Does Basis Change affect Inner Product?

Question

If I have the basis B={b₁=[1,-1], b₂=[0,1] } for ℝ², and I'm assuming an Euclidean inner product, then I understand that:

<b₁, b₁> = 1 + (-1)(-1) = 2

But that is only the case because I chose to represent b₁ using the canonical basis of ℝ². That is, b₁ = [1,-1] because these are its coordinates in the canonical basis.

If I were to represent b₁ using B, then b₁ would be equal to [1,0]_B. So

<b₁,b₁> = 1 + 0 = 1

So, does it mean that the same two vectors can have totally different inner products depending on how which basis I am considering? Or should the inner product always be used for vectors represented canonically?

Thanks!

Answer 1

The Euclidean inner product on $\mathbb{R}^2$ is essentially defined so that the standard basis is orthonormal. If you represent your vectors with respect to a different basis and then use the Euclidean inner product formula on the resulting coordinate vectors then you will typically get incorrect inner products (as your example shows).

The exception to this is if you select an orthonormal basis. For example, if $B=\{[0,-1],[1,0]\}$ then you have that $b_1$ is $[1,1]_B$ in your notation and then if you use the regular formula for inner products you get $<b_1,b_1>=1+1=2$, as you would expect.

More generally, if $\{v_1,\ldots v_n\}$ is an orthonormal basis for $\mathbb{R}^n$ and $w\in \mathbb{R}^n$ then when you write $w$ with respect to this basis you will get

$$w=c_1v_1+\ldots+c_nv_n$$

and the coefficients can be computed as $c_i=<w,v_i>$. More generally, if you have another vector $u$ and you write it as $u=d_1v_1+\ldots+d_nv_n$ and you will find that

$$<w,u>=c_1d_1+\ldots+c_nd_n$$

The moral of the story is that if you write your vectors with respect to any orthonormal basis then the formula for the inner product looks like the standard Euclidean inner product formula. The computations I mentioned above are both fun little exercises that I would highly recommend you try!