# An argument from a blog article of Terence Tao

Let $A_1, A_2, A_3, \ldots , A_m$ be positive semi-definite Hermitian matrices and then consider the polynomial $p(z,z_1,z_2,\ldots,z_m) = \det(z+z_1A_1 + z_2A_2 + \cdots+z_mA_m)$

Now Tao argues that if $z,z_1,z_2,\ldots,z_m$ have a positive imaginary part then the "skew-adjoint part" (what is this?) of $z+z_1A_1 + z_2A_2 + \cdots+z_mA_m$ is strictly positive definite and hence the quadratic form $\operatorname{Im} [ \langle (z+z_1A_1 + z_2A_2 + \cdots+z_mA_m)v, v \rangle ]$ is non-degenerate and hence it follows that $z+z_1A_1 + z_2A_2 + \cdots+z_mA_m$ is non-singular.

Can someone kindly help understand what happened here?

• try it for 2 by 2 matrices with $m=1$ and then $m=2.$ – Will Jagy Nov 8 '14 at 19:42

For a matrix $X$ over $\mathbb{C}$, we can write $X = \frac{1}{2}(X + X^\dagger) + \frac{1}{2}(X - X^\dagger)$ with the first term self-adjoint and the second term skew-adjoint. Put $A = z + z_1 A_1 + \cdots + z_n A_n$, so that $p = \det A$. Then $$A^\dagger = \bar{z} + \bar{z_1} A_1 + \cdots + \bar{z}_m A_m,$$ and so $$\frac{1}{2}(A - A^\dagger) = \text{Im }z+ (\text{Im }z_1) A_1 + \cdots + (\text{Im } z_m)A_m,$$ which is clearly strictly positive-definite.
• Thanks! Firstly are your ellipsis in the wrong place in the last line? Secondly, hows does the positive-definiteness of the skew-adjoint part follow from what you said? [..I believe that a matrix $A$ is called positive-definite if $x^\dagger A x >0$ for all vectors $x \neq 0$ - right?..] Then you say that this skew-adjoint form satisfies this and the self-adjoint form also satisfies this trivially? Is it? (all because of the "z" term that the semi-definiteness of the $A_i$s get lifted to positive-definiteness of the $A$ - right?) – Anirbit Nov 9 '14 at 18:37