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EDIT: This question is actually an attempt to solve this. Please take a look.


Let $A$ be a symmetric postive-definite $n\times n$ matrix, i.e. $A\in\mathbb{S}_{++}^{n}$ Also, let $\mathbf{x}\in\mathbb{R}^n$. Let $Q\colon\mathbf{R}^n\to\mathbb{R}^{*}_{+}$ be the following quadratic form $$ Q(\mathbf{x})=\mathbf{x}^TA\mathbf{x}. $$

If we apply SVD (Singular Value Decomposition) on $A$, we have $$ A=P\Lambda P^T, $$ where $P$ is an orthogonal matrix, and $\Lambda=\operatorname{diag}\{\lambda_1,\ldots,\lambda_n\}$ is the diagonal matrix of the (positive) eigenvalues of $A$, $\lambda_i>0$, $i=1,\ldots,n$.

I would like to express the above quadratic form, $Q(\mathbf{x})$, in terms of the $2$-norm of $\mathbf{x}$, as well as the matrix $A$ (in some way, for instance in terms of the $2$-norm of $\Lambda$, or something else).

What I have thought so far is as follows: $$ Q(\mathbf{x}) = \mathbf{x}^TA\mathbf{x} = \mathbf{x}^T P \Lambda P^T \mathbf{x} = \Big(\mathbf{x}^T P \Lambda^{\frac{1}{2}}\Big)\Big(\Lambda^{\frac{1}{2}} P^T \mathbf{x}\Big) = \Big(\big(P \Lambda^{\frac{1}{2}}\big)^T\mathbf{x}\Big)^T \Big(\Lambda^{\frac{1}{2}} P^T \mathbf{x}\Big). $$

Now, if we set $\mathbf{x}_a=\big(P \Lambda^{\frac{1}{2}}\big)^T\mathbf{x}\in\mathbf{R}^n$, then the quadratic can be rewritten as $$ Q(\mathbf{x}) = \mathbf{x}_a^T\mathbf{x}_a = \big\lVert \mathbf{x}_a \big\rVert^2_2 = \Big\lVert \big(P \Lambda^{\frac{1}{2}}\big)^T\mathbf{x} \Big\rVert^2_2. $$

As far as I know (thanks to @DanielFischer - if I do not misunderstand his words), the following holds true $$ Q(\mathbf{x}) = \Big\lVert \big(P \Lambda^{\frac{1}{2}}\big)^T\mathbf{x} \Big\rVert^2_2 \leq \Big\lVert \big(P \Lambda^{\frac{1}{2}}\big)^T \Big\rVert^2_2 \Big\lVert \mathbf{x} \Big\rVert^2_2. $$

My question is: (a) Are all the above correct? (b) Is there any way of getting rid of the inequality, granted that $A$ is symmetric and positive-definite? Moreover, could we define a function $f\colon\mathbb{R}^n\times\mathbb{S}_{++}^{n}\to\mathbb{R}$, such that $f(\mathbf{x},A)=Q(\mathbf{x})$, where $f$ is expressed in terms of the $2$-norm of $\mathbf{x}$, as well as in terms of $A$ in some way (for instance, in terms of $\Lambda$, etc.)? (c) Any other suggestions?

Thanks in advance!

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(a) the calculation is correct. Your final estimate is nothing else than $Q(x)\le \|A\|_2 \|x\|_2^2$.

As to (b): you cannot get rid of the inequality. It is easy to see [1] $$ \lambda_\min \|x\|_2^2 \le x^TAx \le \lambda_\max \|x\|_2^2 $$ with $\lambda_\min$, $\lambda_\max$ smallest and largest eigenvalue of $A$. Equality in one of these estimates is achieved if $x$ is an eigenvector to the largest/smallest eigenvalue.

It follows that if $$ x^TAx = c \|x\|_2^2 \quad \forall x $$ with some constant $c$, then $A=c\cdot I$. That is, in order to have equality for all $x$ then $A$ must be a multiple of the identity.

[1] Since $A$ is symmetric, there is an orthonormal basis of eigenvectors. This can be used to prove the inequality. I am sure there are already questions & answers addressing this on math.SE.

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  • $\begingroup$ Thanks, but why $Q(x)\le \|A\|_2 \|x\|_2^2$? $\endgroup$ – nullgeppetto Jul 11 '14 at 10:01
  • $\begingroup$ Actually, could you make your answer more informative? Could you explain how did you come up with the first two relations you provide? Thanks! $\endgroup$ – nullgeppetto Jul 11 '14 at 10:13
  • $\begingroup$ $Q(x) \le \|x\|_2 \|Ax\|_2 \le \|A\|_2\|x\|_2^2$ $\endgroup$ – daw Jul 11 '14 at 10:25
  • $\begingroup$ Ok, thanks a lot! This question is actually made because of this one. May have any ideas about? $\endgroup$ – nullgeppetto Jul 11 '14 at 10:47
  • $\begingroup$ Also, it would be nice if it could be $\mathbf{x}^TA\mathbf{x}=\phi(\lambda_1,\ldots,\lambda_n)\|\mathbf{x}\|^2_2$, wouldn't be? @DanielFischer $\endgroup$ – nullgeppetto Jul 11 '14 at 10:59

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