Question

Let \(T: \mathbb{R}^3 \to \mathbb{R}^3\) be a linear transformation satisfying
\(T(1, 0, 0) = (0, 1, 1)\), \(T(1, 1, 0) = (1, 0, 1)\) and \(T(1, 1, 1) = (1, 1, 2)\).
Then

• A. T is one-one but T is NOT onto
• B. T is one-one and onto
• C. T is NEITHER one-one NOR onto
• D. T is NOT one-one but T is onto

Hint:

For a linear map \(T: \mathbb{R}^n \to \mathbb{R}^n\), the properties of being one-one, onto, and having an invertible matrix representation are all equivalent. To check them, you can simply find the matrix for T (or the matrix of the images of a basis) and calculate its determinant. If the determinant is non-zero, all properties hold. If it's zero, none of them hold.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a linear transformation \(T: V \to W\) between finite-dimensional vector spaces, being "one-one" (injective) means the kernel (or null space) contains only the zero vector, i.e., \(\ker(T) = \{0\}\). Being "onto" (surjective) means the range (or image) of \(T\) is equal to the entire codomain \(W\). For a transformation \(T: V \to V\) where the domain and codomain have the same dimension, \(T\) is one-one if and only if it is onto. This is part of the Rank-Nullity Theorem.

Step 2: Key Formula or Approach:
A linear transformation is one-one if and only if it maps a basis of the domain to a set of linearly independent vectors in the codomain. If the dimensions of the domain and codomain are equal, \(T\) is one-one and onto if and only if the image of any basis is also a basis. We can check for linear independence by calculating the determinant of the matrix formed by the image vectors.

Step 3: Detailed Calculation:
The input vectors are \(v_1 = (1,0,0)\), \(v_2 = (1,1,0)\), and \(v_3 = (1,1,1)\). These vectors are linearly independent (they form an upper triangular matrix with non-zero diagonal entries) and thus form a basis for \(\mathbb{R}^3\).
The corresponding image vectors are:
\(w_1 = T(v_1) = (0, 1, 1)\)
\(w_2 = T(v_2) = (1, 0, 1)\)
\(w_3 = T(v_3) = (1, 1, 2)\)
To check if \(T\) is one-one and onto, we check if the set \(\{w_1, w_2, w_3\}\) is linearly independent. We can do this by forming a matrix with these vectors and finding its determinant.
\[ A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 2 \end{pmatrix} \]
The determinant of \(A\) is:
\[ \det(A) = 0 \cdot (0 \cdot 2 - 1 \cdot 1) - 1 \cdot (1 \cdot 2 - 1 \cdot 1) + 1 \cdot (1 \cdot 1 - 0 \cdot 1) \]
\[ \det(A) = 0 - 1 \cdot (2 - 1) + 1 \cdot 1 = -1 \cdot 1 + 1 = 0 \]
Since the determinant is 0, the vectors \(w_1, w_2, w_3\) are linearly dependent.
This means the image of the basis of \(\mathbb{R}^3\) is not a basis for \(\mathbb{R}^3\). The range of \(T\) has a dimension less than 3, so \(T\) is NOT onto.
Since \(T\) is a linear transformation between two vector spaces of the same finite dimension (\(\dim(\mathbb{R}^3) = 3\)), if it is not onto, it cannot be one-one.
Alternatively, we can find a non-zero vector in the kernel. Notice that \(w_3 = w_1 + w_2\), since \((1,1,2) = (0,1,1) + (1,0,1)\).
By linearity of \(T\), since \(T(v_3) = T(v_1) + T(v_2) = T(v_1 + v_2)\), we have:
\[ T(v_3) - T(v_1 + v_2) = 0 \implies T(v_3 - v_1 - v_2) = 0 \]
Let \(u = v_3 - v_1 - v_2 = (1,1,1) - (1,0,0) - (1,1,0) = (-1, 0, 1)\).
Since \(u \neq (0,0,0)\) and \(T(u) = 0\), the kernel of \(T\) is non-trivial. Therefore, \(T\) is NOT one-one.

Step 4: Final Answer:
The transformation \(T\) is neither one-one nor onto.

Step 5: Why This is Correct:
The set of image vectors corresponding to a basis of the domain is linearly dependent, as shown by the determinant being zero. For a linear map between spaces of the same finite dimension, this implies the map is neither injective (one-one) nor surjective (onto).