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I have doubts regarding a statement related to the adjoint linear transformation, here is the statement and its proof:

Proposition: Let $(V,\langle,\rangle)$ be a finite dimensional vector space with inner product and let $f:V \to V$ be a linear transformation. Let $B$ be an orthonormal basis of $V$. Then $|f^*|_B=(|f|_B)^* $

Proof: Suppose $B=\{v_1,…,v_n\}$ is an orthonormal basis given in $V$. Then, for each $1\leq i,j \leq n$, $$(|f^*|_B)_{ij}=\langle f^*(v_j),v_i \rangle=\overline {\langle v_i, f^*(v_j) \rangle}= (\overline {|f|_B})_{ji}=(\overline {{|f|_B}^*})_{ij}.$$

I am trying to understand these two equalities:

  1. $$(|f^*|_B)_{ij}=\langle f^*(v_j),v_i \rangle$$ (first equality)
  2. $$(\overline {|f|_B})_{ji}=(\overline {{|f|_B}^*})_{ij}.$$ (last equality)

With respect to 1), why the entry $ij$ is equal to $\langle f^*(v_j),v_i\rangle$, in other words, why the $i$-th coordinate of $f^*(v_j)$ with respect to the basis $B$ is $\langle f^*(v_j),v_i \rangle$? I can’t figure that out.

According to my linear algebra textbook, $f^*$ evaluated in $w \in V$ is $$f^*(w)=\sum_{i=1}^n \langle w,f(v_i) \rangle v_i$$ where $\{v_1,…,v_n\}$ is an orthonormal basis, I’ve tried to use this definition of self adjoint in order to explain myself equalities 1) and 2) but I couldn’t. I would appreciate if someone could explain these two to me.

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why the $i$-th coordinate of $f^*(v_j)$ with respect to the basis $B$ is $\langle f^*(v_j),v_i \rangle$?

Because $B$ is an orthonormal basis. When a vector $v$ is expanded in the orthonormal basis $(e_k)$, i.e., $v=\sum c_k e_k$, taking the inner product of both sides with $e_i$ results in $\langle v,e_i\rangle = \sum c_k \delta_{ik} = c_i$.

Once you see this relation of matrix entries with inner product, the rest follows: $$\langle f^*(v_j),v_i \rangle = \langle v_j,f(v_i) \rangle \tag{*}$$ by the definition of an adjoint. Then the right hand side of (*) is related to the matrix representing $f$, as it was previously done for $f^*$.

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