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I'm not too sure why $[0;1;0]$ and $ [0;0;1]$ are a basis of $[1;2;3].$ Can anyone explain please?

From the textbook:

$v_1=[1;2;3].$

We find any basis for $\Bbb R^3$ containing $v_1.$

If we take $x_2=[0;1;0]$ and $x_3=[0;0;1],$ then {$v_1,x_2,x_3$} is clearly a basis for $\Bbb R^3.$

J. W. Tanner
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Mr A
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2 Answers2

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One good way of showing if a given set of vectors $S$ form a basis for a vector space $V$ is to compare the dimension $dim(V)$ of the vector space with the number of linearly independent vectors in the set $S$. In case of $V=\mathbb{R}^3$ and $S=\{x_1,e_2,e_3\}$ with $x_1=\begin{bmatrix}1\\2\\3\end{bmatrix},e_2=\begin{bmatrix}0\\1\\0\end{bmatrix},e_3=\begin{bmatrix}0\\0\\1\end{bmatrix}$.

The dimension $dim(V) = 3$. Therefore we only need to show that set $S$ is linearly independent, since it already has three elements, which can be done by forming a $3\times3$ matrix of form $$\begin{bmatrix}x_1\ e_2\ e_3\end{bmatrix}$$ and applying Gauss-Jordan elimination, from which we can get the number of LI elements in set $S$, which turns out to be $3$.

acarturk
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There are many ways to show that three particular elements of $\Bbb R^3$ form a basis of $\Bbb R^3$.

The basic definition is that they are linearly independent and span $\Bbb R^3$.

To show that $(1,2,3), (0,1,0)$, and $(0,0,1)$ are linearly independent,

assume $a_1(1,2,3)+a_2(0,1,0)+a_3(0,0,1)=(0,0,0)$.

Then $(a_1,2a_1+a_2,3a_1+a_3)=(0,0,0)$ so $(a_1,a_2,a_3)=(0,0,0)$;

that means they are linearly independent.

To show they span $\Bbb R^3$,

note that $(b_1,b_2,b_3)=b_1(1,2,3)+(b_2-2b_1)(0,1,0)+(b_3-3b_1)(0,0,1),$

so any element of $\Bbb R^3$ can be expressed as a linear combination of $(1,2,3), (0,1,0), $ and $(0,0,1)$.

For another method, see the answer to this question, and easily calculate that the determinant

$$\begin{vmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 0 & 1 \\ \end{vmatrix}=1\ne0.$$

J. W. Tanner
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