Why is this method valid to show function is PSD?

[Master1022]

Homework Statement

We are looking at the function: $f(x, y) = \frac{x^2}{y}, x \in \mathbb{R} , y > 0$. Is this function positive semi-definite.

Relevant Equations

Hessian

Hi,

I was looking at the following example, and I cannot really understand the justification for why this function is positive semi-definite.

Example Problem: We are looking at the function: $f(x, y) = \frac{x^2}{y}, x \in \mathbb{R} , y > 0$. Is this function positive semi-definite.

Solution provided: We can calculate the Hessian to yield:
\nabla^2 f(x, y) = \frac{2}{y^3} \begin{pmatrix} y^2 & -xy \\ -xy & x^2 \end{pmatrix} = \frac{2}{y^3} \begin{pmatrix} y \\ -x \end{pmatrix} \begin{pmatrix} y \\ -x \end{pmatrix}^T \succcurlyeq 0

Therefore, the function is PSD.

My Question: I am not fully convinced by the last step of the argument. I can understand the decomposition, but it isn't obvious to me why that implied positive semi-definiteness.

From what I understand, a matrix $A$ can be said to be PSD if $x^T Ax \geq 0$ for all $x$. We can also check that the eigenvalues all have real parts $\geq 0$. However, the solution above shows that the Hessian matrix is separable and is multiplied by a pre-factor which is always $\geq 0$ because $y > 0$. I mean, perhaps I could 'transpose' the expression they have and think about the $\frac{2}{y^3}$ as the $A$ matrix, but I am also not sure that is valid to do.Any help would be greatly appreciated as I am sure this is quite a simple question and I am missing something obvious.

 

[Orodruin]

Here is an exercise:

For any column vector $v$, show that $vv^T$ is a positive semi definite matrix.

 

[Master1022]

Here is an exercise:

For any column vector $v$, show that $vv^T$ is a positive semi definite matrix.

Ah yes.

Okay so the $v v^T$ resembles a covariance matrix. It has rank 1 because row can be written as $v_i v$ (that is element $v_i$ multiplied by $v$) and so we can do some row reduction. Therefore because it is rank 1, it has 1 non-zero eigenvalue. We know that the eigenvalues of a matrix sum to the trace so the one non-zero eigenvalue is $\lambda = ||v||_2 ^2$, while the other $n - 1$ eigenvalues are 0.

We can see that the eigenvalues are $\geq 0$ and therefore $v v^T$ is PSD.

Relating it back to my question, we have a $k v v^T$ with some positive constant $k$ and so now that makes sense why the matrix is PSD.

Thank you.

 

[Orodruin]

While that works, the more straightforward way is to directly use the definition of positive semi definite. For any $u$, $a = u^T v$ is a number and so $a = u^T v = v^T u$. Hence, with $A = vv^T$, $u^TAu = u^T v v^T u = a^2 \geq 0$ and thus $vv^T$ is positive semi definite.