Properties of the Main Diagonal Elements of Positive Definite Matrices
Proposition
Let a definite matrix $A = \left( a_{ij} \right) \in \mathbb{C}^{n \times n}$ be given.
Sign of the diagonal entries
The signs of the diagonal entries $a_{ii}$ of $A$ are 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 $a_{ii} \le 0$
Diagonal entries equal to $0$ in symmetric real matrices
Let $A \in \mathbb{R}^{n \times n}$ be a real positive semidefinite matrix and suppose it is a symmetric matrix. If the diagonal entries $a_{ii}$ of $A$ equal $0$ then the $i$-th row and column are the zero vector.
Remark
This property is used in the proof of the Hog–Craig theorem.
Proof
Without loss of generality (generalness not lost), assume $A$ is positive semidefinite.
That $A$ is positive semidefinite means that for every vector $\mathbf{x} \in \mathbb{R}^{n}$ we have $\mathbf{x}^{T} A \mathbf{x} \ge 0$; hence for the standard basis vectors $\mathbf{x} = \mathbf{e}_{1} , \cdots , \mathbf{e}_{n}$ the quadratic form is greater than or equal to $0$. Since $\mathbf{e}_{i}^{T} A \mathbf{e}_{i} \ge 0$, all diagonal entries $\left( A \right)_{ii}$ of $A$ are also greater than or equal to $0$.
Now assume $A$ is $A \in \mathbb{R}^{n \times n}$ and is symmetric, and let a real number $x$ and an index $j \ne i$ be such that $\mathbf{x} := \mathbf{e}_{i} + x \mathbf{e}_{j}$. If $a_{ii} = 0$ equals $0$, then $\mathbf{x}^{T} A \mathbf{x} \ge 0$, and therefore the following holds. $$ \begin{align*} & \mathbf{x}^{T} A \mathbf{x} \\ =& \left( \mathbf{e}_{i} + x \mathbf{e}_{j} \right)^{T} A \left( \mathbf{e}_{i} + x \mathbf{e}_{j} \right) \\ =& \mathbf{e}_{i}^{T} A \mathbf{e}_{i} + x \mathbf{e}_{i}^{T} A \mathbf{e}_{j} + x \mathbf{e}_{j}^{T} A \mathbf{e}_{i} + x^{2} \mathbf{e}_{j}^{T} A \mathbf{e}_{j} \\ =& a_{ii} + x a_{ij} + x a_{ji} + x^{2} a_{jj} \\ =& 2 x a_{ij} + x^{2} a_{jj} \\ \ge & 0 \end{align*} $$
Quadratic formula: For the quadratic equation $ax^{2}+bx+c=0$ (provided $a\neq 0$) we have $$ x=\dfrac{ -b\pm \sqrt { b^{2}-4ac } }{2a} $$
Since $A$ is assumed positive semidefinite, $a_{jj} \ge 0$. The statement $2 x a_{ij} + x^{2} a_{jj} \ge 0$ means that the quadratic function $f(x) = a_{jj} x^{2} + 2 a_{ij} x$, whose graph is a downward-opening parabola, does not meet the $x$-axis or meets it at only one point; in other words, the discriminant satisfies $b^{2} - 4ac \le 0$. Substituting $f$ into the discriminant yields: $$ \left( 2 a_{ij} \right)^{2} - 4 \cdot a_{jj} \cdot 0 \le 0 $$ It follows that the only solution satisfying $4 a_{ij}^{2} \le 0$ is $a_{ij} = 0$. This holds regardless of the choice of $j$, so $a_{i1} = \cdots = a_{in} = 0$, and since $A$ is symmetric, we also have $a_{1i} = \cdots = a_{ni} = 0$.
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