Linear Algebra MT23, Orthogonal sets
Flashcards
Quickly prove that any orthonormal set $\{u _ 1, \cdots, u _ n \}$ is linearly independent.
Suppose
\[\lambda _ 1 u _ 1 + \cdots + \lambda _ m u _ m = 0\]Take $\langle u _ j, \cdot\rangle$ for each $u _ j$, and you see
\[\lambda _ j = 0\]Suppose $\{u _ 1, \cdots, u _ n\}$ is an orthonormal basis, so that every $v$ can be written
\[v = \lambda _ 1 u _ 1 + \cdots + \lambda _ n u _ n\]
How can you write $\langle u _ j, v \rangle$?
Suppose $\{u _ 1, \cdots, u _ n\}$ is an orthonormal basis, so that every $v$ can be written
\[v = \lambda _ 1 u _ 1 + \cdots + \lambda _ n u _ n\]
Can you write $v$ as a sum of inner products with elements of the basis?
Bite-sized
Orthonormal-basis coordinate extraction: if $\{u _ 1, \ldots, u _ n\}$ is an orthonormal basis and $v = \sum _ j \lambda _ j u _ j$, then $\lambda _ j = $ $\langle u _ j, v \rangle$. Coordinates are recovered by inner products — no matrix inversion needed.
Why is every orthonormal set automatically linearly independent?
Suppose $\sum _ j \lambda _ j u _ j = 0$. Taking $\langle u _ k, \cdot \rangle$ of both sides and using orthonormality gives $\lambda _ k \cdot 1 + 0 + \ldots = 0$, hence $\lambda _ k = 0$ for every $k$. So any nontrivial linear combination is nonzero. (This is why orthonormality is so much stronger than mere linear independence: you get coordinate-extraction “for free”.)