Question

Let \(T:\mathbb{R}^3\to\mathbb{R}^3\) be defined by \(T(a,b,c)=(-a-b,a-b,0)\), then ____.

• A. \(\dim N(T)=2\)
• B. \(\dim R(T)=2\)
• C. \(R(T)=N(T)\)
• D. \(\dim R(T)=3\)

Hint:

To find \(\dim R(T)\), write the transformation as a linear combination of column vectors and count the number of linearly independent vectors.

Answer 1

The Correct Option is B

Concept:
For a linear transformation \(T:\mathbb{R}^3\to\mathbb{R}^3\), the range \(R(T)\) is the set of all possible output vectors of \(T\).

Step 1: Write the given transformation.
\[ T(a,b,c)=(-a-b,a-b,0) \] The output vector is: \[ (-a-b,a-b,0) \] Observe that the third component is always: \[ 0 \] So every vector in the range has the form: \[ (x,y,0) \]

Step 2: Express output in matrix form.
We can write: \[ T(a,b,c)= \begin{pmatrix} -a-b\\ a-b 0 \end{pmatrix} \] This can be expressed as: \[ T(a,b,c) = a\begin{pmatrix} -1\\ 1 0 \end{pmatrix} + b\begin{pmatrix} -1\\ -1 0 \end{pmatrix} + c\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \]

Step 3: Find the spanning vectors of the range.
Thus, \[ R(T)=\text{span}\left\{ \begin{pmatrix} -1\\ 1\\ 0 \end{pmatrix}, \begin{pmatrix} -1\\ -1\\ 0 \end{pmatrix} \right\} \] Now check whether these two vectors are linearly independent. Suppose: \[ \alpha \begin{pmatrix} -1\\ 1\\ 0 \end{pmatrix} + \beta \begin{pmatrix} -1\\ -1\\ 0 \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \] Then: \[ -\alpha-\beta=0 \] and \[ \alpha-\beta=0 \] From: \[ \alpha-\beta=0 \] we get: \[ \alpha=\beta \] Substitute in: \[ -\alpha-\beta=0 \] \[ -2\alpha=0 \] \[ \alpha=0 \] So: \[ \beta=0 \] Hence, the two vectors are linearly independent.

Step 4: Find dimension of range.
Since the range is spanned by two linearly independent vectors, \[ \dim R(T)=2 \] \[ \therefore \text{Correct Answer is (B)} \]