Orthonormal basis and matrix.

Question

Let $E = \{e_1,_{...},e_n\} , F = \{f_1,_{...},f_n\} $ be 2 orthonormal basis of V. Does $[Id]_F^E $ the transfer matrix from E to F is orthogonal ? I think that the answer is yes, thought of shoving that $||[Id]_F^E v||= ||v||$ but having difficulties with that.

thanks for helping.

Answer 1

Suppose

$$f_i=\sum_{k=1}^n a_{ki}e_k\;,\;\text{so that the transfer matrix is}\;\;P=(a_{ij})\implies P^t=(b_{ij})\;,\;\text{with}\;\;b_{ij}=a_{ji}$$

and from here:

$$PP^t=(a_{ij})(b_{ij})=\left(\sum_{k=1}^n a_{ik}b_{kj}\right)=\left(\sum_{k=1}^na_{ik}a_{jk}\right)$$

But we can also take the usual inner (dot) product

$$\delta_{ij}=\langle f_i,\,f_j\rangle=\sum_{k,j=1}^na_{ki}a_{ji}\langle e_k,\,e_j\rangle=\sum_{k=1}^n a_{ki}a_{kj}\;\implies\;PP^t=I\implies P\;\;\text{is orthogonal}$$

Answer 2

Joanpemo's proof is fine, but there's an easier (to me) proof. Consider the basis $Q$ consisting of the standard basis vectors $$ q_1 = \begin{bmatrix} 1\\0 \\ \ldots \\ 0 \end{bmatrix} \\ q_2 = \begin{bmatrix} 0\\1 \\ \ldots \\ 0 \end{bmatrix} \\ \ldots \\ q_n = \begin{bmatrix} 0\\0 \\ \ldots \\ 1 \end{bmatrix}. $$ Then the matrix to go from coordinates in the basis $F$ to coordinates in the basis $Q$ is just $$ M_F = \begin{bmatrix} f_{11} & f_{21} & \ldots & f_{n1} \\ f_{12} & f_{22} & \ldots & f_{n2} \\ \ldots \\ f_{1n} & f_{2n} & \ldots & f_{nn} \end{bmatrix}, $$ i.e., the matrix whose columns are the coordinates of the vectors $f_i$ in the standard basis.

The fact that $M_F$ is orthogonal is clear, for $M_F^t M_F$ has, as its $ij$ entry, the inner product $\langle f_i, f_j \rangle$, so the matrix product is the identity.

Now what matrix goes from $F$ coords to $E$ coords? Well, we first apply $M_F$, and the $M_E^{-1}$: $$ H = M_F M_E^{-1} $$ Since this is a product of orthogonl matrices, it's still orthogonal. And so is its inverse, which takes $E$ coords to $F$ coords. QED.