Finding Matrix Representation of a Linear Transformation

To find the matrix representation of a linear transformation $T: V \to W$ relative to given bases $B_1$ for $V$ and $B_2$ for $W$, we apply the transformation $T$ to each vector in the basis $B_1$. Then, we express the resulting vectors as a linear combination of the vectors in the basis $B_2$. The coefficients of these linear combinations form the columns of the matrix representation.

In this problem, the linear transformation is $T: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $T(x, y) = (2y, 3x - y)$. The basis for both the domain and codomain is $B = \{(1, 3), (2, 5)\}$. Let the basis vectors be $v_1 = (1, 3)$ and $v_2 = (2, 5)$. We need to find the matrix representation of $T$ relative to this basis $B$, denoted as $[T]_B$. The columns of this matrix will be the coordinate vectors of $T(v_1)$ and $T(v_2)$ with respect to the basis $B$.

Applying the Transformation to Basis Vectors

First, let's apply the transformation $T$ to each vector in the basis $B$:

• For $v_1 = (1, 3)$: $T(1, 3) = (2 \times 3, 3 \times 1 - 3) = (6, 3 - 3) = (6, 0)$.
• For $v_2 = (2, 5)$: $T(2, 5) = (2 \times 5, 3 \times 2 - 5) = (10, 6 - 5) = (10, 1)$.

Expressing Transformed Vectors in terms of the Basis B

Next, we need to express $T(v_1)$ and $T(v_2)$ as linear combinations of the basis vectors $v_1 = (1, 3)$ and $v_2 = (2, 5)$.

For T(v₁) = (6, 0):

We want to find coefficients $a$ and $b$ such that $T(1, 3) = a(1, 3) + b(2, 5)$.

$(6, 0) = a(1, 3) + b(2, 5)$

$(6, 0) = (a, 3a) + (2b, 5b)$

$(6, 0) = (a + 2b, 3a + 5b)$

This gives us a system of linear equations:

\begin{align*} a + 2b &= 6 \quad &(1) \\ 3a + 5b &= 0 \quad &(2) \end{align*}

From equation (2), $3a = -5b$, which means $a = -\frac{5}{3}b$.

Substitute this into equation (1):

$-\frac{5}{3}b + 2b = 6$

Multiply the entire equation by 3 to clear the fraction:

$-5b + 6b = 18$

$b = 18$

Now substitute the value of $b$ back into the expression for $a$:

$a = -\frac{5}{3}(18) = -5 \times 6 = -30$

So, $T(1, 3) = -30(1, 3) + 18(2, 5)$. The coordinate vector of $T(v_1)$ with respect to $B$ is $\begin{pmatrix} -30 \\ 18 \end{pmatrix}$. This will be the first column of the matrix $[T]_B$.

For T(v₂) = (10, 1):

We want to find coefficients $c$ and $d$ such that $T(2, 5) = c(1, 3) + d(2, 5)$.

$(10, 1) = c(1, 3) + d(2, 5)$

$(10, 1) = (c, 3c) + (2d, 5d)$

$(10, 1) = (c + 2d, 3c + 5d)$

This gives us a system of linear equations:

\begin{align*} c + 2d &= 10 \quad &(3) \\ 3c + 5d &= 1 \quad &(4) \end{align*}

From equation (3), $c = 10 - 2d$.

Substitute this into equation (4):

$3(10 - 2d) + 5d = 1$

$30 - 6d + 5d = 1$

$30 - d = 1$

$d = 30 - 1 = 29$

Now substitute the value of $d$ back into the expression for $c$:

$c = 10 - 2(29) = 10 - 58 = -48$

So, $T(2, 5) = -48(1, 3) + 29(2, 5)$. The coordinate vector of $T(v_2)$ with respect to $B$ is $\begin{pmatrix} -48 \\ 29 \end{pmatrix}$. This will be the second column of the matrix $[T]_B$.

Constructing the Matrix Representation

The matrix representation $[T]_B$ is formed by using the coordinate vectors found above as its columns:

\[ [T]_B = \begin{pmatrix} -30 & -48 \\ 18 & 29 \end{pmatrix} \]

The resulting matrix is $\left( {\begin{array}{*{20}{c}} { - 30}&{ - 48}\\ {18}&{29} \end{array}} \right)$.

Comparison with Options

Comparing our result with the given options, we find that it matches option 3.

Revision Table: Linear Transformation Matrix

Let's summarize the key steps involved in finding the matrix representation of a linear transformation relative to a basis.

Additional Information: Change of Basis

The matrix representation of a linear transformation depends on the choice of basis. If we choose a different basis, we will get a different matrix. The process described above calculates the matrix relative to a specific pair of bases (in this case, the same basis B for both domain and codomain).

The concept of change of basis is related. If we have the matrix representation of $T$ with respect to one basis and want to find it with respect to another basis, we use change-of-basis matrices. The formula for changing the matrix representation from basis $B_1$ to basis $B_2$ for $T: V \to V$ is $[T]_{B_2} = P_{B_2 \leftarrow B_1} [T]_{B_1} P_{B_1 \leftarrow B_2}$, where $P_{B_2 \leftarrow B_1}$ is the change-of-basis matrix from $B_1$ to $B_2$. Note that $P_{B_1 \leftarrow B_2} = (P_{B_2 \leftarrow B_1})^{-1}$.

In this problem, we directly computed the matrix relative to the given basis, which is the standard method when the bases are provided.