# A $2\times 2$ hermitian matrix is positive semidefinite iff both its trace and determinant are nonnegative

I am having trouble proving that a $$2\times 2$$ hermitian matrix is positive semidefinite iff both its trace and determinant are nonnegative.

I have written out the definitions and done algebraic manipulations, but don't seem to be getting much further with any way of the implication. Any help would be appreciated, as I already spend the best part of the evening hitting at it!

• Do you already know that any Hermitian matrix is positive semidefinite iff all its eigenvalues are nonnegative? If not then prove that and come back. – Ian Feb 27 at 20:51
• Sounds useful and I wasn't aware of that, maybe that's the piece I am missing. Thanks! – Megahyttel Feb 27 at 20:56
• Thanks a lot, did the proof you suggested and piecing it together with MachineLearner's hint got me through. Thanks to both of you! – Megahyttel Feb 28 at 15:56

$$\boldsymbol{A} = \begin{bmatrix} a & b\\ c & d\end{bmatrix}$$
$$\chi(\lambda) = \text{det}\begin{bmatrix} a-\lambda & b\\ c & d-\lambda\end{bmatrix}$$ $$=(a-\lambda)(d-\lambda)-bc$$ $$=ad-bc-(a+d)\lambda+\lambda^2$$ $$=\lambda^2-\text{Tr}\boldsymbol{A}\,\lambda+\text{det}\boldsymbol{A}$$