Question-38
Let \(\{x_{1} ,\cdots ,x_{n}\}\) be a set of linearly independent vectors in \(\mathbb{R}^{n}\). Let the \(( i,j)\)-th element of matrix \(A\in \mathbb{R}^{n\times n}\) be given by \(A_{ij} =x_{i}^{T} x_{j} ,1\leqslant i,j\leqslant n\). Which one of the following statements is correct?
We see that \(A\) is a collection of the pair-wise dot product between vectors in the linearly independent set. Introducing a matrix \(B\) whose columns are the linearly independent vectors, we have:
\[ B=\begin{bmatrix} | & & |\\ x_{1} & \cdots & x_{n}\\ | & & | \end{bmatrix} \Longrightarrow A=B^{T} B \]
We see that \(B\) and \(B^{T}\) are invertible. It follows that \(A\) is invertible. To see why the other options are false:
\(0\) cannot be a singular value of an invertible matrix. Recall that the number of non-zero singular values (including multiplicity) is equal to the rank of a matrix. So an invertible matrix cannot have any zero singular values.
The determinant of an invertible matrix is non-zero.
\(A=B^{T} B\) is a positive definite matrix. To see this, consider any \(z\neq 0\):
\[ \begin{aligned} z^{T} Az & =z^{T} B^{T} Bz\\ & =( Bz)^{T}( Bz)\\ & =||Bz||^{2}\\ & >0 \end{aligned} \]
To see how we arrived at the last step, since \(B\) is invertible, \(Bz=0\) if and only if \(z=0\). To put it in another way, \(Bz\) is non-zero whenever \(z\neq 0\).