Question

Let \(V\) be a vector space, and let \(T:V \to V\)be a linear transformation. Given \(z\) in \(V\) , suppose \({x_p}\) in \(V\) satisfies \(T\left( {{x_p}} \right) = z\)and let \(u\) be any vector in the kernel of \(T\) . Show that \(u + {x_p}\)satisfies the nonhomogeneous equation \(T\left( x \right) = z\).

Short answer

It is shown that \(u + {x_p}\) satisfies the nonhomogeneous equation \(T\left( x \right) = z\).

Step-by-step solution

Describe the given statement

It is given that \(z\) is in \(V\), and \({x_p}\) is in \(V\), which satisfies \(T\left( {{x_p}} \right) = z\). Also, \({\rm{u}}\) is in the kernel of \(T\). This implies that \(T\left( u \right) = 0\).

It is given that T is a linear transformation. It means that T satisfies the distributive property.

Use the linearity property and the relation\(T\left( {{x_p}} \right) = z\)

Let \(x\)be \(u + {x_p}\), then according to the linearity property, \(T\left( {u + {x_p}} \right) = T\left( u \right) + T\left( {{x_p}} \right)\).

\(\begin{aligned} T\left( {u + {x_p}} \right) &= T\left( u \right) + T\left( {{x_p}} \right)\\ &= 0 + z\\ &= z\end{aligned}\)

Draw a conclusion

If \(T\left( x \right) = z\), then \(u + {x_p}\) satisfies the nonhomogeneous equation \(T\left( x \right) = z\).