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.

  • 5
    $\begingroup$ Are you sure that $1(d_1) + 1(d_2) = e_2$? Check again. $\endgroup$
    – saulspatz
    Jul 13, 2021 at 17:59
  • 1
    $\begingroup$ 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 $e_2$. $\endgroup$ Jul 13, 2021 at 18:47

1 Answer 1


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 $\|e_2\|=1$


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