Question

Consider two vectors ${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}$and localid="1659542347261" ${\overrightarrow{\mathbf{v}}}_{\mathbf{2}}$in ${\mathit{R}}^{\mathbf{n}}$. Form the matrix

$\mathit{G}\mathbf{=}\left[\begin{array}{cc}{\overrightarrow{v}}_1.{\overrightarrow{v}}_1& {\overrightarrow{v}}_1.{\overrightarrow{v}}_2\\ {\overrightarrow{v}}_2.{\overrightarrow{v}}_1& {\overrightarrow{v}}_2.{\overrightarrow{v}}_2\end{array}\right]$

For which choices of localid="1659542539965" ${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}$and ${\overrightarrow{\mathbf{v}}}_{\mathbf{2}}$is the matrix Ginvertible?

Short answer

The matrix is invertible if and only if ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2$ are not parallel or not lie in the same plane, i.e. ${\overrightarrow{v}}_1$and ${\overrightarrow{v}}_2$will be linearly independent.

Step-by-step solution

Step by step solutionStep 1: Given

Consider two vectors ${\overrightarrow{v}}_1$and ${\overrightarrow{v}}_2$ in $R^n$ to Form the matrix$G=\left[\begin{array}{cc}{\overrightarrow{v}}_1.{\overrightarrow{v}}_1& {\overrightarrow{v}}_1.{\overrightarrow{v}}_2\\ {\overrightarrow{v}}_2.{\overrightarrow{v}}_1& {\overrightarrow{v}}_2.{\overrightarrow{v}}_2\end{array}\right]$

Use Cauchy-Schwarz inequality

The identity is $\mathbf{|}\overrightarrow{\mathbf{v}}\mathbf{·}\overrightarrow{\mathbf{w}}\mathbf{|}\mathbf{⩽}\mathbf{\left|}\mathbf{\right|}\overrightarrow{\mathbf{v}}\mathbf{\left|}\mathbf{\right|}\mathbf{·}\mathbf{\left|}\mathbf{\right|}\overrightarrow{\mathbf{w}}\mathbf{\left|}\mathbf{\right|}$.

Since,

role="math" localid="1659543489932" $\begin{array}{rcl}G& =& \left|\right|{\overrightarrow{v}}_1|{|}^2.|\left|{\overrightarrow{v}}_2\right|{|}^2-\left({\overrightarrow{v}}_1.{\overrightarrow{v}}_2\right)\left({\overrightarrow{v}}_2.{\overrightarrow{v}}_1\right)\\ & =& \left|\right|{\overrightarrow{v}}_1|{|}^2.|\left|{\overrightarrow{v}}_2\right|{|}^2-{\left({\overrightarrow{v}}_1.{\overrightarrow{v}}_2\right)}^2\end{array}$

By using Cauchy-Schwarz inequality:

$$\begin{gathered} {\overrightarrow{v}}_1.{\overrightarrow{v}}_2⩽\left|\right|{\overrightarrow{v}}_1\left|\right|·\left|\right|{\overrightarrow{v}}_2\left|\right| \\ \Rightarrow {\left({\overrightarrow{v}}_1.{\overrightarrow{v}}_2\right)}^2⩽\left|\right|{\overrightarrow{v}}_1|{|}^2·|\left|{\overrightarrow{v}}_2\right|{|}^2 \end{gathered}$$

So, ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2$will be linearly independent.

Thus, the determinant will be non-vanishing, i.e., the matrix is invertible if and only if ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2$are not parallel or not lie in the same plane.

Hence, ${\overrightarrow{v}}_1$and ${\overrightarrow{v}}_2$will be linearly independent. if and only if ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2$are not parallel or not lie in the same plane.