Answer:
the given linear transformation represented by the matrix A = \[ \left(\begin{array}{cc} -3/5 & 4/5 \\ 4/5 & 3/5 \end{array}\right) \] is a linear transformation that maps vectors in the input space to vectors in the output space. It is a linear transformation with linearity properties, transforms basis vectors to specific vectors, and has a determinant of -1 indicating an orientation-reversing behavior.
Explanation:
To understand the properties and behavior of this linear transformation, we can analyze its effect on vectors in the 2D space. Let's consider a vector \[ \mathbf{v} = \left(\begin{array}{l} x \\ y \end{array}\right) \] in the input space.
When this vector is transformed by the matrix A, we obtain the output vector as follows:
\[ A\mathbf{v} = \left(\begin{array}{cc} \frac{-3}{5} & \frac{4}{5} \\ \frac{4}{5} & \frac{3}{5} \end{array}\right) \left(\begin{array}{l} x \\ y \end{array}\right) = \left(\begin{array}{l} -\frac{3}{5}x + \frac{4}{5}y \\ \frac{4}{5}x + \frac{3}{5}y \end{array}\right) = \left(\begin{array}{l} -\frac{3x}{5} + \frac{4y}{5} \\ \frac{4x}{5} + \frac{3y}{5} \end{array}\right) \]
This means that the linear transformation maps the vector (x, y) to the vector (-3x/5 + 4y/5, 4x/5 + 3y/5).
Now, let's analyze some important properties of this linear transformation:
1. Linearity: The given transformation is linear since it satisfies the properties of linearity. For any vectors \[ \mathbf{u} = \left(\begin{array}{l} u_1 \\ u_2 \end{array}\right) \] and \[ \mathbf{v} = \left(\begin{array}{l} v_1 \\ v_2 \end{array}\right) \] and any scalar c, we have:
\[ A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v} \]
\[ A(c\mathbf{u}) = c(A\mathbf{u}) \]
2. Transformation of Basis Vectors: It is important to analyze how the linear transformation affects the standard basis vectors in the input space. The standard basis vectors are:
\[ \mathbf{i} = \left(\begin{array}{l} 1 \\ 0 \end{array}\right) \quad \text{and} \quad \mathbf{j} = \left(\begin{array}{l} 0 \\ 1 \end{array}\right) \]
When we apply the linear transformation to these basis vectors, we obtain:
\[ A\mathbf{i} = \left(\begin{array}{cc} -3/5 & 4/5 \\ 4/5 & 3/5 \end{array}\right) \left(\begin{array}{l} 1 \\ 0 \end{array}\right) = \left(\begin{array}{l} -3/5 \\ 4/5 \end{array}\right) \]
\[ A\mathbf{j} = \left(\begin{array}{cc} -3/5 & 4/5 \\ 4/5 & 3/5 \end{array}\right) \left(\begin{array}{l} 0 \\ 1 \end{array}\right) = \left(\begin{array}{l} 4/5 \\ 3/5 \end{array}\right) \]
Therefore, the linear transformation maps the standard basis vectors to the vectors (-3/5, 4/5) and (4/5, 3/5) respectively.
3. Determinant: The determinant of the matrix A can provide important information about the linear transformation. In this case, the determinant is:
\[ \text{det}(A) = \text{det}\left(\begin{array}{cc} -3/5 & 4/5 \\ 4/5 & 3/5 \end{array}\right) = (-3/5)(3/5) - (4/5)(4/5) = -9/25 - 16/25 = -25/25 = -1 \]
The determinant being -1 indicates that the linear transformation is orientation-reversing, meaning it flips the orientation of vectors in the input space.