Change of basis, am I doing this right?

Question

The basis B in space $\Bbb R^3$, where

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

Give the change of basis $: E\rightarrow B$. E is the standard basis in $\Bbb R^3$.

So the solution I got is:

$$A=\begin{pmatrix}2&1&1 \\ -1&1&0 \\ 1&1&1 \end{pmatrix} $$.

But my friend seem to have different solution. Am I doing something wrong?

Answer 1

Let be $w$ any vector given in the canonical basis and indicate with: $b_1=(1,1,-2)$, $b_2=(0,1,-1)$ and $b_3=(-1,-1,3)$ the three vectors of the basis $\mathcal{B}$, also given with respect to the canonical basis.

We are loking for the coefficient $x_1$, $x_2$ and $x_3$ such that:

$$w=x_1\cdot b_1+x_2\cdot b_2+x_3\cdot b_3$$

or in matrix form:

$$w=B\cdot x$$

$$B=\begin{bmatrix} 1 & 0 & -1\\ 1 & 1 & -1\\ -2 &-1 & 3 \end{bmatrix}$$

Note that $B$ contains as column the vectors of the new basis with respect to the canonical basis. It is important to note that $B$ represent the matrix of change of basis from $\mathcal{B}$ to the canonical.

Thus, the components of any vector $w$ with respect to the new basis are given by:

$$x=B^{-1}\cdot w=A\cdot w$$

$$A=B^{-1}=\begin{bmatrix} 2 & 1 & 1\\ -1 & 1 & 0\\ 1 & 1 & 1 \end{bmatrix}$$

Thus $A=B^{-1}$ represent the matrix of change of basis from the canonical to the new basis $\mathcal{B}$.

So I think you have done well.

Keep in mind that you can always verify the result by the product $AB$.