# Orthogonality and change of basis

As a student in an introductory course in linear algebra, bases are one of the fundamental concepts that I've had the most troubles wrapping my head around.

$$u_1=(1,1,1),\; u_2=(1,2,0),\; u_3=(2,1,3)$$

Now let's say I construct three orthogonal vectors in the basis $u_1,u_2,u_3$

$$w_1=a_1u_1=(a_1,0,0)\\ w_2=a_2u_2=(0,a_2,0)\\ w_3=a_3u_3=(0,0,a_3),$$

$a_i\in\mathbb{R}.$ How can they be orthogonal in one basis but not in another? Can you even say the vectors $w_1,w_2,w_3$ are orthogonal in the basis $u_1,u_2,u_3$? I tried drawing it and drew the conclusion that you can't, but still, if I took the dot products of the vectors $w_1,w_2,w_3$, the conclusion would be that they are orthogonal. What am I missing here?

For a simpler example, if we have the vectors $u_1 = (2, 0)$ and $u_2 = (0,2)$ in $\Bbb R^2$ using the standard basis. Then we have that $u_1\cdot u_1 = 4$. If we change our basis to $u_1, u_2$, then suddenly all naively calculated inner products become four times smaller, such as $u_1\cdot u_1 = 1^2 = 1$.
The remedy (in this example multiply by $4$) is to insert a matrix between the row and column vectors when doing scalar products. Specifically, if $A$ is the change of basis matrix for the right direction (in this example $2I$), then you calculate the inner product between two vectors $v_1, v_2$ (expressed in the new basis) as $v_1^T A^TAv_2$. The effect of this is to transform $v_1$ and $v_2$ into their standard basis representations $Av_1$ and $Av_2$ before transposing one of them and matrix-multiplying them together.
In your example, we have $$A = \begin{bmatrix}1& 1& 2\\1& 2& 1\\1& 0& 3\end{bmatrix}$$ and so the scalar product of $w_1$ and $w_2$ is calculated as $$\left(\begin{bmatrix}a_1&0&0\end{bmatrix}\begin{bmatrix}1&1&1\\1&2&0\\2&1&3\end{bmatrix}\right)\left(\begin{bmatrix}1& 1& 2\\1& 2& 1\\1& 0& 3\end{bmatrix}\begin{bmatrix}0\\a_2\\0\end{bmatrix}\right)\\ = a_1a_2\begin{bmatrix}1&1&1\end{bmatrix}\begin{bmatrix}1\\2\\0\end{bmatrix} = 3a_1a_2$$
Note that the matrix $B = A^TA$ is a symmetric (and positive-definite) matrix. In fact, any symmetric positive-definite matrix $B$ may be thought of as representing the scalar product over some basis in this way.