# Invertible matrix of inner product values

Let $\{v_1,v_2,\ldots,v_k\}$ be a basis in inner product space $V.$

I need to prove that the matrix

$$A= \begin{pmatrix} (v_1,v_1) & (v_1,v_2) & \cdots &(v_1,v_k) \\ \ \vdots & \vdots & \ddots&\vdots\\ (v_k,v_1) & (v_k,v_2) & \cdots&(v_k,v_k) \end{pmatrix}$$

is invertible.

Any hints?

• I changed {$v_1,v_2,...,v_k$} to $\{v_1,v_2,\ldots,v_k\}$. That is standard usage. $\qquad$ May 15, 2018 at 18:00

Hint 1

Take a linear combination of the rows which is zero.

Hint 2

So there are coefficients $a_{i}$ such that $\sum_{i=1}^{k} a_{i} (v_{i}, v_{j}) = 0$ for all $j$.

Hint 3

Your aim is to show that all $a_{i}$ are zero.

Hint 4

So for the vector $v = \sum_{i=1}^{k} a_{i} v_{i}$ you have $(v, v_{j}) = 0$ for all $j$.

• Why is the linear combination of the rows is zero? May 15, 2018 at 18:13
• You want to prove that if a linear combination is zero, then the coefficients have all to be zero. May 16, 2018 at 8:58

start with an orthonormal basis $e_i$ of $V$ and create the matrix $W$ where the entries of column $j$ are the elements of $v_j$ in that basis ($w_{ij} = e_i \cdot v_j$ )

Then $$W^T W = A$$

Meanwhile, $W$ is just a change of basis matrix. Is it allowed to be singular?

Your construction is often called the Gram matrix, sometimes Grammian, although I suspect Grammian should refer to the determinant.

See any lattice in LATTICES. Note that, in order to keep the entries of $W$ integers for an integral lattice, their $W$ may not be square; let's see, the matrix they call BASIS is my $W^T$