Question

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.]

Short answer

The equation has a nontrivial solution.

Step-by-step solution

Write the condition for mapping and the set to be linearly dependent

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 \({c_1}T\left( {{{\bf{v}}_1}} \right) + {c_2}T\left( {{{\bf{v}}_2}} \right) + ... + {c_p}T\left( {{{\bf{v}}_p}} \right) = 0\).

Apply the linear transformation definition

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

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

Check the solution of the transformation

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.