Question

For every subspace V of$R^3,$ there exists a 3 x 3 matrix A such that data-custom-editor="chemistry" $V=Im\left(A\right)$.

Short answer

The above statement is true.

For every subspace V of$R^3,$there exists a 3 x 3 matrix A such that V = Im(A).

Step-by-step solution

Definition of a basis

Any subset S of a vector space V(K) is called as a basis of V(K), if -

1. S is linearly independent
2. S generates V

i.e. L(S) = V

Considering an example

Let us consider a matrix$A=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]$

Then the set$S=\left\{\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]\right\}$ Let us suppose that the set S is linearly dependent, then there exists scalars $a_1,a_{2}and{a}_3\in V,$not all of them 0,such that

$a_1\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]+a_2\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]+a_3\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

$$\begin{gathered} \Rightarrow a_1+2a_2+3a_3=0 \\ {a}_2+2a_3=0 \\ {a}_3=0 \end{gathered}$$

$\Rightarrow a_1=a_2=a_3=0$

Hence, the vectors are L.I. and only solution of these equation is

$a_1=0,a_2=0,a_3=0$

To prove V = Im (A)

$S=\left\{\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]\right\}$Since the set are linearly independent and so they forms the

basis of V.

$\Rightarrow$Span (S) = V

$\Rightarrow$ker (A) = 0

$\Rightarrow$Im (A) = V

Final Answer

Consider the matrix $A=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]$ then the subset$S=\left\{\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]\right\}$ of subspace V of $R^3,$are linearly independent and so, they forms the basis of V of$R^3,$

$\Rightarrow$span (S) = V

$\Rightarrow$ker (A) = 0

$\Rightarrow$Im (A) = V

Thus for every subspace V of $R^3,$there exists a 3 x 3 matrix A such that V = Im (A).