how do I change the basis of orthogonal vector? I'm not getting the correct answer

Question

Given vector $\mathbf{v} = \begin{bmatrix} 5 \\ -1 \end{bmatrix}$, $\mathbf{b_1} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and $\mathbf{b_2} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$ all written in the standard basis, what is $\mathbf v$ in the basis defined by $\mathbf{b_1}$ and $\mathbf{b_2}$? You are given that $\mathbf{b_1}$ and $\mathbf{b_2}$ are orthogonal to each other.

a) $\mathbf{v_b} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$

b) $\mathbf{v_b} = \begin{bmatrix} -3 \\ 2 \end{bmatrix}$

c) $\mathbf{v_b} = \begin{bmatrix} 2 \\ 3 \end{bmatrix} $

THESE are the options to choose from , need written soln for reference

Answer 1

If $\bf{b}_1,\bf{b}_2$ are orthogonal, then if $\bf{v}=\alpha \bf{b}_0 + \beta \bf{b}_1$ we can find the coefficients by inner products:

$$\langle \bf{v},\bf{b}_1\rangle = \beta \langle\bf{b}_1, \bf{b}_1\rangle$$ so

$$\beta = \frac{\langle \bf{v},\bf{b}_1\rangle}{\langle\bf{b}_1, \bf{b}_1\rangle}$$

and similarly for $\alpha$. For orthonormal vectors the denominator would have been $1$, here it is $2$ for both.

So for $\begin{bmatrix} 5 \\ -1\end{bmatrix}$ we get $\alpha= \frac{4}{2}=2$ and $\beta= \frac{6}{2}=3$ etc.

Answer 2

Since $\mathcal{B}=\{b_1,b_2\}$ is orthogonal, the coordinates $\alpha_1,\alpha_2$ of any vector $v$ in this base are given by: $$\alpha_1=\frac{\langle v,b_{1}\rangle}{\langle b_1,b_1\rangle}\qquad\text{e}\qquad\alpha_2=\frac{\langle v,b_{2}\rangle}{\langle b_2,b_2\rangle},$$ where $\langle\cdot,\cdot\rangle$ denote the inner product (scalar product). Then $$[v]_\mathcal{B}=\left[\begin{eqnarray}\alpha_1\\\alpha_2\end{eqnarray}\right].$$

Answer 3

You don't need orthogonality; just $\mathbf{v_b} = \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}$ where $a_1\begin{bmatrix} 1 \\ 1 \end{bmatrix}+a_2 \begin{bmatrix} 1 \\ -1 \end{bmatrix}=\begin{bmatrix} 5 \\ -1 \end{bmatrix}.$

Can you solve $a_1+a_2=5$ and $a_1-a_2=-1$ for $a_1$ and $a_2$?