19
$\begingroup$

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. $Ax$)?

$\endgroup$
1
  • 4
    $\begingroup$ 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 $x$? $\endgroup$ Nov 28, 2011 at 9:05

3 Answers 3

20
$\begingroup$

Matrix being positive definite is like a (real) number being positive.

Say you have a positive (real) number $a$. It has a property such that you can multiply any real number $b \in R$ by $a$ and the result of multiplication $a \cdot b$ will preserve $b$'s sign. In another words, $a$ scales $b$ but does not reflect it (or change its sign).

Now let's apply this concept to matrices.

First, recall that for any vectors $a, b$: $$ a^T b = |a||b|cos\theta $$ where $\theta$ is the angle between $a$ and $b$. The value of $a^T b$ is positive iff $cos\theta > 0$ (i.e. $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$)

Second, recall the definition of a positive-definite matrix $A \in R^{n \times n}$: $$ x^T A x > 0 $$ for all $x \in R^n \backslash \{ 0 \}$.

The meaning of $x^T A x$ is: Take any vector $x$ and transform it into $Ax$ and then take a dot product of $x$ with its "transformed self". Since the result is always positive this must mean that the angle between $x$ and $Ax$ must have been such that $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$. This means that $A$ does not "reflect" any vector, thus behaving like a real (positive) number $a$.

$\endgroup$
1
  • 3
    $\begingroup$ Awesome explanation. $\endgroup$
    – apen
    Oct 2, 2021 at 22:21
5
$\begingroup$

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 $x^TA^{-1}x$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – Shira
    Nov 28, 2011 at 9:52
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 28, 2011 at 10:33
  • $\begingroup$ 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. $\endgroup$
    – fabee
    Nov 28, 2011 at 10:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .