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Matrix$A+{\mathrm{kl}}_n$

symmetric$n脳n$ matrix A

Any n times n symmetric matrix A determines the quadratic form$q\left(x\right)=x^T\mathrm{Ax}$in a unique way. Let${位}_1,{位}_2,鈥�,{位}_n$represent${\mathrm{A}}^{\text{'}}\mathrm{s}$eigenvalues and B represent${\mathrm{A}}^{\text{'}}\mathrm{s}$eigen basis. The quadratic form $q\left(x\right)=x^T\mathrm{Ax}$.

$q\left(x\right)=x^{T}Ax={位}_1{c}_1^2+{位}_2{c}_2^2+L+{位}_n{c}_n^2-\left(1\right)$

thus be written as, where $(c_1,c_2,鈥�,c_n{)}^T$are the coordinate vectors of x in relation to B. As a result of (1), we get.

${位}_{m}in{x}^{T}x鈮�x^{T}Ax鈮�{位}_{m}ax{x}^{T}x$

That is,

${位}_{m}in鈥�x{鈥�}^2鈮�x^{T}Ax鈮�{位}_{m}ax鈥�x{鈥�}^2$

Consider the following to see if $A+{\mathrm{kl}}_n$is definite.

$$\begin{gathered} x^T(A+k{I}_n)x=x^{T}Ax+k{x}^{T}x \\ 鈮�{位}_{m}in鈥�x{鈥�}^2+k鈥�x{鈥�}^2 \\ =({位}_{m}in+k)鈥�x{鈥�}^2 \end{gathered}$$

For$A+{\mathrm{kl}}_n$ to be positive definite, we must have$x^T(A+k{I}_n)x>0$ for all $x鈮�0$. Thus , by choosing$k>\left|{位}_{m}in\right|,({位}_{m}in+k)鈥�x{鈥�}^2>0$ for all $x鈮�0$.

As a result, there exists k such that$A+{\mathrm{kl}}_n$ is a positive definite matrix for any symmetric matrices A of order n.

there exists such that$A+{\mathrm{kl}}_n$ is a positive definite matrix for any symmetric matrices , using the fact that${\mathrm{位}}_{\min }{\mathrm{x}}^{\mathrm{T}}\mathrm{x}鈮�{\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}鈮�{\mathrm{位}}_{\max }{\mathrm{x}}^{\mathrm{T}}$ and choosing$k>\left|{位}_m\right|$