[Math] Proof: $\det\pmatrix{\langle v_i , v_j \rangle}\neq0$ $\iff \{v_1,\dots,v_n\}~\text{l.i.}$

Question

Let $V$ be a real inner product space and $S=\{v_1,v_2, \dots, v_n\}\subset V$. How am I to prove that $S$ is linearly independent if and only if the determinant of the matrix
$$ a_{ij}=\pmatrix{\langle v_i , v_j \rangle}$$
is nonzero?

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Just to be clear, the matrix we're talking about is this one:

$$\pmatrix{\langle v_1,v_1\rangle & \langle v_1,v_2\rangle &\langle v_1,v_3\rangle & \cdots & \langle v_1,v_{n-1}\rangle & \langle v_1, v_n\rangle \\\langle v_2,v_1\rangle & \langle v_2,v_2\rangle &\langle v_2,v_3\rangle & \cdots & \langle v_2,v_{n-1} \rangle & \langle v_2,v_n\rangle \\\langle v_3,v_1\rangle & \langle v_3,v_2\rangle &\langle v_3,v_3\rangle & \cdots & \langle v_3,v_{n-1}\rangle & \langle v_3,v_n \rangle \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\\langle v_{n-1}, v_1\rangle & \langle v_{n-1},v_2\rangle &\langle v_{n-1},v_3\rangle & \cdots & \langle v_{n-1},v_{n-1}\rangle & \langle v_{n-1},v_n\rangle \\\langle v_n,v_1\rangle & \langle v_n,v_2\rangle &\langle v_n,v_3\rangle & \cdots & \langle v_n,v_{n-1}\rangle & \langle v_n,v_n\rangle \\ }$$

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I highly doubt that anybody here has Roman's Advanced Linear Algebra, or maybe you do, but I think on page $261$ there is a small note on something which looks similar.

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Should anybody need the code (C++) in their research, here is a gadget which streams LaTeX code to a file named "matrix.txt" for an $n \times n$ matrix such as this with some value for $n$:

ofstream fout;
fout.open("matrix.txt");

int n;
cout << "Enter your desired n: ";
cin >> n;

fout << endl;

fout << "$\\begin{pmatrix}" << endl;

for( int j = 1 ; j <= n ; j++ )
{
    for( int i = 1 ; i<= n ; i++ )
    {
        if( j == i && i == n )
        {
            fout << "\\langle v_" << j << "," << "v_" << i << " \\rangle" << endl;
        }
        else
        {
            if( i == n )
            {
                fout << "\\langle v_" << j << "," << "v_" << i << " \\rangle\\\\" << endl;
            }
            else
            {
                fout << "\\langle v_" << j << "," << "v_" << i << " \\rangle&" << endl;
            }
        }
    }
    //fout << endl;
}

fout << "\\end{pmatrix}$" << endl << endl;

For n equal to 5 you get this result:

$\begin{pmatrix}
\langle v_1,v_1 \rangle&
\langle v_1,v_2 \rangle&
\langle v_1,v_3 \rangle&
\langle v_1,v_4 \rangle&
\langle v_1,v_5 \rangle\\
\langle v_2,v_1 \rangle&
\langle v_2,v_2 \rangle&
\langle v_2,v_3 \rangle&
\langle v_2,v_4 \rangle&
\langle v_2,v_5 \rangle\\
\langle v_3,v_1 \rangle&
\langle v_3,v_2 \rangle&
\langle v_3,v_3 \rangle&
\langle v_3,v_4 \rangle&
\langle v_3,v_5 \rangle\\
\langle v_4,v_1 \rangle&
\langle v_4,v_2 \rangle&
\langle v_4,v_3 \rangle&
\langle v_4,v_4 \rangle&
\langle v_4,v_5 \rangle\\
\langle v_5,v_1 \rangle&
\langle v_5,v_2 \rangle&
\langle v_5,v_3 \rangle&
\langle v_5,v_4 \rangle&
\langle v_5,v_5 \rangle
\end{pmatrix}$

As is lengthily explained here.

Answer 1

Let $V=[v_1,v_2,\dots,v_n]$. Then $A=V^TV$. So

$$\det(A)=\det(V^TV)=\det(V^T)\det(V)=\det(V)^2.$$

The determinant of a square matrix is non-zero if and only if its columns form a linearly independent set of vectors.