Some Questions of Linear Algebra: Linear Transformation

 

Some Questions of Linear Algebra: Linear Transformation

Definition: Linear Transformation

A linear transformation \(T: \mathbb{R}^n \to \mathbb{R}^m\) is a function that satisfies:

  • Additivity: \(T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})\) for all \(\mathbf{u}, \mathbf{v} \in \mathbb{R}^n\).
  • Homogeneity: \(T(c \mathbf{v}) = c T(\mathbf{v})\) for all \(\mathbf{v} \in \mathbb{R}^n\) and scalars \(c\).

Numerical Example: Linear Transformation to Matrix

Consider \(T: \mathbb{R}^2 \to \mathbb{R}^2\) defined by \(T(x,y) = (2x + y, x – 3y)\).

Using the basis \( \{(1,1), (1,-1)\} \), compute:

\[
T(1,1) = \begin{bmatrix}3\\-2\end{bmatrix}, \quad T(1,-1) = \begin{bmatrix}1\\4\end{bmatrix}
\]

Expressing in the new basis and solving, we get:

\[
A’ = \begin{bmatrix} \frac{1}{2} & \frac{5}{2} \\ \frac{5}{2} & -\frac{3}{2} \end{bmatrix}
\]

Practice Question

Let \(W\) be the vector space of all real polynomials of degree at most 3. Define \(T: W \to W\) by \((Tp)(x) = p'(x)\), where \(p’\) is the derivative of \(p\). The matrix of \(T\) in the basis \(\{1, X, X^2, X^3\}\) is:

  1. \(\begin{bmatrix}0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0\end{bmatrix}\)
  2. \(\begin{bmatrix}0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0\end{bmatrix}\)
  3. \(\begin{bmatrix}0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}\)
  4. \(\begin{bmatrix}0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3\end{bmatrix}\)

Which option is correct? Solve and comment your answer below!

Solution

Applying differentiation on the basis polynomials:

\[
T(1) = 0, \quad T(X) = 1, \quad T(X^2) = 2X, \quad T(X^3) = 3X^2
\]

Thus, the transformation matrix is:

\[
\begin{bmatrix}0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0\end{bmatrix}
\]

The correct answer is (b).

PDF

Q_29_35_JUNE_2011

Video

 

Connect with PostNetwork Academy

Thank You!

 

©Postnetwork-All rights reserved.