Question

Consider two vectors$\overrightarrow{\mathbf{v}}$ and $\overrightarrow{\mathbf{w}}$ in ${\mathit{R}}^{\mathbf{n}}$ . Form the matrix $\mathit{A}\mathbf{=}\left[\begin{array}{cc}\overrightarrow{v}& \overrightarrow{w}\end{array}\right]$ . Express$\mathit{d}\mathit{e}\mathit{t}\left(A^{T}A\right)$ in terms of $\left|\left|\overrightarrow{v}\right|\right|\mathbf{,}\left|\left|\overrightarrow{w}\right|\right|$, and$\overrightarrow{\mathbf{v}}\mathbf{.}\overrightarrow{\mathbf{w}}$. What can you say about the sign of the result?

Short answer

Therefore, the determinant of $\det \left(A^{T}A\right)$is given by,

$\det \left(A^{T}A\right)={\left|\left|\overrightarrow{v}\right|\right|}^2{\left|\left|\overrightarrow{w}\right|\right|}^2-(\overrightarrow{v}·\overrightarrow{w}{)}^2⩾0$

Step-by-step solution

Step by Step Solution: Step 1: Definition

A determinant is a unique number associated with asquare matrix.

A determinant is a scalar value that is a function of the entries of a square matrix.

It is the signed factor by which areas are scaled by this matrix. If the sign is negative the matrix reverses orientation.

Given

Given matrix,

$A=\left[\begin{array}{cc}\overrightarrow{v}& \overrightarrow{w}\end{array}\right]$

To find det(ATA)

Let $\overrightarrow{v},\overrightarrow{w}\in {\mathrm{ℝ}}^n$. If

$A=\left[\begin{array}{cc}\overrightarrow{v}& \overrightarrow{w}\end{array}\right]$

$A^T=\left[\begin{array}{c}\overrightarrow{v}\\ \overrightarrow{w}\end{array}\right]$then,

$A^{T}A=\left[\begin{array}{cc}\overrightarrow{v}·\overrightarrow{v}& \overrightarrow{v}·\overrightarrow{w}\\ \overrightarrow{v}·\overrightarrow{w}& \overrightarrow{w}·\overrightarrow{w}\end{array}\right]$

So, $\det \left(A^{T}A\right)={\left|\left|\overrightarrow{v}\right|\right|}^2{\left|\left|\overrightarrow{w}\right|\right|}^2-(\overrightarrow{v}·\overrightarrow{w}{)}^2⩾0$, due to the Cauchy-Schwarz-Bunyakowski inequality.