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!


1 Answer 1


(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.

  • $\begingroup$ Thanks, but why $Q(x)\le \|A\|_2 \|x\|_2^2$? $\endgroup$ Jul 11, 2014 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! $\endgroup$ Jul 11, 2014 at 10:13
  • $\begingroup$ $Q(x) \le \|x\|_2 \|Ax\|_2 \le \|A\|_2\|x\|_2^2$ $\endgroup$
    – daw
    Jul 11, 2014 at 10:25
  • $\begingroup$ Ok, thanks a lot! This question is actually made because of this one. May have any ideas about? $\endgroup$ Jul 11, 2014 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 $\endgroup$ Jul 11, 2014 at 10:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.