A linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ rotates vectors counter-clockwise by $\theta = 45^\circ$ and subsequently scales them by a factor of $\sqrt{2}$. The standard matrix of the transformation is $[T] = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. The value of $a^2 + b^2 + c^2 + d^2$ is ________ (answer in integer).
Answer Key:
4
Step-by-Step Solution & Explanation
**Method 1: Direct Coordinate Matrix**
The standard rotation matrix for counter-clockwise rotation by $45^\circ$ is:
$R = \begin{pmatrix} \cos(45^\circ) & -\sin(45^\circ) \\ \sin(45^\circ) & \cos(45^\circ) \end{pmatrix} = \begin{pmatrix} 1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix}$
Scaling by $\sqrt{2}$ gives:
$[T] = \sqrt{2} R = \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$
Here $a = 1, b = -1, c = 1, d = 1$.
$a^2 + b^2 + c^2 + d^2 = 1^2 + (-1)^2 + 1^2 + 1^2 = 1 + 1 + 1 + 1 = 4$
**Method 2: Frobenius Norm Shortcut**
$\|[T]\|_F^2 = \operatorname{trace}([T]^T [T]) = \operatorname{trace}((\sqrt{2}R)^T (\sqrt{2}R)) = 2 \operatorname{trace}(R^T R) = 2 \operatorname{trace}(I_2) = 2 \times 2 = 4$