Let \(F\) be a subfield of the complex numbers and let \(T\) be the function from \(F^3\) into \(F^3\) defined by \[ T(x_1, x_2, x_3) = (x_1 - x_2 + 2x_3,\; 2x_1 + x_2,\; -x_1 - 2x_2 + 2x_3). \]
Verify that \(T\) is a linear transformation.
If \((a, b, c)\) is a vector in \(F^3\), what are the conditions on \(a\), \(b\), and \(c\) that the vector be in the range of \(T\)? What is the rank of \(T\)?
What are the conditions on \(a\), \(b\), and \(c\) that \((a, b, c)\) be in the null space of \(T\)? What is the nullity of \(T\)?
Solution
1. Part (a): Verification of linearity:
1.1. The map \(T: F^3 \to F^3\) can be expressed as matrix multiplication \(T(x) = Ax\), where:\[ A = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & 0 \\ -1 & -2 & 2 \end{pmatrix}. \]
evaluating \(A(x_1, x_2, x_3)^T\) reproduces the coordinate formulas.
1.2. For any \(u, v \in F^3\) and \(c \in F\):\[ T(u + c v) = A(u + c v) = Au + c(Av) = T(u) + c T(v). \]Thus \(T\) is a linear transformation.
linearity of matrix-vector multiplication.
2. Part (b): Range condition and rank of \(T\):
2.1. A vector \((a, b, c)^T \in F^3\) lies in the range of \(T\) if and only if the linear system \(Ax = (a, b, c)^T\) is consistent.Perform elementary row operations on the augmented matrix:\[ \begin{pmatrix} 1 & -1 & 2 & a \\ 2 & 1 & 0 & b \\ -1 & -2 & 2 & c \end{pmatrix} \xrightarrow{R_2 - 2R_1,\, R_3 + R_1} \begin{pmatrix} 1 & -1 & 2 & a \\ 0 & 3 & -4 & b - 2a \\ 0 & -3 & 4 & c + a \end{pmatrix} \xrightarrow{R_3 + R_2} \begin{pmatrix} 1 & -1 & 2 & a \\ 0 & 3 & -4 & b - 2a \\ 0 & 0 & 0 & -a + b + c \end{pmatrix}. \]
Gaussian elimination.
2.2. The system is consistent if and only if the bottom row satisfies:\[ -a + b + c = 0 \quad (\text{or equivalently } a = b + c). \]Thus \((a, b, c) \in \operatorname{Range}(T) \iff a = b + c\).
consistency criterion for augmented linear systems.
2.3. The row echelon form has 2 pivots (in columns 1 and 2), so \(\operatorname{rank}(T) = 2\).
number of pivot columns in echelon form.
3. Part (c): Null space condition and nullity of \(T\):
3.1. \((a, b, c)^T \in \operatorname{Null}(T)\) if and only if \(A(a, b, c)^T = 0\).Using the echelon form with \((0, 0, 0)^T\) on the right-hand side:- Row 2: \(3b - 4c = 0 \implies b = \frac{4}{3}c\).- Row 1: \(a - b + 2c = 0 \implies a = b - 2c = \frac{4}{3}c - 2c = -\frac{2}{3}c\).
back substitution.
3.2. Setting \(c = 3t\) for \(t \in F\), the vectors in the null space are of the form:\[ (a, b, c) = (-2t, 4t, 3t) = t(-2, 4, 3) \quad \text{for } t \in F. \]Equivalently, \((a, b, c) \in \operatorname{Null}(T) \iff 2a + b = 0 \text{ and } 3b - 4c = 0\).
parametrization of solution space.
3.3. The null space is 1-dimensional with basis \(\{(-2, 4, 3)\}\), so \(\operatorname{nullity}(T) = 1\) (which also follows from the Rank-Nullity Theorem: \(\operatorname{rank}(T) + \operatorname{nullity}(T) = 2 + 1 = 3\)).
Rank-Nullity Theorem.
4. Conclusion:\(T\) is linear, \((a, b, c) \in \operatorname{Range}(T) \iff a = b + c\) with \(\operatorname{rank}(T) = 2\), and \((a, b, c) \in \operatorname{Null}(T) \iff (a, b, c) \in \operatorname{span}\{(-2, 4, 3)\}\) with \(\operatorname{nullity}(T) = 1\). Q.E.D.
1 through 3.