# How to express the Euclidean basis vector $(0, 1)$ under this new basis?

Changing our basis. Suppose we are working in the $$\mathbb{R}^2(\mathbb{R})$$ vector space under the standard Euclidean basis and dot product. Suppose we then transform the standard basis vectors $$e_1 = (1,0) \mapsto (1/\sqrt{2}, 1/\sqrt{2}) = d_1$$ and $$e_2 = (0,1) \mapsto (-1/\sqrt{2}, 1/\sqrt{2})= d_2$$. Note that this is an orthogonal change in basis (in fact a $$90$$ degree rotation), so norms should be preserved.

How to re-express $$e_2$$ under the new $$d_i$$ basis? Now let us consider $$e_2$$ under the new $$d_i$$ basis. It seems to me $$e_2 = (1, 1)$$ under this new basis, since

$$1(d_1) + 1(d_2) = e_2$$

Yielding a contradiction. But doesn't this then imply that

$$||e_2|| \ne 1$$

which is false? To see why, consider the expansion of $$||e_2||$$:

$$||e_2|| = \sqrt{e_2 \cdot e_2} = \sqrt{(1,1) \cdot (1,1)} = \sqrt{1*1 + 1*1} = \sqrt{2} \ne 1$$

So what I have done wrong? Note that under the standard basis, $$||e_2|| = 1$$ since

$$||e_2|| = \sqrt{e_2 \cdot e_2} = \sqrt{(0,1) \cdot (0, 1)} = \sqrt{0*0+1*1} = 1$$

...so this is clearly not right.

• Are you sure that $1(d_1) + 1(d_2) = e_2$? Check again. Jul 13, 2021 at 17:59
• First, $1(d_1) + 1(d_2) = (0,\sqrt{2})$. Second, this is in the new basis. $(0,1)$ in the new basis is not the same thing as $e_2$. Jul 13, 2021 at 18:47

Your basis change is $$d_1=\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2,$$ $$d_2=-\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2.$$ Solving for $$e_2$$ requires to do $$\begin{eqnarray*} d_1+d_2&=&\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2 +\left(-\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2\right),\\ &=&\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2 -\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2,\\ &=&\frac{2}{\sqrt 2}e_2. \end{eqnarray*}$$ Then $$e_2=\frac{\sqrt 2}{2}d_1+\frac{\sqrt 2}{2}d_2.$$
Also, here we can see $$\|e_2\|=1$$