An set of vectors is an orthogonal set if each pair of vectors is orthogonal to eachother such that

(1)In addition, if there is a subspace S formed by an orthogonal set, that set is linearly independent and is thus a basis for S. Therefore, if there exists a vector $\vec{y}$ is S, then $\vec{y}$ can be formed by the orthogonal set

(2)We can therefore multiply each term by $\vec{u_1}$ for instance, and since all other u's are perpendicular to $u_1$, all other terms drop out except $\vec{y} \cdot \vec{u_1} = c_1 \vec{u_1}^2$. This can be simplified and $c_1$ can be found using

(3)This is true for all c's so we now have the tools to find any vector in S by a linear combination of the orthogonal basis.

## Orthogonal Projection

Now, if we're looking to express $\vec{y}$ as a sum of two vectors, one which is a multiple of $\vec{u}$ and one perpendicular to $\vec{u}$, we must apply some of what we just learned. We know the scalar $c_1$ to create a vector in S that is a component of $\vec{y}$ is $\frac{\vec{y} \cdot \vec{u_1}}{\vec{u_1}^2}$. Using this to scale $\vec{u_1}$ which we'll call $\hat{y}$, we find

(4)where $\vec{v}$ is the perpendicular vector such that

(5)**Orthonormal Sets**

An Orthonormal set is a set of orthogonal vectors whose magnitudes are all 1.

To normalize a vector, just divide each component by the magnitude of the vector so…

(6)Then the orthonormal set has the property $U^TU = I$. Thus, $U^T = U^{-1}$.