Question

Consider an orthonormal basis B of the inner product space V. For an element fof V, what is the relationship between $||f||$and $||{\left[f\right]}_B||$(the norm in ${\mathbf{R}}^{\mathbf{n}}$defined by the dot product)?

Short answer

$||f||={||f||}_B$

Step-by-step solution

Consider the solution.

Let a basis will be $B=\left\{f_1,f_2,...,f_3\right\}$.

So, it gives $f={}_{i=1}{}^{n}a_i{f}_i$. Observe that

${\left[f\right]}_B=\left[\begin{array}{c}{\alpha }_1\\ .\\ .\\ .,{\alpha }_n\end{array}\right]$

Now, with the help of Pythagorean Theorem.

$$\begin{gathered} {||f||}^2={||{}_{i=1}{}^{n}a_i{f}_i||}^2 \\ ={}_{i=1}{}^n||f_i||^2 \\ ={}_{i=1}{}^n||{\alpha }_i{f}_i||^2 \\ ={}_{i=1}{}^n\left|{\alpha }_i\right|^2 \\ ={||f||}_B^2 \end{gathered}$$

Hence, it is proved $||f||={||f||}_B$.