Linear Transformations (LT).C41

Week04 Discussion

Solution 1:

\(\begin{aligned} {T}\bigg(\left[\begin{matrix} 2 \\ 3 \end{matrix}\right]\bigg)\end{aligned} = \left[\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}\right]\)

\(\begin{aligned} {T}\bigg(\left[\begin{matrix} 3 \\ 4 \end{matrix}\right]\bigg)\end{aligned} = \left[\begin{matrix} -1 \\ 0 \\ 2 \end{matrix}\right]\)

Write standard basis vectors \(e_1\) and \(e_2\) as linear combinations of \(\left[\begin{matrix} 2 \\ 3 \end{matrix}\right]\) and \(\left[\begin{matrix} 3 \\ 4 \end{matrix}\right]\)

\(e_1 = -4 \left[\begin{matrix} 2 \\ 3 \end{matrix}\right] + 3 \left[\begin{matrix} 3 \\ 4 \end{matrix}\right]; e_2 = 3 \left[\begin{matrix} 2 \\ 3 \end{matrix}\right] - 2 \left[\begin{matrix} 3 \\ 4 \end{matrix}\right]\)

\(\begin{aligned} {T}(e_1) = T \bigg(-4 \left[\begin{matrix} 2 \\ 3 \end{matrix}\right] + 3 \left[\begin{matrix} 3 \\ 4 \end{matrix}\right]\bigg) = -4 T \bigg(\left[\begin{matrix} 2 \\ 3 \end{matrix}\right]\bigg) + 3 T \bigg(\left[\begin{matrix} 3 \\ 4 \end{matrix}\right]\bigg)\end{aligned}\)

\(\begin{aligned} {T}(e_1) = -4 \left[\begin{matrix} 2 \\ 2 \\ 1\end{matrix}\right] + 3 \left[\begin{matrix} -1 \\ 0 \\ 2 \end{matrix}\right] = \left[\begin{matrix} -8 \\ -8 \\ -4\end{matrix}\right] + \left[\begin{matrix} -3 \\ 0 \\ 6 \end{matrix}\right] = \left[\begin{matrix} -11 \\ -8 \\ 2\end{matrix}\right] \end{aligned}\)

\(\begin{aligned} {T}(e_2) = T \bigg(3 \left[\begin{matrix} 2 \\ 3 \end{matrix}\right] - 2 \left[\begin{matrix} 3 \\ 4 \end{matrix}\right]\bigg) = 3 T \bigg(\left[\begin{matrix} 2 \\ 3 \end{matrix}\right]\bigg) - 2 T \bigg(\left[\begin{matrix} 3 \\ 4 \end{matrix}\right]\bigg)\end{aligned}\)

\(\begin{aligned} {T}(e_2) = 3 \left[\begin{matrix} 2 \\ 2 \\ 1\end{matrix}\right] -2 \left[\begin{matrix} -1 \\ 0 \\ 2 \end{matrix}\right] = \left[\begin{matrix} 6 \\ 6 \\ 3\end{matrix}\right] - \left[\begin{matrix} -2 \\ 0 \\ 4 \end{matrix}\right] = \left[\begin{matrix} 8 \\ 6 \\ -1\end{matrix}\right] \end{aligned}\)

The matrix representation of \(T\) is \(A_T\) = \(\begin{aligned} \left[\begin{matrix} -11 & 8 \\ -8 & 6 \\ 2 & -1 \end{matrix}\right] \end{aligned}\)

Solution 2:

Let \(\begin{aligned} {T}=\left[\begin{matrix} x_1 & x_2 \\ y_1 & y_2 \\ z_1 & z_2 \end{matrix}\right]\end{aligned}\)

\(\begin{aligned} (2x_1 + 3x_2 = 2) \end{aligned}\) and \(\begin{aligned} (3x_1 + 4x_2 = -1) \end{aligned}\) \(\begin{aligned} \Rightarrow (x_1=-11; x_2= 8) \end{aligned}\)

\(\begin{aligned} (2y_1 + 3y_2 = 2) \end{aligned}\) and \(\begin{aligned} (3y_1 + 4y_2 = 0) \end{aligned}\) \(\begin{aligned} \Rightarrow (y_1=-8; y_2=6) \end{aligned}\)

