📂Matrix Algebra

Properties of the Main Diagonal Elements of Positive Definite Matrices

Theorem

Suppose a positive definite matrix is given as $A = \left( a_{ij} \right) \in \mathbb{C}^{n \times n}$.

Sign of the Main Diagonal Elements

The sign of the main diagonal elements in $A$, denoted as $a_{ii}$, is the same as the sign of $A$.

• If $A$ is positive definite, then $a_{ii} > 0$
• If $A$ is positive semidefinite, then $a_{ii} \ge 0$
• If $A$ is negative definite, then $a_{ii} < 0$
• If $A$ is negative semidefinite, then $a_{ii} \le 0$

Main Diagonal Elements in Symmetric Real Matrix satisfying $0$

Suppose a semidefinite matrix $A \in \mathbb{R}^{n \times n}$ is a symmetric matrix composed of real numbers. If the main diagonal components $a_{ii}$ of $A$ satisfy $0$, then the $i$-th row and column are zero vectors.

Explanation

This property is used in the proof of the Hogg-Craig theorem.

Proof

Without loss of generality, let us assume $A$ is positive semidefinite.

The fact that matrix $A$ is positive semidefinite means that for every vector $\mathbf{x} \in \mathbb{R}^{n}$, $\mathbf{x}^{T} A \mathbf{x} \ge 0$ holds, hence the quadratic form with respect to standard basis vectors $x = \mathbf{e}_{1} , \cdots , \mathbf{e}_{n}$ must also be greater than or equal to $0$. Therefore, since $\mathbf{e}_{i}^{T} A \mathbf{e}_{i} \ge 0$, all main diagonal elements $\left( A \right)_{ii}$ of $A$ must also be greater than or equal to $0$.

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Now, assume $A$ is $A \in \mathbb{R}^{n \times n}$ and a symmetric matrix, and let the real number be $x$ and the index be $j \ne i$ such that $\mathbf{x} := \mathbf{e}_{i} + x \mathbf{e}_{j}$. If $a_{ii} = 0$ is $0$, then $\mathbf{x}^{T} A \mathbf{x} \ge 0$ must hold, leading to: $$\begin{align*} & \mathbf{x}^{T} A \mathbf{x} \\ =& \left( \mathbf{e}_{i} + x \mathbf{e}_{j} \right)^{T} A \left( \mathbf{e}_{i} + x \mathbf{e}_{j} \right) \\ =& \mathbf{e}_{i}^{T} A \mathbf{e}_{i} + x \mathbf{e}_{i}^{T} A \mathbf{e}_{j} + x \mathbf{e}_{j}^{T} A \mathbf{e}_{i} + x^{2} \mathbf{e}_{j}^{T} A \mathbf{e}_{j} \\ =& a_{ii} + x a_{ij} + x a_{ji} + x^{2} a_{jj} \\ =& 2 x a_{ij} + x^{2} a_{jj} \\ \ge & 0 \end{align*}$$

Quadratic Formula: For the quadratic equation $ax^{2}+bx+c=0$ (where $a\neq 0$), $$x=\dfrac{ -b\pm \sqrt { b^{2}-4ac } }{2a}$$

Assuming $A$ is a positive semidefinite matrix implies $a_{jj} \ge 0$. The statement $2 x a_{ij} + x^{2} a_{jj} \ge 0$ implies that the graph of the downward convex parabola represented by the quadratic function $f(x) = a_{jj} x^{2} + 2 a_{ij} x$ does not meet the $x$-axis or meets it at only one point, which, through the discriminant, indicates $b^{2} - 4ac \le 0$. Substituting $f$ into the discriminant gives: $$\left( 2 a_{ij} \right)^{2} - 4 \cdot a_{jj} \cdot 0 \le 0$$ According to this, the only case satisfying $4 a_{ij}^{2} \le 0$ is $a_{ij} = 0$. This holds regardless of the choice of $j$, therefore $a_{i1} = \cdots = a_{in} = 0$, and since $A$ is a symmetric matrix, $a_{1i} = \cdots = a_{ni} = 0$ as well.

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