How to compute coordinates of three points in the standard basis?

Question

I have the orthonormal basis for $b_1 = (-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$, $b_2 =(-\frac{\sqrt{2}}{2}), (-\frac{\sqrt{2}}{2})$ for $\mathbb{R}^2$.

I need to compute the coordinates of the following three points (in the standard basis) in the basis $\{b_1,b_2\}$:

$(0,0)$, $(1,0)$, $(2,3)$

Can anyone help me get going with this? I dont know how to compute the coordinates.

Thanks.

Answer 1

By definition, the coordinates of a point $(x,y)$ in a base $(b_1, b_2)$ are the coefficients $\lambda$ and $\mu$ such that $(x,y)=\lambda b_1 + \mu b_2$.

Let's take $(1,0)$. You have to find $\lambda$ and $\mu$ such that $$(1,0)=\lambda \left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) + \mu \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$$

This is equivalent to the system $$1 = \lambda \times \left( - \frac{\sqrt{2}}{2} \right) + \mu \times \left( - \frac{\sqrt{2}}{2} \right) \quad \text{and} \quad 0 = \lambda \times \left( \frac{\sqrt{2}}{2} \right) + \mu \times \left( - \frac{\sqrt{2}}{2} \right)$$

You can solve this system and find the solution $$\lambda = \mu = - \frac{\sqrt{2}}{2}$$

Those are the coordinates of $(1,0)$ in the basis $(b_1, b_2)$.

You can do the same with the other vectors !

(Hint : for $(0,0)$, the answer is straightforward !)

Answer 2

Recall that in general the matrix

$$M=[b_1\, b_2]$$

represents the change of basis from the basis $\{b_1,b_2\}$ to the standard basis therefore

$$v_S=Mv_B \iff v_B=M^{-1}v_S$$

and $M^{-1}$ represents the change of basis from the standard basis to the basis $\{b_1,b_2\}$.

In that particular case since $\{b_1,b_2\}$ is an orthonormal basis we have $M^{-1}=M^T$.

Answer 3

For a point $(x_1,x_2)$, you want $c_1b_1+c_2b_2=(x_1\ x_2)^T$. This is a system of linear equations. Thus place the vectors $b_1,b_2$ in the columns of a matrix $\mathbf{B}$, and solve $\mathbf{B}(c_1\ c_2)^T=(x_1\ x_2)^T$. The solution $(c_1\ c_2)^T$ then gives the coordinates you are looking for.