Question

Let T:Rn→Rm be a linear transformation. Show that there exists a number M>0 such that ∥Tx∥ ≤ M∥x∥

Answer 1

To show that there exists a number M>0 such that ∥Tx∥ ≤ M∥x∥, we need to prove that the linear transformation T satisfies the boundedness property. In other words, there exists a constant M such that the norm of the image of any vector x under T is less than or equal to the norm of x multiplied by M.

Step-by-step explanation:

To show that there exists a number M>0 such that ∥Tx∥ ≤ M∥x∥, we need to prove that the linear transformation T satisfies the boundedness property. In other words, there exists a constant M such that the norm of the image of any vector x under T is less than or equal to the norm of x multiplied by M.

Let's suppose T is a linear transformation from Rn to Rm. Then, for any vector x in Rn, we have:

∥Tx∥ ≤ M∥x∥

To prove this, we can consider the fact that T is a linear transformation means it satisfies the following properties:

1. Linearity under vector addition: T(u + v) = T(u) + T(v) for any vectors u and v in Rn.
2. Linearity under scalar multiplication: T(ku) = kT(u) for any scalar k and vector u in Rn.

Using these properties, we can show that there exists a constant M such that the inequality holds.