Quadratic form if corresponding matrix is not symmetric matrix . We know that every real quadratic form is of the form $X^t AX$ where $A$ is real symmetric matrix . What if $A$ is not a symmetric matrix ? In my book it is written that if $A$ is not symmetric then replace $A$ by $\frac{A+A^t}{2}$. What is meaning of this ? Why to such to replacement ? How it  has same effect by replacement ? Thanks.
 A: Note that
$\left (\dfrac{A + A^T}{2} \right )^T = \dfrac{A + A^T}{2}, \tag 1$
that is, the matrix
$A_\Sigma = \dfrac{A + A^T}{2} \tag 2$
is symmetric; further note that
$\left (\dfrac{A - A^T}{2} \right )^T = -\dfrac{A - A^T}{2}, \tag 3$
whence
$A_K =  \dfrac{A - A^T}{2} \tag 4$
is skew-symmetric; and yet further observe that
$A = A_\Sigma + A_K, \tag 5$
which easily follows from (2) and (4); thus
$X^T A X = X^T (A_\Sigma + A_K) X = X^TA_\Sigma X + X^T A_K X; \tag 6$
now $X^TA_K X$ is a $1 \times 1$ matrix, whence
$(X^TA_K X)^T = X^TA_K X; \tag 7$
but we also have
$(X^TA_K X)^T = X^TA_K^T X(X^T)^T = X^T(-A_K)X = -X^T A_K X; \tag 8$
combining (7) and (8) yields
$X^TA_KX = -X^TAX \Longrightarrow X^TA_KX = 0; \tag 9$
now (6) becomes
$X^T A X = X^TA_\Sigma X, \tag{10}$
which shows that $X^T AX$ only depends on $A_\Sigma$, the symmetric part of $A$; therefore we may replace $A$ by $A_\Sigma$ and obtain the same quadratic form $X^TAX$.
A: If $A$ is an $n \times n$ matrix and $v$ is a column-vector with $n$ entries, then $ \frac {A+A^T}{2}$ is a symmetric matrix and $$v^TAv=v(\frac {A+A^T}{2})v$$Thus if we have a theorem abou quadratic forms and symmetric matrices, we can use that theorem to obtain a result about a quadratic form with a matrix that is not symmetric. Note that we can proceed in this fashion with real or complex coefficients or over any field whose characteristic is not 2. Even more generally, the result is true over any commutative ring with identity in which the element 2 is invertible.
