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Let $q$ be a quadratic form defined on an n-dimensional real vector space. Let $A$ be the matrix of the quadratic form $q$ in a given basis. This means that $A$ is a symmetric $n \times n$ matrix such that

$q(v)=x^\mathrm{T} Ax,$

where $x$ is the column vector of coordinates of $v$ in the chosen basis. Under a change of basis, the column $x$ is multiplied on the left by an $n \times n$ invertible matrix $S$, and the symmetric square matrix $A$ is transformed into another symmetric square matrix $B$ of the same size according to the formula

$ A\to B=SAS^\mathrm{T}.$


Two n-ary quadratic forms $φ$ and $ψ$ over $K$ are equivalent if there exists a nonsingular linear transformation $T \in GL(n, K)$ such that

$ \psi(x)=\varphi(Tx). $

Let us assume that the characteristic of $K$ is different from 2. (The theory of quadratic forms over a field of characteristic 2 has important differences and many definitions and theorems have to be modified.) The coefficient matrix $A$ of $q$ may be replaced by the symmetric matrix $(A + A^T)/2$ with the same quadratic form, so it may be assumed from the outset that $A$ is symmetric. Moreover, a symmetric matrix $A$ is uniquely determined by the corresponding quadratic form. Under an equivalence $T$, the symmetric matrix $A$ of $φ$ and the symmetric matrix $B$ of $ψ$ are related as follows:

$ B=T^\mathrm{T}AT. $

These two seem to talk about the same thing, yet they have a different product ($B = SAS^T$ while the second one is $B = T^TAT$. Can anyone explain why these two result in a different form?

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I didn't fully get the first one, $SAS^T$.

But it doesn't really matter, as we can exchange $S$ and $T$ by taking $T:=S^T$, then $SAS^T=T^TAT$ as wished, and of course $S$ is invertible iff $S^T$ is invertible.

Let $e_1,e_2,...$ denote the standard basis of $\Bbb R^n$, and let $A$ be a symmetric matrix. Then the bilinear form it determines is $$(x,y)\mapsto x^TAy$$ and observe that $A_{i,j}=e_i^TAe_j$, so, multiplying by the basis elements will give the matrix coefficients. Now, if we have another basis, $s_1,...,s_n$, and put them (as column vectors) in the matrix $S:=[s_1|s_2|..|s_n]$, then the rows of $S^T$ are $s_1^T,s_2^T,...$, and so $$S^TAS=(s_i^TAs_j)_{i,j}\,,$$ that is, the matrix of the bilinear (quadratic) form, w.r.t. the basis $s_1,s_2,..$, is $S^TAS$.

Well, if one writes (in a rather unusual way) the basis vectors in the rows of a matrix, i.e. starts with $S^T$ instead of $S$, then of course, it would lead to $SAS^T$...

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