6.3 Orthogonal bases and projections
Reading
Try out the Preview Activity and read Orthogonal bases and projections in Understanding Linear Algebra by David Austin.
If a basis is also an orthogonal set, we call it an orthogonal basis.
An orthogonal basis allows us to quickly and easily express a vector as a linear combination of the basis elements, without using row reduction. We will work with this idea in this course, but it will also be an important idea in Math 251, when we work with Fourier series — we will express functions as linear combinations of a basis of sine and cosine functions.
Proposition 6.3.4: Why is it so great to have an orthogonal basis?
If [latex]\vec{b}[/latex] is a linear combination of an orthogonal set of vectors, [latex]\vec{v}_1, \dots, \vec{v}_k[/latex], then [latex]\vec{b} = c_1 \vec{v}_1+ \dots + c_k \vec{v}_k[/latex], where [latex]c_i = \frac{\vec{b}\cdot \vec{v}_i}{\vec{v}_i \cdot \vec{v}_i}.[/latex]
Proposition 6.3.5: Orthogonal sets are always linearly independent.
Suppose that [latex]\{\vec{v}_1, \dots, \vec{v}_k \}[/latex] is a set of orthogonal vectors. Consider any linear combination of these vectors [latex]c_1\vec{v}_1+\dots+c_k\vec{v}_k[/latex] with at least one non-zero coefficient, [latex]c_i[/latex]. Then [latex](c_1\vec{v}_1+\dots+c_k\vec{v}_k)\cdot \vec{v}_i = c_1 (\vec{v}_1\cdot \vec{v}_i) +\dots+ c_i (\vec{v}_i\cdot \vec{v}_i)+ \dots+ c_k (\vec{v}_k\cdot \vec{v}_i).[/latex] By orthogonality of the set, only one of these dot-products is non-zero, so this is equal to [latex]c_i (\vec{v}_i \cdot \vec{v}_i) \neq 0[/latex]. In particular, this linear combination could not be the zero-vector, as it has a non-zero dot-product with a non-zero vector. Therefore the set of vectors is linearly independent.