Question

T(X,Y,Z)=(Z,X,-Y,-Z) dimension of the kernel of this operator

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Miles

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1. Let the vector \mathbf{v} = \begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} in \( \mathbb{R}^4 \). The transformation \( T \) is defined as:

T(x, y, z, w) = (z, x, -y, -z)

2. Find the kernel of \( T \), i.e., solve for \( \mathbf{v} \) such that:

T(x, y, z, w) = \mathbf{0}

This means:

(z, x, -y, -z) = (0, 0, 0, 0)

3. This gives the system of equations:

z = 0

x = 0

-y = 0 \implies y = 0

-z = 0

4. Since \( z = 0 \), \( x = 0 \), and \( y = 0 \), the only free variable is \( w \).

5. Therefore, any vector in the kernel has the form:

\begin{pmatrix} 0 \\ 0 \\ 0 \\ w \end{pmatrix}

6. This indicates that the kernel is spanned by \( \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \) and any multiple of it.

7. The dimension of the kernel is therefore:

\text{dimension of the kernel} = 1

2. Find the kernel of \( T \), i.e., solve for \( \mathbf{v} \) such that:

This means:

3. This gives the system of equations:

4. Since \( z = 0 \), \( x = 0 \), and \( y = 0 \), the only free variable is \( w \).

5. Therefore, any vector in the kernel has the form:

6. This indicates that the kernel is spanned by \( \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \) and any multiple of it.

7. The dimension of the kernel is therefore:

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