\(\begin{aligned} (2z_1 + 3z_2 = 1) \end{aligned}\) and \(\begin{aligned} (3z_1 + 4z_2 = 2) \end{aligned}\) \(\begin{aligned} \Rightarrow (z_1=2; z_2=-1) \end{aligned}\)

\(\begin{aligned} {T}=\left[\begin{matrix} x_1 & x_2 \\ y_1 & y_2 \\ z_1 & z_2 \end{matrix}\right] \Rightarrow \end{aligned}\) \(\begin{aligned} {T}=\left[\begin{matrix} -11 & 8 \\ -8 & 6 \\ 2 & -1 \end{matrix}\right] \end{aligned}\)

Explanation of Solution 1:

The key here is to use the two standard basis vectors. They are \(\begin{aligned} {e1} = \left[\begin{matrix} 1 \\ 0 \end{matrix}\right] \end{aligned}\) and \(\begin{aligned} {e2} = \left[\begin{matrix} 0 \\ 1 \end{matrix}\right] \end{aligned}\)

Any vector \(\begin{aligned} {x} = \left[\begin{matrix} x_1 \\ x_2 \end{matrix}\right] \end{aligned} \epsilon C^2\) is a linear combination of \(e_1\) and \(e_2\) because

\(\begin{aligned} {x} = \left[\begin{matrix} x_1 \\ x_2 \end{matrix}\right] = \left[\begin{matrix} x_1 \\ 0 \end{matrix}\right] + \left[\begin{matrix} 0 \\ x_2 \end{matrix}\right] = x_1.\left[\begin{matrix} 1 \\ 0 \end{matrix}\right] + x_2.\left[\begin{matrix} 0 \\ 1 \end{matrix}\right] = x_1.e_1 + x_2.e_2 \end{aligned}\)

Since T is a linear transformation, we know that:

\(\begin{aligned} T(x) = T (x_1.e_1 + x_2.e_2) \end{aligned}\)

\(\begin{aligned} T(x) = x_1.T(e_1) + x_2.T(e_2) \end{aligned}\)

\(\begin{aligned} T(x) = \left[\begin{matrix} T(e_1) & T(e_2) \end{matrix}\right].\left[\begin{matrix} x_1 \\ x_2 \end{matrix}\right] \end{aligned}\)

\(\begin{aligned} T(x) = A.X \end{aligned}\)

where A is a 3 X 2 matrix \(\begin{aligned} A = \left[\begin{matrix} T(e_1) & T(e_2) \end{matrix}\right] \end{aligned}\)

We need to determine the matrix \(A\)

\(\begin{aligned} A = \left[\begin{matrix} T(e_1) & T(e_2) \end{matrix}\right] \end{aligned} = \left[\begin{matrix} {T}\bigg(\left[\begin{matrix} 1 \\ 0 \end{matrix}\right]\bigg) & {T}\bigg(\left[\begin{matrix} 0 \\ 1 \end{matrix}\right]\bigg) \end{matrix}\right]\)

We need to determine \(\begin{aligned} {T}\bigg(\left[\begin{matrix} 1 \\ 0 \end{matrix}\right]\bigg)\end{aligned}\) and \(\begin{aligned} {T}\bigg(\left[\begin{matrix} 0 \\ 1 \end{matrix}\right]\bigg)\end{aligned}\) from the given values of \(\begin{aligned} {T}\bigg(\left[\begin{matrix} 2 \\ 3 \end{matrix}\right]\bigg)\end{aligned}\) and \(\begin{aligned} {T}\bigg(\left[\begin{matrix} 3 \\ 4 \end{matrix}\right]\bigg) \end{aligned}\).

\(\begin{aligned} \left[\begin{matrix} 1 \\ 0 \end{matrix}\right] = \left[\begin{matrix} 9 - 8 \\ 12 - 12 \end{matrix}\right] = \left[\begin{matrix} 3.3 - 4.2 \\ 3.4 - 4.3 \end{matrix}\right] = \left[\begin{matrix} 3.3 \\ 3.4 \end{matrix}\right] - \left[\begin{matrix} 4.2 \\ 4.3 \end{matrix}\right] = 3.\left[\begin{matrix} 3 \\ 4 \end{matrix}\right] - 4.\left[\begin{matrix} 2 \\ 3 \end{matrix}\right] \end{aligned}\)

