}}_{i}$ is a linear combination of the preceding vectors ${\stackrel{\to }{v}}_{1},\stackrel{\to }{{v}_{2}},\dots ,{\stackrel{\to }{v}}_{i-1}$
2. The vectors localid="1659362216543" ${\stackrel{\to }{v}}_{1},\stackrel{\to }{{v}_{2}},\dots ,{\stackrel{\to }{v}}_{n}\in {\mathrm{ℝ}}^{n}$ 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 ${\stackrel{\to }{v}}_{1},{\stackrel{\to }{v}}_{2},\dots ,{\stackrel{\to }{v}}_{n}$ in a subspace V of ${\mathrm{ℝ}}^{n}$ form a basis of V if they span V and are linearly independent.

## Step 2:   Finding the redundant vectors of the given matrix, if any

Let us consider

Let ${\stackrel{\to }{u}}_{1}=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}],{\stackrel{\to }{u}}_{2}=\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}],{\stackrel{\to }{u}}_{3}=\left[\begin{array}{c}2\\ 0\\ 0\\ 0\end{array}\right],{\stackrel{\to }{u}}_{4}=\left[\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right],{\stackrel{\to }{u}}_{5}=\left[\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right\right],{\stackrel{\to }{u}}_{6}=\left[\begin{array}{c}3\\ 4\\ 5\\ 0\end{array}\right\right]$

Here we have

${\stackrel{\to }{u}}_{1}=0.{\stackrel{\to }{u}}_{2},{\stackrel{\to }{u}}_{3}=2.{\stackrel{\to }{u}}_{2}$ and ${\stackrel{\to }{u}}_{6}=3{\stackrel{\to }{u}}_{2}+4{\stackrel{\to }{u}}_{4}+5{\stackrel{\to }{u}}_{5}$

$⇒{\stackrel{\to }{u}}_{1},{\stackrel{\to }{u}}_{3},{\stackrel{\to }{u}}_{6}$ are redundant vectors and ${\stackrel{\to }{u}}_{2},{\stackrel{\to }{u}}_{4},{\stackrel{\to }{u}}_{5}$ are linearly independent vectors.

## Step 3: Finding basis of the image of the given matrix

Since the vectors ${\stackrel{\to }{u}}_{2},{\stackrel{\to }{u}}_{4},{\stackrel{\to }{u}}_{5}$ are linearly independent and the vectors ${\stackrel{\to }{u}}_{1},{\stackrel{\to }{u}}_{3},{\stackrel{\to }{u}}_{6}$ 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 $\left\{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right]\right\}$.

Since the vectors ${\stackrel{\to }{u}}_{2},{\stackrel{\to }{u}}_{4},{\stackrel{\to }{u}}_{5}$ are linearly independent and the vectors ${\stackrel{\to }{u}}_{1},{\stackrel{\to }{u}}_{3},{\stackrel{\to }{u}}_{6}$ are redundant vectors

Hence, basis of image of A is $\left\{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right]\right\}$. ### Want to see more solutions like these? 