I've been working on convex optimization and got stuck.

What exactly does a positive definite(p.d) matrix represent geometrically ? what kind of vector space it forms ?

If I have a p.d matrix which represent a convex cone (which I can't understand why), how do I prove the convexity for that matrix ? What's the input variable say X should be ?

Say if I have a plane, $$W^TX = B$$

at least I know I should put X into the equation, but for a p.d matrix...

it's just a matrix, why does that even represent a function ?

I am totally confused. Any hint helps a lot.

  • 1
    $\begingroup$ It is not the matrix that represent a convex cone. The space off all positive definite matrix is a convex cone. You have to prove the convexity of the space, i.e. if $\alpha\in [0,1]$ and $X,Y$ are p.d.m. then $\alpha X+(1-\alpha)Y$ is an p.d.m. $\endgroup$
    – Tomás
    Commented Oct 14, 2012 at 14:49
  • 1
    $\begingroup$ Can you elaborate more? it seems that the proof start from any p.d matrix is convex, so their convex combination is convex too, which then goes back to my original question is that corect? $\endgroup$
    – Jing
    Commented Oct 15, 2012 at 8:07

2 Answers 2


Ok. Let $P$ be the set of all positive definite matrix. Im gonna show that if $X,Y\in P$ and $\alpha,\beta>0$ then $\alpha X+\beta Y\in P$. Note \begin{eqnarray} x^{\top}(\alpha X+\beta Y)x &=& \alpha x^{\top} Xx+\beta x^{\top} Yx \nonumber \\ &>& 0\end{eqnarray}

Above, i have used algebraic properties of matrix product and the positive definitiness of $X$ and $Y$. With this you can conclude that $\alpha X+\beta Y\in P$

  • $\begingroup$ But positive definite matrix does not include the origin. Thus, not a cone! $\endgroup$
    – Mohammad
    Commented Oct 7, 2017 at 19:20

Let us take a simple positive definite matrix: $A = \bigl(\begin{smallmatrix} 1&0 \\ 0&1 \end{smallmatrix} \bigr)$

Let $\vec{x}=(x_1, x_2)$ be a vector such as $\vec{x}\in \mathcal{R}^2$, $\vec{x} \neq \vec{0}$

Now compute the quadratic form: $\vec{x}^\top A \vec{x} = x_1^2 + x_2^2$; which is definitely convex, since the square term is a convex function and the sum of two convex terms is convex. You can verify by plotting the above function


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