Question

Let \({\bf{u}} = \left( {{u_1},{u_2},{u_3}} \right)\). Explain why \({\bf{u}} \cdot {\bf{u}} \ge 0\). When is \({\bf{u}} \cdot {\bf{u}} = 0\)?

Short answer

The expression \({\bf{u}} \cdot {\bf{u}}\) means the sum of squares of element which will always be greater than zero. Their product will be zero if and only if the elements themselves are zero.

Step-by-step solution

Find the product \({\bf{u}} \cdot {\bf{u}}\) 

The given vector can be written as,\({\bf{u}} = \left( {\begin{aligned}{*{20}{c}}{{u_1}}\\{{u_2}}\\{{u_3}}\end{aligned}} \right)\).

Find \({\bf{u}} \cdot {\bf{u}}\).

\(\begin{aligned}{c}{\bf{u}} \cdot {\bf{u}} &= {\left( {\begin{aligned}{*{20}{c}}{{u_1}}\\{{u_2}}\\{{u_3}}\end{aligned}} \right)^T} \cdot \left( {\begin{aligned}{*{20}{c}}{{u_1}}\\{{u_2}}\\{{u_3}}\end{aligned}} \right)\\ &= u_1^2 + u_2^2 + u_3^2\end{aligned}\)

 Verification of statement

Since, the product \({\bf{u}} \cdot {\bf{u}}\) is the sum of squares of element which will always be greater than zero.

Therefore,

\({\bf{u}} \cdot {\bf{u}} \ge 0\)

The equation \({\bf{u}} \cdot {\bf{u}} = 0\) will be valid only when all the elements of the vector will be equal to zero.