0
$\begingroup$

Let $f:H \times H \to \mathbb{R}$ be a mapping with $H$ a Hilbert space.

Let $A$ be a matrix with entries $a_{ij}=f(b_i, b_j)$ with $$a_{ii}=f(b_i, b_i) \geq C\lVert b_i\rVert_{H}^2.$$ Suppose $b_i \neq b_j$ and $(b_i, b_j)_H = 0$ for $i \neq j$ . How do I show that $A$ is invertible?

$\endgroup$
9
  • $\begingroup$ I don't think $f$ can be constant by the coercivity condition on it. $\endgroup$
    – matt.w
    May 20 '13 at 12:57
  • $\begingroup$ Oops, my bad. I'll delete that comment. But what if all $b_j$ are equal? $\endgroup$ May 20 '13 at 13:00
  • $\begingroup$ Good point, I added more detail. $\endgroup$
    – matt.w
    May 20 '13 at 13:04
  • $\begingroup$ If $(b_i)$ is orthonormal, for instance, you could consider the function $f\equiv 1$. Then the matrix is not invertible. Did you miss some assumptions? $\endgroup$
    – Julien
    May 20 '13 at 15:59
  • $\begingroup$ @julien $f$ is coercive. $\endgroup$ May 20 '13 at 15:59
1
$\begingroup$

If $f(u,v)$ is given by scalar product $(Bu,v)_H$, $B\in\mathcal L(H,H)$ - symetric continuous linear operator which is positive definite (because $f$ is coercive). If $b_j$ are linearly independent, then the matrix $A$ is a metric tensor on $\text{span} \{b_j\}$ and it should be invertible.

Edit I'll develop a little on this case. Suppose my hypothesis is true and $a_{i,j}=(Bb_i,b_j)_H$, $i,j=1..n$. Suppose that $A$ is singular, then there exists $u\in\mathbb R^n$ such that $(Au,u)_{\mathbb R^n} =0 $, but

$\displaystyle (Au,u)_{\mathbb R^n} =\sum_i\sum_j (Bb_i,b_j)_Hu_i u_j = \left( B\left(\sum_i b_i u_i\right),\left(\sum_j b_j u_j\right)\right)_H \ge C\left\|\sum_i b_i u_i\right\|_H^2>0$. Hence $A$ is invertible. As it's easy to see, this proof relies heavily on the fact that $f$ is given by a scalar product.

$\endgroup$
3
  • $\begingroup$ Thanks. Your sentence "If $b_j$ are linearly independent, then the matrix $A$ is a metric tensor on span$\{b_j\}$ and it should be invertible" is stand alone, right? I don't need $f$ to be given by a scalar product? Do you know where I can get a proof? $\endgroup$
    – matt.w
    May 20 '13 at 15:32
  • $\begingroup$ Well "continuous&bilinear" implies "given by a scalar product". If you study just a coercive function, then you need some additional hypothesis. Take, for example, $f(u,v) = \|u\|_H\cdot\|v\|_H$, it's coercive. Take $b_j$ - elements of orthonormal basis of $H$. Clearly, all elements of $A$ are equal to $1$, and hence $A$ is singular. $\endgroup$ May 20 '13 at 15:43
  • $\begingroup$ And no, "If $b_j$ are linearly independent, then the matrix $A$ is a metric tensor on $\text{span}\{b_j\}$ and it should be invertible" is not standalone (cf. comment above). $\endgroup$ May 20 '13 at 15:47

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.