I got a positive definite matrix $B,$ that is, $V(x) = x^TBx > 0$ for any vector $ x \neq 0.$ I want to show that $ \lambda_\min \|x\|_2^2 \leq V(x) \leq \lambda_\max \|x\|_2^2$ for any $x \neq 0,$ where $\lambda_\min$ and $ \lambda_\max$ are defined by $$ \lambda_\min = \min \lbrace | \lambda|: \lambda \text{ is an eigenvalue of } B \rbrace$$ and $$ \lambda_\max = \max \lbrace | \lambda|: \lambda \text{ is an eigenvalue of } B \rbrace$$ Any hint please?
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You can start with the fact that $||x||_2^2= x^T I x$ where $I$ is the Identity matrix. Together with the answer of @RobertIsrael this should do the trick – Thomas Nov 27 '12 at 19:34
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Can you show that $B - \lambda_\min I$ is a positive semi-definite? Thanks – Zizo Nov 27 '12 at 19:35
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I know it works if $B$ has a full set of eigenvectors but I am not sure otherwise? – Zizo Nov 27 '12 at 19:36
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See my comment on @RobertIsrael's Answer – Thomas Nov 27 '12 at 19:39
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I did, thanks. I wrote back. – Zizo Nov 27 '12 at 19:42
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Define $B=Q^TDQ$ as the eigendecomposition of $B$. Note that diagonal entries of diagonal matrix $D$ will be positive (since $B$ is positive definite) and will be the eigenvalues of $B$ (denoted as $\lambda_i$ hereafter). In the following, I use the substituitions , $u=\frac{x}{\sqrt{x^Tx}}$ and $z=Qu$. Note that $u$ and $z$ are unit norm vectors. \begin{align} \frac{V(x)}{||x||^2}=\frac{x^TBx}{x^{T}x}=u^{T}Bu=u^{T}Q^{T}DQu=z^TDz=\sum_{i=1}^{N}\lambda_i|z_i|^2 \end{align} where $z_i$'s are entries of the vector $z$. Now what is the minimum value of the above expression. Also what is its maximum value?. In essence, what is the minimum and maximum of a sum of weighted combination of positive numbers ($\lambda_i$'s) such that their (non-negative) weights ($|z_i|^2$) sum to one. (Answer will be the minimum of those numbers and the maximum of them respectively). Infact the above argument works for hermitian matrices also where some of the eigenvalues can be negative. In general, what you have stated is courant fischer minmax theorem for highest and smallest eigenvalues.
dineshdileep
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Hi @dineshdileep, could you take a look here? I think that your approach could be applied to my case. Could you confirm? Many thanks! – nullgeppetto Aug 25 '14 at 13:24
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I don't think I have anything to say more than the accepted answer. The inequality is due to a generalized cauchy-inequality... – dineshdileep Aug 27 '14 at 09:17
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Yes, I see. your approach is -of course- correct, but unfortunately does not apply to my problem! However, it's a nice way to show why it does not apply! – nullgeppetto Aug 27 '14 at 09:56
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@dineshdileep could you please give me a good reference that contain that proof of inequalities ? – SAKLY Sep 17 '21 at 09:27
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@SAKLY see results and references here https://en.wikipedia.org/wiki/Min-max_theorem – dineshdileep Sep 20 '21 at 08:25
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Since $B$ is positive definite, you can forget the absolute value signs in the definitions of $\lambda_{max}$ and $\lambda_{min}$. Note that $B - \lambda_{min} I$ and $\lambda_{max} I - B$ are positive semidefinite.
Robert Israel
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Thanks! I see why I should forget the sign and I also see that $ \lambda_ \min |x|2^2 \leq V(x) \leq \lambda\max |x|2^2$ for any vector $x,$ if $B$ has a full set of eigenvectors that span the space and I don't have information about this. In other words, I can't see why $B - \lambda \min I$ and $\lambda_ \max I - B$ are positive definite? – Zizo Nov 27 '12 at 19:34
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A Matrix is positive definite, iff all its eigenvalues are positive. Now what do you know about the eigenvalues of $B-\lambda_{min}I$ ? – Thomas Nov 27 '12 at 19:38
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I don't have the proof of result that if the eigenvalues are positive then the matrix must be positive definite. I'll think about it – Zizo Nov 27 '12 at 19:41
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think about a diagonizable matrix (http://en.wikipedia.org/wiki/Diagonalizable_matrix) – Thomas Nov 27 '12 at 19:50
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@Zizo: http://math.stackexchange.com/questions/4336/if-eigenvalues-are-positive-is-the-matrix-positive-definite?rq=1 – Martin Argerami Nov 27 '12 at 21:16