• 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?

  • $\begingroup$ 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. $\endgroup$
    – Berci
    Commented Nov 29, 2020 at 9:01
  • $\begingroup$ @ Berci ,Do you mean the second part? it would be appreciated if you post your solution with more details. $\endgroup$
    – user852833
    Commented Nov 29, 2020 at 9:25

1 Answer 1


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 $E_k\dots E_2E_1I=A^{-1}$.


You must log in to answer this question.