7
$\begingroup$

I am looking to the review document for linear algebra (Zico Kolter (updated by Chuong Do), Linear Algebra Review and Reference), 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 reason to assume that the matrix describing a quadratic form (over $\mathbb{R}$) is symmetric? I also do not get the idea proposed by the argument? Can someone light me about?

$\endgroup$
  • 10
    $\begingroup$ 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. $\endgroup$ – Rahul Feb 18 '13 at 19:59
  • $\begingroup$ what does "symmetric part" mean? $\endgroup$ – erogol Feb 18 '13 at 20:43
  • 5
    $\begingroup$ 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 $C=-C^T$. $\endgroup$ – Rahul Feb 18 '13 at 20:46
9
$\begingroup$

Any matrix A can be written as sum of $(A+A^{T})/2$ and $(A-A^{T})/2$. You can verify that the quadratic form of second term (i.e; $x^{T}(A-A^{T})x$) turns out to be zero (Try to evaluate the second term and check for yourself).

Hence, we can assume that we begin with a symmetric matrix. If not, it is anyway possible to convert to an equivalent quadratic form with a symmetric matrix.

$\endgroup$
5
$\begingroup$

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 $Ux$.

$\endgroup$
2
$\begingroup$

We can show that the Matrix $\bf{W}$ of the quadratic form $x^T {\bf{W}} x$ can be assumed to be symmetric in the following way:

In general a matrix can be decomposed into a symmetric and an anti-symmetric part as follows:

$$M = \frac{1}{2} ( {\bf{M}} + {\bf{M}}^\top) + \frac{1}{2} ({\bf{M}} - {\bf{M}}^\top) = M_s + M_a$$

Further we note that the symmetric part is invariant to the transpose, $$ {\bf{M}}_s^\top = {\bf{M}}_s $$ while the antisymmetric part changes its sign under transposition: $$ {\bf{M}}_a^\top = - {\bf{M}}_a$$

Let us examine the quadratic form of a purely antisymmetric matrix: $$q = {\bf x}^\top {\bf M}_a {\bf x} = ({{\bf x}^\top {\bf M}_a^\top {\bf x}})^\top = -({\bf x}^\top {\bf M}_a {\bf x})^\top = -q$$ Which implies that $q=0$ has to hold true.

Now for a general Matrix $\bf W$ we can see that: $${\bf x}^\top {\bf M} {\bf x} = {\bf x}^\top ({\bf M}_s + {\bf M}_a){\bf x} = {\bf x}^\top {\bf M}_s {\bf x} + {\bf x}^\top {\bf M}_a {\bf x}$$ As we know that $${\bf x}^\top {\bf M}_a {\bf x} = 0$$ we finally find $${\bf x}^\top {\bf M} {\bf x} = {\bf x}^\top {\bf M}_s {\bf x}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.