Prove that Gramian matrix is Invertible iff $(v_1,...,v_k) $ is linearly independent
$G=G(v_1,...,v_k) = [\left\langle v_i,v_j\right\rangle ]_{i,j=1}^k $
I have great idea to calculate $\det G$ and show that if $(v_1,...,v_k) $ is linearly independent then $\det G \neq0$ and vice versa.
Plan sounds good (?) but how to calculate $\det$ of this?
\begin{vmatrix} \langle v_1,v_1\rangle & \langle v_1,v_2\rangle &\dots & \langle v_1,v_n\rangle\\ \langle v_2,v_1\rangle & \langle v_2,v_2\rangle &\dots & \langle v_2,v_n\rangle\\ \vdots&\vdots&\ddots&\vdots\\ \langle v_n,v_1\rangle & \langle v_n,v_2\rangle &\dots & \langle v_n,v_n\rangle\end{vmatrix}. Probably I should use some scalar product propeties but it isn't clear for me how can I do that.

  • $\begingroup$ Since $G=V^T V$, $\det(G)=\det(V)^2$ is non-zero iff $\det(V)$ is non-zero, i.e. iff the convex hull of $0,v_1,\ldots,v_n$ has a non-zero (oriented) volume. $\endgroup$ – Jack D'Aurizio Jan 2 '19 at 14:25

Let $v_i=\sum_{k=1}^n a_{ki}e_k$, where $\{e_1,\dots,e_n\}$ is an orthonormal basis. Then $$ \langle v_i,v_j\rangle=\sum_{k}\sum_{l}a_{ki}a_{lj}\langle e_k,e_l\rangle= \sum_{k}a_{ki}a_{kj} $$ and therefore, if $A=[a_{ij}]$, we have $G=A^TA$.

If $Gx=0$, then also $x^TA^TAx=(Ax)^T(Ax)=0$, so $Ax=0$.

The matrix $A$ is the matrix (with respect to the basis $\{e_1,\dots,e_n\}$) of the linear map defined by $e_i\mapsto v_i$, for $i=1,\dots,n$. The rank of this matrix is $n$ if and only if $\{v_1,\dots,v_n\}$ is linearly independent.

If the set is linearly independent, then $Ax=0$ implies $x=0$; thus $Gx=0$ implies $x=0$ and $G$ has rank $n$.

If the set is not linearly independent, then there is $x\ne0$ with $Ax=0$, so also $Gx=0$ and $G$ is not invertible.

Notes. If the inner product is over the complex numbers, then $G=A^HA$ (the Hermitian transpose) and the argument goes through using the Hermitian transpose instead of the transpose. An orthonormal basis exists because of Gram-Schmidt algorithm.

  • $\begingroup$ Thanks for clear solution $\endgroup$ – user617243 Jan 2 '19 at 18:39

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