# Show that the set of the given vectors form a basis in $\mathbb R^3$ and represent the standard basis as a linear combination of these vectors

• Show that the vectors $$\langle 1,0,-1\rangle$$,$$\langle 1,2,1\rangle$$,$$\langle 0,-3,2\rangle$$ from a basis in $$\mathbb R^3$$.
• Represent standard basis in $$\mathbb R^3$$ as a linear combination of the given vectors.

By the definition a nonempty linearly independent subset of vectors $$S=\langle v_1,v_2,...,v_n \rangle$$ from a vector space $$V$$ is a basis of $$V$$ if it spans $$V$$, in other words if every element of $$V$$ can be represented as a linear combination of the vectors in $$S$$.

We've given three vectors in $$\mathbb R^3$$,so we are allowed to use the fact that The columns of an invertible matrix are linearly independent ,thus the matrix is given by:

$$\begin{pmatrix} 1 & 1 & 0 \\ 0& 2 & -3 \\ -1 & 1 & 2 \end{pmatrix}$$

With a nonzero determinant,so it follows that the vectors are linearly independent.

Now it's needed to show that every vector $$\langle x,y,z \rangle$$ in $$\mathbb R^3$$ can be defined as a linear combination of the given vectors,let $$\lambda_1,\lambda_2,\lambda_3$$ be some scalars,then :

$$\begin{bmatrix} 1 & 1 & 0 \\ 0& 2 & -3 \\ -1 & 1 & 2 \\ \end{bmatrix} \left[ \begin{array}{c} \lambda_1 \\ \lambda_2 \\ \lambda_3 \end{array} \right]= \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]$$

Reduce it by elementary row operations gives:

$$\begin{bmatrix} 1 & 0 & 0 \\ 0& 1& 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \left[ \begin{array}{c} \lambda_1 \\ \lambda_2 \\ \lambda_3 \end{array} \right]= \left[ \begin{array}{c} x' \\ y' \\ z' \end{array} \right]$$

Where $$x',y',z'$$ are expressed as $$x,y,z$$.

Thus for every vector $$\langle x,y,z \rangle$$ in $$\mathbb R^3$$ it's possible to find scalars for which the vector can be defined as a linear combination of the given vectors which implies that the vectors span $$\mathbb R^3$$ and so the set containing them is a basis.

Another way would be using the fact that The set of three linearly independent vectors in $$\mathbb R^3$$ is a spanning set.

For the second part I thought that we need to find scalars and solve an equation ,but I guess for each one of the three standard basis we need to solve one system of equations which takes much time,so what else can I do? Does there exist a faster method?

• Do the same row operations on the identity matrix, you'll receive the inverse of your matrix, and basically that's what you're looking for. Commented Nov 29, 2020 at 9:01
• @ Berci ,Do you mean the second part? it would be appreciated if you post your solution with more details.
– user852833
Commented Nov 29, 2020 at 9:25

Let's call your matrix $$A$$.
Showing $$\det A\ne 0$$ is indeed enough to get linear independence of the column vectors, and since they are three, they form a basis of $$\Bbb R^3$$.
For the second part, you're basically looking for the inverse of $$A$$, as its $$i$$th column contains the coordinates of the $$i$$th standard basis vector in the given basis, as we can read it from the matrix product $$AA^{-1}=I$$.
If you already performed a full row reduction to obtain $$I$$ from $$A$$, then the exact same steps in the same order, but starting out from $$I$$ will produce $$A^{-1}$$.
This is because every elementary row operation can be expressed by a left multiplication of a corresponding elementary matrix $$E_i$$, and thus we have $$E_k\dots E_2E_1A=I$$, so $$E_k\dots E_2E_1I=A^{-1}$$.