Linear algebra - change of basis - clarification

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I got the following question. I would really appreciate if someone can have a look at it and let me know what I am doing wrong.

Question: Given vectors

$V = \begin{bmatrix}5\\-1\end{bmatrix}, b_1 = \begin{bmatrix}1\\1\end{bmatrix}, b_2 = \begin{bmatrix}1\\-1\end{bmatrix} $

All written in the standard basis, what is v in the basis defined $b_1$ and $b_2$? $b_1$ and $b_2$ are orthogonal to each other.

My steps:

Project V onto $b_1$

$V_b1 = \frac{V*b_1}{\lvert b_1\rvert \lvert b_1\rvert}*b_1$

$V_b1 = \frac{(5*1)+(-1*1)}{\lvert \sqrt(2)\rvert \lvert \sqrt(2)\rvert}*b_1$

$V_b1 = \frac{4}{2}*b_1$

$V_b1 = 2*b_1 = \begin{bmatrix}2\\2\end{bmatrix}$

Project V onto $b_2$

$V_b2 = \frac{V*b_2}{\lvert b_2\rvert \lvert b_2\rvert}*b_2$

$V_b2 = \frac{(5*1)+(-1)(-1)}{\lvert \sqrt(2)\rvert \lvert \sqrt(2)\rvert}*b_2$

$V_b2 = \frac{6}{2}*b_2$

$V_b2 = 3*b_2 = \begin{bmatrix}3\\-3\end{bmatrix}$

Add them together: $V = \begin{bmatrix}3\\-3\end{bmatrix} + \begin{bmatrix}2\\2\end{bmatrix} = \begin{bmatrix}5\\-1\end{bmatrix}$

Is my answer correct? Please let me know.

Thanks!

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While your answer is correct, I don't think your approach is correct.

I would begin with

$\begin{bmatrix} 5\\-1 \end{bmatrix} = c_1\mathbf b_1 + c_2\mathbf b_2$

Which then yields the system of equations

$c_1 + c_2 = 5\\ c_1 - c_2 = -1$

And solve for $c_1, c_2$

Or you can do this with matrix notation.

$B = \begin{bmatrix} 1&1\\1&-1 \end{bmatrix}\begin{bmatrix} c_1\\c_2\end{bmatrix}$ would transform a vector represented in the basis $\mathbf b_1,\mathbf b_2$ to the standard basis

$B^{-1}\begin{bmatrix} 5\\-1 \end{bmatrix} = \begin{bmatrix} c_1\\c_2\end{bmatrix} $ will transform the vector in the standard basis into the basis $B$

In this case $B^{-1} = \frac 12 B$ which is keeping you from getting into trouble.