How can I determine if a complex matrix is positive semidefinite?

Question

By definition, a matrix $A$ is positive semidefinite if $\langle A\vec x, \vec x \rangle \ge 0 $ for all $\vec x \in V$ where $V$ is an inner-product space. Consider the matrix

$$ A=\begin{bmatrix} 1&i\newline i&1\end{bmatrix} $$

How can I determine if this matrix is positive semidefinite?

Answer 1

A $n\times n$ Hermitian complex matrix $A$ is said to be positive-definite (positive semi-definite) if $\langle Ax,x \rangle > 0$ $\left(\langle Ax,x \rangle \geq 0\right)$.

A matrix is hermitian if $A=A^{H}$ ($A$ is equal to it's conjugate transpose), which is not satisfied by your matrix.

I therefore simply assume you are given a hermitian matrix. In that case definition does make sense since $\langle Ax,x \rangle \in \mathbb{R}$ which can be proved using the fact that hermitian matrices are self-adjoint ($A=A^{H}=A^{\ast}$). Using the properties of the inner product it follows that $$\alpha := \langle Ax,x\rangle = \overline{\langle x,Ax\rangle} = \overline{\langle A^{\ast}x,\rangle} = \overline{\langle Ax,x\rangle} =\overline{\alpha} $$ and hence $\alpha=\overline{\alpha}$ which is only true for $\alpha \in \mathbb{R}$.

Assuming $\langle \cdot, \cdot \rangle $ is the canonical inner product on $\mathbb{C}^{n}$ given by $$\langle x,y \rangle := y^{H}x=\sum\limits_{i=1}^{n}x_{i}\overline{y}_{i}$$ you can can check if $A$ is positive definite (semi-definit) by calculating whether $$\langle Ax,x \rangle = \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}a_{ij}x_{i}\overline{x_{j}} >0 (\geq 0) $$ for all $x\in \mathbb{C}^{n}$.

For example for the matrix $$ A = \left( \begin{array}{cc} 1 & i \\ -i & 1 \end{array} \right) $$ we get $$\begin{align} \langle Ax, x \rangle &= \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}a_{ij}x_{i}\overline{x_{j}}\\ &= a_{11}x_{1}\overline{x_{1}}+a_{22}x_{2}\overline{x_{2}}+a_{12}x_{1}\overline{x_{2}}+a_{21}x_{2}\overline{x_{1}}\\ &= a_{11}|x_{1}|^{2}+a_{22}|x_{2}|^{2}+a_{12}x_{1}\overline{x_{2}}+\overline{a_{12}}x_{2}\overline{x_{1}}\\ &= a_{11}|x_{1}|^{2}+a_{22}|x_{2}|^{2}+a_{12}x_{1}\overline{x_{2}}+\overline{a_{12}x_{1}\overline{x_{2}}}\\ &= a_{11}|x_{1}|^{2}+a_{22}|x_{2}|^{2}+2Re(a_{12}x_{1}\overline{x_{2}}) \end{align} $$. Plugin in the values of $A$ this yields $\begin{align} \langle Ax, x \rangle=|x_{1}|^{2}+|x_{2}|^{2}+2Re(ix_{1}\overline{x_{2}}) \end{align} $ which can be further simplified by writing $x_{1}:=a+bi$ and x_{2}=c+di$.

Then $Re(ix_{1}x_{2})=Re(i(a+bi)(c+di)=i(ac-bd+adi+bci))=-(ad+bc)$ and $|x_{1}|^2=a^2+b^2$, $|x_{2}|^2=c^2+d^2$. Plugging in yields

$$ \begin{align} \langle Ax, x\rangle &= |x_{1}|^{2}+|x_{2}|^{2}+2Re(ix_{1}=a^2+b^2+c^2+d^2-2ad-2bc\\ &= a^{2}-2ad+c^{2}+b^{2}-2bc+d^{2}\\ &= |(a-c)+(b-c)i|^{2}\\ &= |(x_1 - x_2)|^2 \geq 0 \end{align} $$

Answer 2

A matrix is positive (semi)definite if all of its eigenvalue values are positive (nonnegative)reals.

A matrix’s complex quadratic form will always gives real values iff it is Hermitian.

If your restrict your vectors to real numbers, then it’s possible to have a non-symmetric positive definite matrix, but that doesn’t apply to this instance.

A Hermitian matrix is positive (semi)definite iff all of its eigenvalues are positive (nonnegative). This is clear by diagonalizing the matrix and then considering what happens to the unit vectors.

Your matrix $A$ is not Hermitian, so it’s not semidefinite.

If you did have a 2x2 Hermitian matrix, then the two eigenvalues are nonnegative iff their sum is nonnegative and their product is nonnegative, so it suffices to check whether the trace and determinant are both nonnegative. For larger matrices, this is necessary but not sufficient (ex: 3x3 matrix with eigenvalues $-1,-1,3$).

For example, if you replaced one of the $i$’s with $-i$ in your matrix $A$, then it would be Hermitian with trace $1+1=2$ and determinant $1\cdot 1-i\cdot(-i)=0$, so it would be positive semidefinite. Solving the eigenvalues explicitly gives $0,2 \geq 0$.