Question

Show that T(x1, x2, x3, x4) = 3x1 −7x2 + 5x4 is a linear transformation by finding the

matrix for the transformation. Then find a basis for the null space of the transforma￾tion.

Answer 1

T(x1, x2, x3, x4) = 3x1 −7x2 + 5x4 can be expressed as

$T\begin{bmatrix} x1 \\ x2\\ x3\\ x4 \end{bmatrix} = \begin{bmatrix} 3x1 \\ -7 x2\\ 5x4\\ \end{bmatrix}$

T: R⁴$\to$ R³

So the domain of T is R³. To find the column of the standard matrix of transformation we find T(e₁), T(e₂), T(e₃), and T(e₄)

Using the rule for transformation

T(e₁) = $\begin{bmatrix} 1 \\ 0\\ 0\\ 0 \end{bmatrix} = \begin{bmatrix} 3(1) \\ -7(0)\\ 5(0)\\ \end{bmatrix} = \begin{bmatrix} 3 \\ 0\\ 0\\ \end{bmatrix}$

T(e₂) = $\begin{bmatrix} 0 \\ 1\\ 0\\ 0 \end{bmatrix} = \begin{bmatrix} 3(0) \\ -7(1)\\ 5(0)\\ \end{bmatrix} = \begin{bmatrix} 0 \\ -7\\ 0\\ \end{bmatrix}$

T(e₃) = $\begin{bmatrix} 0 \\ 0\\ 1\\ 0 \end{bmatrix} = \begin{bmatrix} 3(0) \\ -7(0)\\ 5(0)\\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0\\ 0\\ \end{bmatrix}$

T(e4) = $\begin{bmatrix} 0 \\ 0\\ 0\\ 1 \end{bmatrix} = \begin{bmatrix} 3(0) \\ -7(0)\\ 5(1)\\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0\\ 5\\ \end{bmatrix}$

Therefore the transformation matrix can be expressed as

$\begin{bmatrix} 3 & 0&0&0 \\ 0&-7&0&0\\ 0&0&0&5\\ \end{bmatrix}$

Finding the Basis for the null space of the transformation

$\begin{bmatrix} 3 & 0&0&0 \\ 0&-7&0&0\\ 0&0&0&5\\ \end{bmatrix} \begin{bmatrix} x1\\ x2\\ x3\\ x4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0\\ 0\\ \end{bmatrix}$

Reduced Row Echelon Matrix for the augmented matrix

$\begin{bmatrix} 3 & 0 & 0 & 0 &\bigm| & 0 \\ 0 & -7 & 0 & 0 &\bigm| & 0 \\ 0 & 0 & 0 & 0 &\bigm| & 0 \\ 0 & 0 & 0 & 5 &\bigm| & 0 \end{bmatrix}$

Multiply row 1 by 1/3:

$\begin{bmatrix} 1 & 0 & 0 & 0 &\bigm| & 0 \\ 0 & -7 & 0 & 0 &\bigm| & 0 \\ 0 & 0 & 0 & 0 &\bigm| & 0 \\ 0 & 0 & 0 & 5 &\bigm| & 0 \end{bmatrix}$

Multiply row 2 by -1/7:

$\begin{bmatrix} 1 & 0 & 0 & 0 &\bigm| & 0 \\ 0 & 1 & 0 & 0 &\bigm| & 0 \\ 0 & 0 & 0 & 0 &\bigm| & 0 \\ 0 & 0 & 0 & 5 &\bigm| & 0 \end{bmatrix}$

Multiply row 3 by 1/5:

$\begin{bmatrix} 1 & 0 & 0 & 0 &\bigm| & 0 \\ 0 & 1 & 0 & 0 &\bigm| & 0 \\ 0 & 0 & 0 & 0 &\bigm| & 0 \\ 0 & 0 & 0 & 1 &\bigm| & 0 \end{bmatrix}$

Convert the matrix equation back to an equivalent system:

$\begin{array}{rcrcrcr} x_{1}&&&&&=&\phantom{-}0 \\ &&x_{2}&&&=&\phantom{-}0 \\ &&&&x_{4}&=&\phantom{-}0 \\ \end{array}$

Add an equation for each free variable:

$\begin{array}{rcrcrcrcr} x_{1}&&&&&&&=&\phantom{-}0 \\ &&x_{2}&&&&&=&\phantom{-}0 \\ &&&&x_{3}&&&=&x_{3} \\ &&&&&&x_{4}&=&\phantom{-}0 \\ \end{array}$

Solve for each variable in terms of the free variables:

$\begin{array}{rcrcrcrcr} x_{1}&&&&&&&=&\phantom{-}0 \\ &&x_{2}&&&&&=&\phantom{-}0 \\ &&&&x_{3}&&&=&x_{3} \\ &&&&&&x_{4}&=&\phantom{-}0 \\ \end{array}$

Collect terms into vectors:

$\left[ \begin{array}{c}x_{1}\\x_{2}\\x_{3}\\x_{4}\\\end{array} \right] = \left[ \begin{array}{c} 0\\0\\x_{3}\\0\\\end{array} \right]$

Factor out variables on the right side:

$\left[ \begin{array}{c}x_{1}\\x_{2}\\x_{3}\\x_{4}\\\end{array} \right] = x_{3}\left[ \begin{array}{c} \phantom{-}0\\\phantom{-}0\\\phantom{-}1\\\phantom{-}0\\\end{array} \right]$

Thus, a basis for the null space is:

$\displaystyle \left\{\left[ \begin{array}{c} \phantom{-}0\\ \phantom{-}0\\ \phantom{-}1\\ \phantom{-}0\\ \end{array} \right]\right\}$ [Answer]