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What is the geometrical interpretation of positive definite matrix ? (not necessarily symmetric)

if $A$ is positive definite, what does it do to a vector $x$ (i.e. $Ax$)?

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Do you mean positive definite? If not necessarily symmetric, what exactly do you mean by positive definite? Just $xAx^\mathrm{T}>0$ for all row vectors $x$? – Jonas Meyer Nov 28 '11 at 9:05

If the matrix is not symmetric, there is no notion of positivity other than having eigenvalues all positive. For symmetric matrices, you get a partial ordering such as $A\succ B$ which means $A-B$ is positive definite.

Moreover, every symmetric positive definite matrix defines an ellipsoid. The principal axes are given by the eigenvectors of $A$ define and the square root of the eigenvalues are the radii of the corresponding axes. The usual convention is to use the inverse of the matrix to define the ellipsoid. This is to define the unit ball as the image of the mapping $x^TA^{-1}x$.

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welcome to math.stackexchange. Assuming that you mean that $A$ is positive definite, one intuition is a parabola: From the definition of positive definite, we get that $x^\top A x > 0$ for all $x\not=0$. For example, if $$A=\left(\begin{array}{cc} 1 & 0\\ 0 & 1\end{array}\right)$$ then $x^\top A x = x_1^2 + x_2^2$.

For a general p.d. $A$ you can first transform $x$ into the basis spanned by the eigenvectors of $A$. In that space $A$ becomes diagonal with the positive eigenvalues on the diagonal (positive since $A$ is p.d.). Then, the same intuition holds again.

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thank you for your fast replay. If I have two orthogonal vectors in R^2, v1 v2, that their length is not necessarily 1. now we define A=[v1 v2]. can we say something about A being positive definite? – Shira Nov 28 '11 at 9:52
no. take $v_1 = (0,1)$ and $v_2 = (1,0)$. The matrix $A = [v_1,v_2]$ is not positive definite. The matrix $B = [v_2,v_1]$ is. – Willie Wong Nov 28 '11 at 10:33
If your two vectors are orthogonal the action of the matrix is positive scaling and permutation/rotation. The dot product between $x$ and $Ax$ is proportional to the cosine between $x$ and $Ax$. Therefore, if the matrix with the normalized column vectors $v_1$ and $v_2$ produces a vector that is not off by more than $(-\pi/2,\pi/2)$, then the cosine will be greater than zero and $A$ will be p.d. – fabee Nov 28 '11 at 10:37

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