Question

Let \(A\) be an \(m \times n\) matrix and let \({\mathop{\rm u}\nolimits} \) be a vector in \({\mathbb{R}^n}\) that satisfies the equation \(Ax = 0\). Show that for any scalar \(c\), the vector \(c{\mathop{\rm u}\nolimits} \) also satisfies \(Ax = 0\). (That is, show that \(A\left( {c{\mathop{\rm u}\nolimits} } \right) = 0\).)

Short answer

The vector \(cu\) also satisfies that \(Ax = 0\), i.e., \(A\left( {cu} \right) = 0\) is

Step-by-step solution

Consider the given part

Let \(A\) be an \(m \times n\) matrix and let \(u\) be a vector in \({\mathbb{R}^n}\) that satisfies the equation \(Ax = 0\).

Show that vector \(cu\) satisfies \(Ax = 0\)

Theorem 5tells that \(A\) is an \(m \times n\) matrix; \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) are vectors in \({\mathbb{R}^n}\), and \(c\) is a scalar, then

a. \(A\left( {u + v} \right) = Au + Av\)

b. \(A\left( {cu} \right) = c\left( {Au} \right)\)

If \(u\) satisfies the equation \(Ax = 0\), then \(Au = 0\). For any scalar \(c\),

\(\begin{array}{c}A\left( {cu} \right) = c \cdot Au\\ = c \cdot 0\\ = 0.\end{array}\)

Thus, the vector \(cu\) also satisfies \(Ax = 0\), i.e., \(A\left( {cu} \right) = 0\) is proved.