WeBWorK Standalone Renderer

Suppose $f : \mathbb{R}^2 \to \mathbb{R}^2$ is a linear transformation. The two pictures on top in the figure use standard $\mathcal{S}$ -coordinates, where $\mathcal{S} = \lbrace \mathbf{e_1}, \mathbf{e_2} \rbrace$ . The two figures on bottom in the figure use $\mathcal{B}$ -coordinates, where $\mathcal{B} = \lbrace \mathbf{b_1}, \mathbf{b_2} \rbrace$ . The figure shows the vectors $\mathbf{b_1}$ and $\mathbf{b_2}$ in blue and the vectors $f(\mathbf{e_1})$ and $f(\mathbf{e_2})$ in red.

Standard basis $\mathcal{S} = \lbrace \mathbf{e_1}, \mathbf{e_2} \rbrace$		Standard basis $\mathcal{S} = \lbrace \mathbf{e_1}, \mathbf{e_2} \rbrace$
	$\Large \stackrel{\displaystyle \lbrack f \rbrack_\mathcal{S}^\mathcal{S}}{\to}$

$\Large \lbrack id \rbrack_\mathcal{B}^\mathcal{S} \ \uparrow$		$\Large \lbrack id \rbrack_\mathcal{B}^\mathcal{S} \ \uparrow$

	$\Large \stackrel{\displaystyle \lbrack f \rbrack_\mathcal{B}^\mathcal{B}}{\to}$
Custom basis $\mathcal{B} = \lbrace \mathbf{b_1}, \mathbf{b_2} \rbrace$		Custom basis $\mathcal{B} = \lbrace \mathbf{b_1}, \mathbf{b_2} \rbrace$

Part 1: Eigenvectors and eigenvalues

Find the matrix $A$ for the linear transformation $f$ relative to the standard basis in the domain and in the codomain. That is, find the matrix $A$ such that $\lbrack f \rbrack_\mathcal{S}^\mathcal{S} (\mathbf{x}) = A \mathbf{x}$ .
$A =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$ $\left]\Rule{0pt}{2.4em}{0pt}\right.$

Write $f(\mathbf{b_1})$ and $f(\mathbf{b_2})$ as linear combinations of the vectors in the basis $\mathcal{B}$ . Enter a vector sum of the form 5 b1 + 6 b2.
$f(\mathbf{b_1}) =$
$f(\mathbf{b_2}) =$

The vector $\mathbf{b_1}$ an eigenvector for the linear transformation $f$ with eigenvalue (enter a number or DNE).
The vector $\mathbf{b_2}$ an eigenvector for the linear transformation $f$ with eigenvalue (enter a number or DNE).

Part 2: Diagonalization

You can earn partial credit on this problem.