In each case, a linear transformation T : M22 M22 is defined. Give the matrix for T with

Chapter 4, Problem 1

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QUESTION:

In each case, a linear transformation \(T:\ M_{2\ \times\ 2} \rightarrow\ M_{2\ \times\ 2}\) is defined. Give the matrix for T with respect to the “standard basis” \(\mathbf{v}_{1}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right],\ \mathbf{v}_{2}=\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right],\ \mathbf{v}_{3}=\left[\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right],\ \mathbf{v}_{4}=\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]\) for \(M_{2\ \times\ 2}\). In each case, determine ker(T) and image (T).

a. T(X) = \(X^T\)

b. \(T(X)=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\ X\)

c. \(T(X)=\left[\begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right]\ X\)

d. \(T(X)=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\ X-X\ \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\)

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QUESTION:

In each case, a linear transformation \(T:\ M_{2\ \times\ 2} \rightarrow\ M_{2\ \times\ 2}\) is defined. Give the matrix for T with respect to the “standard basis” \(\mathbf{v}_{1}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right],\ \mathbf{v}_{2}=\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right],\ \mathbf{v}_{3}=\left[\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right],\ \mathbf{v}_{4}=\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]\) for \(M_{2\ \times\ 2}\). In each case, determine ker(T) and image (T).

a. T(X) = \(X^T\)

b. \(T(X)=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\ X\)

c. \(T(X)=\left[\begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right]\ X\)

d. \(T(X)=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\ X-X\ \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\)

ANSWER:

Step 1 of 8

(a)

The given matrix  is          

Find Ker(T) and Image(T)

The given basis are

The Equation for Kernel of T is:

Ker(T)=

The Equation for Image of T:

Image(T)=

Here, is basis of V

The Equation for Image of T is:

Image(T)

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