Vector projection is equivalent to the dot product.
The dot product of r and s is the projection of the vector \(r\) on the vector \(s\).
For two n component vectors, the dot product is defined as: $$ a . b = a_1b_1 + a_2b_2 \dots + a_nb_n $$ Geometrically, for two vectors r and s, the projection appears as follows:
It is equivalent to: $$ r . s = |r| |s| \cos \theta $$
Basis
To be considered as a basis, a set of vectors must: - Be linearly independent - Span the space Basis vectors can be orthogonal because orthogonal vectors are independent. However, the converse is not necessarily true: non-orthogonal vectors can be linearly independent and thus form a basis (but not a standard basis). ![[Pasted image 20250429102423.png]]
Change of Basis
Suppose we have a vector \(\begin{pmatrix} 1 \cr 3 \end{pmatrix}\) or \(1\hat{i} + 3\hat{j}\). Here \(\hat{i}\) and \(\hat{j}\) are the basis vectors. These vectors are also known as a standard basis. Here, \(\hat{i}\) is a unit vector in the \(x\) direction and \(\hat{j}\) is a unit vector in the \(y\) direction.
The vector \(\begin{pmatrix} 5 \cr -1 \end{pmatrix}\) can be written as a linear combination of the basis vectors \(\hat{i}\) and \(\hat{j}\) as follows: $$\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} \times \begin{pmatrix} 5 \cr -1 \end{pmatrix} $$ Now suppose we want to change the basis vectors to \(\begin{pmatrix} 1 \cr 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \cr -1 \end{pmatrix}\). Now instead of \(\hat{i}\) and \(\hat{j}\), we have \(\hat{i'}\) and \(\hat{j'}\). We can write the vector \(\begin{pmatrix} 1 \cr 3 \end{pmatrix}\) as a linear combination of the new basis vectors as follows: $$\begin{pmatrix}1 & 1\\ 1 & -1\end{pmatrix} \times \begin{pmatrix} 5 \cr -1 \end{pmatrix} = 2 \cdot \begin{pmatrix}2 \cr 3\end{pmatrix} $$
Useful Links to Consult:
- https://www.nagwa.com/en/explainers/792181370490/