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?

  • 1
    $\begingroup$ I changed {$v_1,v_2,...,v_k$} to $\{v_1,v_2,\ldots,v_k\}$. That is standard usage. $\qquad$ $\endgroup$ May 15 '18 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 $j$.

  • $\begingroup$ Why is the linear combination of the rows is zero? $\endgroup$ May 15 '18 at 18:13
  • $\begingroup$ You want to prove that if a linear combination is zero, then the coefficients have all to be zero. $\endgroup$ May 16 '18 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 $W^T$


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.