91Ó°ÊÓ

Since $\mathrm{A}={\left[{\mathrm{a}}_{\mathrm{ij}}\right]}_{\mathrm{n}×\mathrm{n}}$Because ${\mathrm{e}}_{\mathrm{i}}={(0,....,0,1,0,...,)}^{\mathrm{T}}∈{\mathrm{R}}^{\mathrm{n}}$(a vector with 1 at the i-th coordinate and 0 elsewhere) is positive definite, we have $0$0<{\mathrm{e}}_{\mathrm{i}}^{\mathrm{T}}{\mathrm{Ae}}_{\mathrm{i}}={\mathrm{a}}_{\mathrm{ii}}$for each $1≤\mathrm{i}≤\mathrm{n}$. As a result, each of A 's diagonal entries, namely ${\mathrm{a}}_{\mathrm{ii}}$, is positive. Let ${\mathrm{a}}_{\mathrm{kk}}$represent the greatest of A 's diagonal entries. It is now necessary to demonstrate that ${\mathrm{a}}_{\mathrm{kk}}$is also the greatest of the off diagonal elements.

Consider i and j as two integers with the property that $\mathrm{i}≠\mathrm{j},1≤\mathrm{i}≤\mathrm{j}≤\mathrm{n}$, , and the function $\mathrm{p}\left(\stackrel{→}{\mathrm{x}}\right)$as defined in Exercise 53. We know that is positive definite when the quadratic form defined by the symmetric matrix is positive definite because of Exercise 53(b). As a result, we must have$1≤\mathrm{i}≤\mathrm{j}≤\mathrm{n}$

$\mathrm{B}=\left[\begin{array}{cc}{\mathrm{a}}_{\mathrm{ii}}& {\mathrm{a}}_{\mathrm{ij}}\\ {\mathrm{a}}_{\mathrm{ji}}& {\mathrm{a}}_{\mathrm{jj}}\end{array}\right]$

To be positive definite, that is,

$0<\det \left(\mathrm{B}\right)={\mathrm{a}}_{\mathrm{ii}}{\mathrm{a}}_{\mathrm{jj}}-{\mathrm{a}}_{\mathrm{ij}}^2<{\mathrm{a}}_{\mathrm{kk}}^2-{\mathrm{a}}_{\mathrm{ij}}^2=({\mathrm{a}}_{\mathrm{kk}}-{\mathrm{a}}_{\mathrm{ij}})({\mathrm{a}}_{\mathrm{kk}}+{\mathrm{a}}_{\mathrm{ij}})$

Now, role="math" localid="1660724884116" $({\mathrm{a}}_{\mathrm{kk}}-{\mathrm{a}}_{\mathrm{ij}})({\mathrm{a}}_{\mathrm{kk}}+{\mathrm{a}}_{\mathrm{ij}})>0$implies the following two cases:

Case 1:

$({\mathrm{a}}_{\mathrm{kk}}-{\mathrm{a}}_{\mathrm{ij}})({\mathrm{a}}_{\mathrm{kk}}+{\mathrm{a}}_{\mathrm{ij}})>0⇒({\mathrm{a}}_{\mathrm{kk}}-{\mathrm{a}}_{\mathrm{ij}})>0\mathrm{and}({\mathrm{a}}_{\mathrm{kk}}+{\mathrm{a}}_{\mathrm{ij}})>0$

That is ${\mathrm{a}}_{\mathrm{kk}}>{\mathrm{a}}_{\mathrm{ij}}\mathrm{and}{\mathrm{a}}_{\mathrm{kk}}>-{\mathrm{a}}_{\mathrm{ij}}\mathrm{for}\mathrm{each}1≤\mathrm{i}≤\mathrm{j}≤\mathrm{n}$, which further implies ${\mathrm{a}}_{\mathrm{kk}}>\left|{\mathrm{a}}_{\mathrm{ij}}\right|â‰\yen 0\mathrm{for}\mathrm{each}1≤\mathrm{i}≤\mathrm{j}≤\mathrm{n}$.

Case 2:

$({\mathrm{a}}_{\mathrm{kk}}-{\mathrm{a}}_{\mathrm{ij}})({\mathrm{a}}_{\mathrm{kk}}+{\mathrm{a}}_{\mathrm{ij}})>0⇒({\mathrm{a}}_{\mathrm{kk}}-{\mathrm{a}}_{\mathrm{ij}})<0\mathrm{and}({\mathrm{a}}_{\mathrm{kk}}+{\mathrm{a}}_{\mathrm{ij}})<0\mathrm{that}\mathrm{is}{\mathrm{a}}_{\mathrm{kk}}<{\mathrm{a}}_{\mathrm{ij}}$for each $1≤\mathrm{i}≤\mathrm{j}≤\mathrm{n}$, which further implies ${\mathrm{a}}_{\mathrm{kk}}<-\left|{\mathrm{a}}_{\mathrm{ij}}\right|<0$for each $1≤\mathrm{i}≤\mathrm{j}≤\mathrm{n}$which contradicts that ${\mathrm{a}}_{\mathrm{kk}}>0$and thus, this case is not possible for a positive definite A.

As a result, the greatest entry in the matrix A, which is a diagonal element, is ${\mathrm{a}}_{\mathrm{kk}}$. As a result, the diagonal contains the greatest entry of a positive definite matrix.

Proof based on A 's positive definiteness and the result of Exercise53(b)