Let $X$ be nay basis of $\scrH$ and let $Y$ be the basis biorthogonal to it. Using matrix multiplication, $X$ gives a linear transformation from $\bbC^n$ to $\scrH$. The inverse of this is given by $Y^*$. In the special case when $X$ is orthonormal (so that $Y=X$), this transformation is inner-product preserving if the standard inner product is used on $\bbC^n$. \eex$$

解答: $$\beex \bea \sex{\ba{c} a_1\\ \vdots\\ a_n \ea}\in\bbC^n&\ra X\sex{\ba{c} a_1\\ \vdots\\ a_n \ea}\in \scrH,\\ X\sex{\ba{c} a_1\\ \vdots\\ a_n \ea}=\sex{\ba{c} b_1\\ \vdots\\ b_k \ea}&\ra \sex{\ba{c} a_1\\ \vdots\\ a_n \ea}=Y^*\sex{\ba{c} b_1\\ \vdots\\ b_k \ea},\\ \sef{X\sex{\ba{c} a_1\\ \vdots\\ a_n \ea},Y\sex{\ba{c} b_1\\ \vdots\\ b_n \ea}}&=\sex{\bar a_1,\cdots,\bar a_n}X^*Y\sex{\ba{c} b_1\\ \vdots\\ b_n \ea}=\sum_{i=1}^n \bar a_ib_i. \eea \eeex$$

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.1.2