Question

Let V be the space of all n×1n \times 1 matrices over C and f the bilinear form on V given by f(X,Y)=XtYf(X, Y)=X^{t} Y. Let Ab belong to O(n, C). What is the matrix off in the basis of I’ consisting of the columns M1,M2,,MnM_{1}, M_{2}, \ldots, M_{n} of M?

Solution

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Let VV is the space of all n×1n \times 1 matrices over CC and ff the bilinear form on VV given by f(X,Y)=XtYf(X,Y)=X^{t}Y and let MM belong to O(n,C)O(n,C), that is

MTM=I.\begin{align*} M^{T}M&=I. \end{align*}

Let M1,M2,,MnM_{1}, M_{2}, \cdots, M_{n} are columns of matrix MM, such that

M=[M1M2Mn],\begin{align*} M&=\begin{bmatrix} M_{1}&M_{2}&\cdots&M_{n} \end{bmatrix}, \end{align*}

each MiM_{i} is n×1n \times 1 column vector,

MT=[M1TM2TMnT]\begin{align*} M^{T}&=\begin{bmatrix} M^{T}_{1}\\ M^{T}_{2}\\ \vdots\\ M^{T}_{n}\\ \end{bmatrix} \end{align*}

each MiM_{i} is 1×n1 \times n row vector, then consider MTM=IM^{T}M=I,

[M1TM2TMnT][M1M2Mn]=I[M1TM1M1TM2M1TMnM2TM1M2TM2M2TMnMnTM1MnTM2MnTMn]=I\begin{align*} \begin{bmatrix} M^{T}_{1}\\ M^{T}_{2}\\ \vdots\\ M^{T}_{n}\\ \end{bmatrix}\begin{bmatrix} M_{1}&M_{2}&\cdots&M_{n} \end{bmatrix}&=I\\ \begin{bmatrix} M_{1}^{T}M_{1}&M_{1}^{T}M_{2}&\cdots&M_{1}^{T}M_{n}\\[7pt] M_{2}^{T}M_{1}&M_{2}^{T}M_{2}&\cdots&M_{2}^{T}M_{n}\\[7pt] \cdots&\cdots&\ddots&\cdots\\ M_{n}^{T}M_{1}&M_{n}^{T}M_{2}&\cdots&M_{n}^{T}M_{n} \end{bmatrix}&=I\\ \end{align*}

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