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

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?

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

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

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.

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.

• I have Roman's book but, as far as I could see reading it very quickly, it only mentions the matrix of bilinear forms in general (which is what we want, of course), but says nothing about the particular case you're interested in Jun 1, 2013 at 9:03
• You seem interested in a rather particular, restricted case: when we have a square matrix, though with the appropiate changes the claim's true in general, just taking $\,A^tA\,$ , say Jun 1, 2013 at 9:04
• You did not say that $\dim V=n$ and even less that $V=\Bbb R^n$; that makes interpreting $v_1,\ldots,v_n$ as columns of a matrix (as is done in the two answers so far) inappropriate, and taking their determinants impossible. It might be thought that this is the case you have in mind (as @DonAntonio suggests), in which case you should say so. Jun 1, 2013 at 9:19
• Related Wikipedia link: Gramian matrix. A related question at MSE: Gram matrix invertible iff set of vectors linearly independent. Jun 1, 2013 at 11:44
• I've noticed that you have asked quite a few questions recently. I wanted to make sure that you are aware of the quotas 50 questions/30 days and 6 questions/24 hours, so that you can plan posting your questions accordingly. (If you try to post more questions, stackexchange software will not allow you to do so.) For more details see meta. Jun 1, 2013 at 11:49

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.

• A problem is that you assume the $V$ is square. Nov 28, 2017 at 9:22
• Another thread deals with exactly the same problem: math.stackexchange.com/questions/36580/… . . . Note that as pointed out in a comment, $A = V^T V$ isn't quite general.
– Ryo
Jan 3 at 15:45

Let $V=(v_1,\dots,v_n)$, then $(a_{ij})=V^TV$, so $\det (a_{ij})=\det V^TV=(\det V)^2$.

• $V$ is what now? Is $V = S$? Please, if you don't mind, explain your symbols. Jun 2, 2013 at 0:53
• $V$ is a matrix with column vectors $v_i$, right? Jun 2, 2013 at 2:04

I'll give an answer that does not assume that $\def\R{\Bbb R}V=\R^n$ or even that $\dim V=n$, even though the essential argument boils down to that case anyway.

One direction is easy: if the $v_i$ satisfy a nontrivial linear dependence relation, then so do the corresponding rows of your matrix (and the columns as well) which forces the determinant to be zero.

In the other direction, suppose the $v_i$ are linearly independent. Now $W=\left<v_1,\ldots,v_n\right>\subseteq V$ is a subspace of dimension$~n$. Then the linear map $f:W\to\R^n$ given by $f(w)=(\left<v_1,w\right>,\ldots,\left<v_n,w\right>)$ is injective: a vector $w$ in the kernel of $f$ is othogonal to each $v_i$, hence by linearity to each vector of $W$, and this implies $w=0$ because (the restriction to$~W$ of) the bilinear form is non-degenerate. The image by the injective linear map $f$ of the linearly independent $n$-tuples of vectors $v_1,\ldots,v_n$ in$~W$ gives a linearly independent $n$-tuple of vectors in$~\R^n$, whose determinant is nonzero.

A less abstract way to finish off the second part is to choose an orthonormal basis of$~W$, and express the $v_i$ on this basis. The matrix of $f$ with respect to this basis has as row$~i$ the list of coordinates of$~v_i$ in the chosen basis, in other words it is the transpose of the matrix whose column$~j$ gives the coordinates of$~v_j$, and the two matrices have the same nonzero determinant$~d$. The matrix you are interested in the the product of those two transpose matrices, and its determinant is therefore $d^2\neq0$.

Hints (for you to understand, complete and, eventually, prove):

Suppose we have $\,\{v_1,...,v_m\}\subset V\;,\;\;\text{with}\;\;m\le\dim V\;$, so we have that the Gramian is

$$G:=A^tA\;,\;\;\text{with}\;\;A=\left(v_1\;v_2\,\ldots\;v_m\right)=\text{ the matrix with columns \,v_i\,}$$

Note that $\,G\,$ is your matrix and this is an $\,m\times m\;$ square matrix , so:

$$(1)\;\;\;\exists\, 0\neq u\in V\;\;s.t.\;\;Gu=0\implies 0=u^tA^tAu=\langle\,(Au)^t\,,\,Au\,\rangle=\left\|Au\right\|\implies Au=0$$

and since $\,u\neq 0\,$ this means the rank of $\;a\;$ isn't full

(2) OTOH, if $\;\text{rk}(A)\;$ isn't full then $\,\exists\,0\neq u\in V\;\;s.t.\,\,Au=0\;$, so that

$$Gu=A^tAu=0\ldots\ldots$$

• Are you assuming $V=\mathbb R^n$? Otherwise I can't make sense of the statement $A=\left(v_1\;v_2\,\ldots\;v_m\right)=$ the matrix with columns $\,v_i\,$. Nov 15, 2020 at 10:54
• @Filippo As usual, any $\;n-$ dimensional vector space over a field $\;\Bbb F\;$ is isomorphic with $\;\Bbb F^n\;$ once we've chosen a basis in $\;V\;$ . Observe ithis is more or less the same is assumed in most other answers. Nov 15, 2020 at 11:52