Question

Find a basis of the image of the matrix $\mathbf{\left[}\begin{array}{cc}\mathbf{1}& \mathbf{5}\\ \mathbf{2}& \mathbf{6}\\ \mathbf{3}& \mathbf{7}\\ \mathbf{5}& \mathbf{8}\end{array}\mathbf{\right]}$.

Short answer

The basis of the image of the matrix $\left[\begin{array}{cc}1& 5\\ 2& 6\\ 3& 7\\ 4& 8\end{array}\right]$is $\left\{\left[\begin{array}{c}1\\ 2\\ 3\\ 5\end{array}\right],\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right]\right\}$.

Step-by-step solution

  Concept of redundant vectors; linear independence; basis

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

a. 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 vecto${\overrightarrow{v}}_1.\overrightarrow{v_2}\dots ,{\overrightarrow{v}}_1$

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

c. We say that the vectors ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,....,{\overrightarrow{v}}_n$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$A=\left[\begin{array}{cc}1& 5\\ 2& 6\\ 3& 7\\ 5& 8\end{array}\right]$

Applying,

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

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

Apply$R_3\to 2R_2-R_3$

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

Finding basis of the image of the given matrix

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 $\left\{\left[\begin{array}{c}1\\ 2\\ 3\\ 5\end{array}\right],\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right]\right\}$.

  Final Answer

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 $\left\{\left[\begin{array}{c}1\\ 2\\ 3\\ 5\end{array}\right],\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right]\right\}$.