# How to show this matrix is invertible?

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?

• I don't think $f$ can be constant by the coercivity condition on it. May 20 '13 at 12:57
• Oops, my bad. I'll delete that comment. But what if all $b_j$ are equal? May 20 '13 at 13:00
• Good point, I added more detail. May 20 '13 at 13:04
• 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? May 20 '13 at 15:59
• @julien $f$ is coercive. May 20 '13 at 15:59

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.
• 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? May 20 '13 at 15:32
• 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. May 20 '13 at 15:43
• 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). May 20 '13 at 15:47