\(\begin{aligned} \left[\begin{matrix} 1 \\ 0 \end{matrix}\right] =- 4.\left[\begin{matrix} 2 \\ 3 \end{matrix}\right] + 3.\left[\begin{matrix} 3 \\ 4 \end{matrix}\right] \end{aligned}\)

\(\begin{aligned} e_1 =- 4.\left[\begin{matrix} 2 \\ 3 \end{matrix}\right] + 3.\left[\begin{matrix} 3 \\ 4 \end{matrix}\right] \end{aligned}\)

\(\begin{aligned} {T(e_1)} = {T}\bigg(\left[\begin{matrix} 1 \\ 0 \end{matrix}\right]\bigg) = - 4.{T}\bigg(\left[\begin{matrix} 2 \\ 3 \end{matrix}\right]\bigg) + 3.{T}\bigg(\left[\begin{matrix} 3 \\ 4 \end{matrix}\right]\bigg) \end{aligned}\)

We are given the values of \(\begin{aligned} {T}\bigg(\left[\begin{matrix} 2 \\ 3 \end{matrix}\right]\bigg)\end{aligned}\) and \(\begin{aligned} {T}\bigg(\left[\begin{matrix} 3 \\ 4 \end{matrix}\right]\bigg) \end{aligned}\), so we substitute to get \(T(e_1)\).

\(\begin{aligned} {T(e_1)} = \left[\begin{matrix} -11 \\ -8 \\ 2\end{matrix}\right] \end{aligned}\)

\(\begin{aligned} \left[\begin{matrix} 0 \\ 1 \end{matrix}\right] = \left[\begin{matrix} 6 - 6 \\ 9 - 8 \end{matrix}\right] = \left[\begin{matrix} 3.2 - 2.3 \\ 3.3 - 2.4 \end{matrix}\right] = \left[\begin{matrix} 3.2 \\ 3.3 \end{matrix}\right] - \left[\begin{matrix} 2.3 \\ 2.4 \end{matrix}\right] = 3.\left[\begin{matrix} 2 \\ 3 \end{matrix}\right] - 2.\left[\begin{matrix} 3 \\ 4 \end{matrix}\right] \end{aligned}\)

\(\begin{aligned} \left[\begin{matrix} 0 \\ 1 \end{matrix}\right] = 3.\left[\begin{matrix} 2 \\ 3 \end{matrix}\right] - 2.\left[\begin{matrix} 3 \\ 4 \end{matrix}\right] \end{aligned}\)

\(\begin{aligned} e_2 = 3.\left[\begin{matrix} 2 \\ 3 \end{matrix}\right] - 2.\left[\begin{matrix} 3 \\ 4 \end{matrix}\right] \end{aligned}\)

\(\begin{aligned} {T(e_2)} = {T}\bigg(\left[\begin{matrix} 0 \\ 1 \end{matrix}\right]\bigg) = 3.{T}\bigg(\left[\begin{matrix} 2 \\ 3 \end{matrix}\right]\bigg) - 2.{T}\bigg(\left[\begin{matrix} 3 \\ 4 \end{matrix}\right]\bigg) \end{aligned}\)

We are given the values of \(\begin{aligned} {T}\bigg(\left[\begin{matrix} 2 \\ 3 \end{matrix}\right]\bigg)\end{aligned}\) and \(\begin{aligned} {T}\bigg(\left[\begin{matrix} 3 \\ 4 \end{matrix}\right]\bigg) \end{aligned}\), so we substitute to get \(T(e_2)\).

\(\begin{aligned} {T(e_2)} = \left[\begin{matrix} 8 \\ 6 \\ 1\end{matrix}\right] \end{aligned}\)

\(\begin{aligned} A = \left[\begin{matrix} T(e_1) & T(e_2) \end{matrix}\right] \end{aligned} = \left[\begin{matrix} {T}\bigg(\left[\begin{matrix} 1 \\ 0 \end{matrix}\right]\bigg) & {T}\bigg(\left[\begin{matrix} 0 \\ 1 \end{matrix}\right]\bigg) \end{matrix}\right]\)

\(\begin{aligned} A = \left[\begin{matrix} T(e_1) & T(e_2) \end{matrix}\right] = \left[\begin{matrix} -11 & 8 \\ -8 & 6 \\ 2 & -1 \end{matrix}\right] \end{aligned}\)