slimbluejays
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- Jan 31, 2021
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I would like to prove that for an [MATH]n \times n[/MATH] symmetric matrix [MATH]P[/MATH] there is always a scalar [MATH]a \in \mathbb{R}[/MATH] such that [MATH]P+aI[/MATH] is negative definite.
Firstly, we prove that the eigenvalues of [MATH]P+aI[/MATH] are given by [MATH]\lambda_{n} + a, n = 1, ..., n[/MATH] assuming the eigenvalues of P are [MATH]\lambda_{n}, n = 1, ..., n[/MATH].
[MATH] Px=\lambda x \\ Px + aIx = \lambda x + aIx \\ (P+aI)x = (\lambda + a)x [/MATH]Then, for a negative definite [MATH]P+aI[/MATH], all eigenvalues must be negative, which means that
[MATH]\lambda_n + a < 0[/MATH] for all [MATH]n = 1,...,n[/MATH]
This means that we just need to find a value of [MATH]a[/MATH] such that
[MATH]a < -\lambda_{max}[/MATH]
Does this proof suffice?
Firstly, we prove that the eigenvalues of [MATH]P+aI[/MATH] are given by [MATH]\lambda_{n} + a, n = 1, ..., n[/MATH] assuming the eigenvalues of P are [MATH]\lambda_{n}, n = 1, ..., n[/MATH].
[MATH] Px=\lambda x \\ Px + aIx = \lambda x + aIx \\ (P+aI)x = (\lambda + a)x [/MATH]Then, for a negative definite [MATH]P+aI[/MATH], all eigenvalues must be negative, which means that
[MATH]\lambda_n + a < 0[/MATH] for all [MATH]n = 1,...,n[/MATH]
This means that we just need to find a value of [MATH]a[/MATH] such that
[MATH]a < -\lambda_{max}[/MATH]
Does this proof suffice?