In Exercises 32–36, column vectors are written as rows, such as \({\bf{x}} = \left( {{x_1},{x_2}} \right)\), and \(T\left( {\bf{x}} \right)\) is written as \(T\left( {{x_1},{x_2}} \right)\).

34.Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\) be a linear transformation. Show that if \(T\) maps two linearly independent vectors onto a linearly dependent set, then the equation \(T\left( {\bf{x}} \right) = 0\) has a nontrivial solution. [Hint:Suppose u and v in \({\mathbb{R}^n}\) are linearly independent and yet \(T\left( {\bf{u}} \right)\) and \(T\left( {\bf{v}} \right)\) are linearly dependent. Then \({c_1}T\left( {\bf{u}} \right) + {c_2}T\left( {\bf{v}} \right) = 0\) for some weights \({c_1}\) and \({c_2}\), not both zero. Use this equation.]

The vectors are said to be linearly dependent if the equation is in the form of \({c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + ... + {c_p}{{\bf{v}}_p} = 0\), where \({c_1},{c_2},...,{c_p}\) are scalars.

For the vectors \({{\bf{v}}_1},{{\bf{v}}_2},...,{{\bf{v}}_p}\) and the weights \({c_1},{c_2},...,{c_p}\), if \(T\) transforms \(n\) linearly dependent vectors, then \({c_1}T\left( {{{\bf{v}}_1}} \right) + {c_2}T\left( {{{\bf{v}}_2}} \right) + ... + {c_p}T\left( {{{\bf{v}}_p}} \right) = 0\).

Let the two vectors be u and v, and \({\bf{x}}\) be the linear combination of these vectors. Thus, xcan be written as shown below:

\({\bf{x}} = {c_1}{\bf{u}} + {c_2}{\bf{v}}\)

Apply the transformation on both sides for \({\bf{x}} = {c_1}{\bf{u}} + {c_2}{\bf{v}}\).

\(\begin{array}{l}T\left( {\bf{x}} \right) = T\left( {{c_1}{\bf{u}} + {c_2}{\bf{v}}} \right)\\T\left( {\bf{x}} \right) = {c_1}T\left( {\bf{u}} \right) + {c_2}T\left( {\bf{v}} \right)\end{array}\)

It is known that the condition for linearly dependent\(T\left( {\bf{x}} \right)\)must be 0.

For linearly dependent vectors, the equation is shown below:

\(\begin{aligned}{c}{c_1}T\left( {\bf{u}} \right) + {c_2}T\left( {\bf{v}} \right) &= 0\\T\left( {{c_1}{\bf{u}} + {c_2}{\bf{v}}} \right) &= 0\\T\left( {\bf{x}} \right) &= 0\end{aligned}\)

Thus, the equation has a nontrivial solution.