# Why do we assume that a matrix in quadratic form is Symmetric?

I am looking to the review document for linear algebra and the part of the quadratic form (pg17) mentions about an assumption of being symmetric for a matrix in quadratic form. It also includes some declarative equality for that proposed argument.

What is the practical concern of assuming a matrix as symmetric in quadratic form? I also do not get the idea proposed by the argument? Can someone light me about?

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The argument shows that if $A$ were not symmetric, you could replace it with its symmetric part $\frac12(A+A^T)$ and the quadratic form $x^TAx$ would not change. – Rahul Feb 18 '13 at 19:59
what does "symmetric part" mean? – Erogol Feb 18 '13 at 20:43
You can write any matrix $A$ as the sum of a symmetric matrix and an antisymmetric matrix. Let's call them $B$ and $C$, where $B$ is symmetric and $C$ is antisymmetric. Then it turns out that $B$ is actually equal to $\frac12(A+A^T)$, and $C$ is $\frac12(A-A^T)$. These are called the symmetric and antisymmetric parts of the matrix $A$. Exercise: Verify that $B=B^T$ and $C=-C^T$. – Rahul Feb 18 '13 at 20:46

The main reason for getting the matrix of a real quadratic form symmetric by replacing the original matrix with its symmetric part ${A+A^T}\over 2$ is that any symmetric matrix is orthogonally diagonalizable and all eigenvalues are real. Then for symmetric $A$ you have some orthogonal matrix $U$ (that is, $U^T=U^{-1}$) and $U^T A U=D$ is real diagonal, and $x^T A x=(Ux)^T(UAU^T)(Ux)=(Ux)^TD(Ux)$ is a sum of squares with real coefficients where the length $\Vert x \Vert$ of the vector $x$ is the same as the length $\Vert Ux\Vert$ of the vector $Ux$.