Linear Transformation: Determining the Standard Matrix

The question asks us to find the standard matrix of a linear transformation \(T : \mathbb{R}^2 \rightarrow \mathbb{R}^2\) that reflects points through the line \(x_1 = -x_2\). Understanding how linear transformations are represented by matrices is key to solving this problem.

Standard Basis Vectors of ℝ²

The standard ordered basis for \(\mathbb{R}^2\) consists of two vectors:

• \(e_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\), which represents the point \((1,0)\).
• \(e_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\), which represents the point \((0,1)\).

To find the standard matrix of a linear transformation \(T\), we apply \(T\) to each of these standard basis vectors. The images \(T(e_1)\) and \(T(e_2)\) will form the columns of the standard matrix.

Reflection Through the Line \(x_1 = -x_2\)

The linear transformation \(T\) performs a reflection through the line \(x_1 = -x_2\). This line can also be written as \(x_1 + x_2 = 0\). We can use the general formula for reflecting a point \((a,b)\) across a line \(Ax + By + C = 0\). The reflected point \((a',b')\) is given by:

$$ \frac{a' - a}{A} = \frac{b' - b}{B} = -2 \frac{Aa + Bb + C}{A^2 + B^2} $$

For our line \(x_1 + x_2 = 0\), we have \(A=1\), \(B=1\), and \(C=0\).

Transforming Basis Vector \(e_1 = (1,0)\)

Let's find the image of \(e_1 = (1,0)\) under the reflection. Here, \(a=1\) and \(b=0\).

$$ \frac{a' - 1}{1} = \frac{b' - 0}{1} = -2 \frac{1(1) + 1(0) + 0}{1^2 + 1^2} $$

$$ \frac{a' - 1}{1} = \frac{b'}{1} = -2 \frac{1}{2} $$

$$ a' - 1 = -1 \implies a' = 0 $$

$$ b' = -1 $$

So, the reflected point \(T(e_1)\) is \((0, -1)\). In vector form, \(T(e_1) = \begin{pmatrix} 0 \\ -1 \end{pmatrix}\).

Transforming Basis Vector \(e_2 = (0,1)\)

Now, let's find the image of \(e_2 = (0,1)\) under the reflection. Here, \(a=0\) and \(b=1\).

$$ \frac{a' - 0}{1} = \frac{b' - 1}{1} = -2 \frac{1(0) + 1(1) + 0}{1^2 + 1^2} $$

$$ \frac{a'}{1} = \frac{b' - 1}{1} = -2 \frac{1}{2} $$

$$ a' = -1 $$

$$ b' - 1 = -1 \implies b' = 0 $$

So, the reflected point \(T(e_2)\) is \((-1, 0)\). In vector form, \(T(e_2) = \begin{pmatrix} -1 \\ 0 \end{pmatrix}\).

Standard Matrix of T

The standard matrix of \(T\) has \(T(e_1)\) as its first column and \(T(e_2)\) as its second column:

$$ \text{Standard Matrix of T} = \begin{pmatrix} T(e_1) & T(e_2) \end{pmatrix} $$

$$ = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} $$

Summary of Steps for Matrix Calculation

This result matches option 2 provided in the question.