Let $B=\{v_1,…,v_n\}$ be an orthonormal basis for $\Bbb R^n$ and $T:R^n \to R^n$ be a linear transformation. Let $A=(a_{ij})$ a representing matrix of the transformation $T$ by basis $B$ (I am sorry if this is not the correct term of this matrix "representing matrix" I tried translating but could not find the correct term.) Prove that the element $a_{ij}$ is equal to $T(v_j)\cdot v_i$.
My try:
we know that the columns of the matrix are coordinate vectors of $T(v_j)$ so the coloumn $j$ of A is the coordinate $j$ vector of $T(v_j)$ and the row $i$ of A is the row $i$ of $T(v_j)$
we also know that if we have an Orthonormal basis of $\Bbb R^n$ and let $a,b \in \Bbb R^n$ be any vectors in that space so we can represent $a$ as a linear combination of $B$ this way:
$a=\sum_{i=1}^n (a \cdot u_i)\cdot u_i$ so we can do that same in our case $T(v_j)=\sum_{b=1}^n (T(v_j) \cdot v_b)\cdot v_b$ and since it is an Orthonormal basis we get that $T(v_j) \cdot v_b=1$ and I got stuck here
According to the textbook (they do not show the full solution and not the first parts) after this part : $T(v_j)=\sum_{b=1}^n (T(v_j) \cdot v_b)\cdot v_b$ it says that the Coefficient of vector $v_b$ is the coordinate $b$ of $T(v_j)$ and according to this we get that element $a_{ij}$ is equal to $T(v_j)\cdot v_i$
I did not understand the textbooks explanation can anyone help me understand and tell me if my way is correct? or if there is another way to suggest?
I am sorry for the English translations I wasn't able to find a translation for everything
Best Answer
You have $$T(v_{j})=\sum_{i=1}^{n}a_{ij}v_{i}$$. by definition of the matrix of $[T]_{\mathcal{B}}$.
So $$\langle T(v_{j}),v_{k}\rangle=\left\langle\sum_{i=1}^{n}a_{ij}v_{i},v_{k}\right\rangle=\sum_{i=1}^{n}a_{ij}\langle v_{i},v_{k}\rangle=\sum_{i=1}^{n}a_{ij}\delta_{ik}=a_{kj}$$. Where $\delta_{ij}$ is the Kronecker Delta. such that $\delta_{ij}=\begin{cases} 1\,,i=j\\0\,,i\neq j\end{cases}$. Basically all other terms become $0$ by Orthogonality and only $\langle v_k,v_k\rangle$ survives and equals $1$.
Thus the $kj$-th entry of the matrix is given by $\langle T(v_{j}),v_{k}\rangle$.