How does orthogonality relate to the notion of a basis?

Question

Let's say you have standard basis $\beta = \{ (1,0), (1,0) \}$ for $\mathbb{R}^2$ and linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ that is diagonalizable with two distinct eigenvalues, so you have a basis of eigenvectors $\gamma = \{ w_1, w_2 \}$.

The eigenvectors are not guaranteed to be orthogonal, and if you use G-S, they may no longer be eigenvectors. However, you can always write the eigenvectors in terms of the basis $\gamma$ and they will be orthonormal.

To be concrete, I mean that $[w_1]_\gamma = (1, 0)_\gamma$ and likewise for $w_2$.

So does orthogonality fully relate to the notion of what basis you are using, and any basis you choose can be made orthogonal simply by choosing that basis as reference?

This doesn't make complete sense to me, since we have the G-S process, and this seems to make the notion of orthogonality almost useless if this is the case.

So when we say that for a finite dimensional inner product space $(V, \langle \cdot , \cdot \rangle)$ over some $\mathbb{F}$, do we define this notion of orthogonality (when $\langle x , y \rangle = 0$ for $x, y \in V$) as independent of basis?

Answer 1

Arbitrary linear transformation do not preserve orthogonality. "Expressing a vector in the basis $\gamma$" is a linear transformation $\phi_\gamma : \mathbb R^2 \to \mathbb R^2$ with $\phi_\gamma(v) = [v]_\gamma$. Linear transformations which preserve the inner product (and hence orthogonality) are called orthogonal transformations, and are represented by orthogonal matrices relative to an orthogonal basis. $\phi_\gamma$ is not orthogonal because it takes a nonorthogonal basis ($\gamma$) to an orthogonal one (the standard basis).

As you say, orthogonality is defined via $\langle x,y\rangle = 0$. This has nothing to do with a basis. "Preserving the inner product" for linear $T : \mathbb R^2 \to \mathbb R^2$ means $$ \langle Tx, Ty\rangle = \langle x,y\rangle $$ for all $x,y\in\mathbb R^2$, so this is the formal definition for "$T$ is orthogonal". This also has nothing to do with a basis. You will find that $\phi_\gamma$ does not satisfy this condition for the standard inner product.

Just because nonorthogonal transformations exist does not mean that orthogonality is useless.

Answer 2

The eigenvectors are not guaranteed to be orthogonal, and if you use G-S, they may no longer be eigenvectors. However, you can always write the eigenvectors in terms of the basis γ and they will be orthonormal.

You are confusing a vector in $V = \mathbb R^2$ with it's coordinates (also in $W = \mathbb R^2$)! Consider a general vector space $V$ with an inner product which is a map $ V \times V \to V$. This map is defined for $V$ and not for the space $ W = \mathbb R^{\dim V}$ obtained by fixing a basis. You can only take the dot product between vectors in the vector space $V$. That is, the dot product is base independent !

What you are doing is taking two basis vectors $w_1,w_2$ looking at their coordinates in the basis $\gamma$ and taking the dot product of their coordinates as if they were points of the vector space $V$. But the coordinates are in the space $W$ which is not equipped with a dot product.

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If you want to equip $\mathbb R^{\dim V}$ with a dot product then the only natural way to do so is to define a map $\phi_\gamma : V \to \mathbb R^{\dim V}$ which maps $v$ to it's coordinates in the basis $\gamma$ and define the inner product on $\mathbb R^{\dim V}$ by

$$ \langle \phi_\gamma(v), \phi_\gamma(v')\rangle_{\mathbb R^{\dim V}} := \langle v,v' \rangle_{V}$$

Doing so in your example you will see that the dot product of $[w_1]_\gamma = (1,0)$ and $[w_2]_\gamma = (0,1)$ is not $1 \times 0 + 0 \times 1 = 0$ but is by definition the dot product of the vectors $w_1,w_2$. This is because the dot product is by definition

$$ \langle [w_1]_\gamma, [w_2]_\gamma \rangle := \langle w_1, w_2 \rangle $$

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In order to avoid confusion it is probably best to use a different notation for coordinates. You might want to write the coordinates of a vector as a column vector.

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For a specific example suppose that $w_1 = (0,1)$ and $w_2 = (1,1)$ and $V = \mathbb R^2$ defined with the usual dot product. In the basis $\{w_1,w_2\}$ the basis vectors have the coordinates $(1,0)^T$ and $(0,1)^T$. And a point of coordinate $(x,y)^T$ cooresponds to the point in $V$ defined as $ xw_1 + yw_2 = (y,y+x)$. It follows that the dot product on the coordinate space is defined by \begin{align*} \langle (x,y)^T,(x',y')^T \rangle &= \langle (y,y+x), (y',y'+x') \rangle \\ &= yy' + (y+x)(y'+x'). \end{align*} This means that $\langle (1,0)^T,(0,1)^T \rangle \neq 1 \times 0 + 0 \times 1$. Rather $$ \langle (1,0)^T,(0,1)^T \rangle = 0 \times 1 + (0+1)(1+0) = 1. $$