Question

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

Solution

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Let $V$ is the space of all $n \times 1$ matrices over $C$ and $f$ the bilinear form on $V$ given by $f(X,Y)=X^{t}Y$ and let $M$ belong to $O(n,C)$, that is

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

Let $M_{1}, M_{2}, \cdots, M_{n}$ are columns of matrix $M$, such that

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

each $M_{i}$ is $n \times 1$ column vector,

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

each $M_{i}$ is $1 \times n$ row vector, then consider $M^{T}M=I$,

\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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