Let $\bold{v} = \begin{pmatrix} x \\ y \end{pmatrix}$.  Then
\[\|\bold{v}\| = \left\| \begin{pmatrix} x \\ y \end{pmatrix} \right\| = \sqrt{x^2 + y^2},\]and
\begin{align*}
\left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \bold{v} \right\| &= \left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \right\| \\
&= \left\| \begin{pmatrix} 2x + 3y \\ -2y \end{pmatrix} \right\| \\
&= \sqrt{(2x + 3y)^2 + (-2y)^2} \\
&= \sqrt{4x^2 + 12xy + 13y^2},
\end{align*}so the given inequality becomes
\[\sqrt{4x^2 + 12xy + 13y^2} \le C \sqrt{x^2 + y^2},\]or
\[\sqrt{\frac{4x^2 + 12xy + 13y^2}{x^2 + y^2}} \le C.\]Thus, we can think of $C$ as the maximum value of the expression in the left-hand side.

Maximizing the expression in the left-hand side is equivalent to maximizing its square, namely
\[\frac{4x^2 + 12xy + 13y^2}{x^2 + y^2}.\]Let $k$ be a possible value of this expression, which means the equation
\[\frac{4x^2 + 12xy + 13y^2}{x^2 + y^2} = k\]has a solution in $x$ and $y$.  We can re-write this equation as
\[(4 - k) x^2 + 12xy + (13 - k) y^2 = 0.\]For this quadratic expression to have a solution in $x$ and $y$, its discriminant must be nonnegative.  In other words,
\[12^2 - 4 (4 - k)(13 - k) \ge 0,\]or $4k^2 - 68k + 64 \le 0$.  This inequality factors as $4(k - 1)(k - 16) \le 0$.  The largest value of $k$ that satisfies this inequality is 16, so the value of $C$ we seek is $\sqrt{16} = \boxed{4}$.  Note that equality occurs for
\[\bold{v} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}.\]
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Find the smallest positive real number $C$ for which
\[\left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \bold{v} \right\| \le C \|\bold{v}\|\]for all two-dimensional vectors $\bold{v}.$

Note that for a two-dimensional vector $\mathbf{a},$ $\|\mathbf{a}\|$ is the magnitude of $\mathbf{a}.$
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
4
Find the smallest positive real number $C$ for which
\[\left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \bold{v} \right\| \le C \|\bold{v}\|\]for all two-dimensional vectors $\bold{v}.$

Note that for a two-dimensional vector $\mathbf{a},$ $\|\mathbf{a}\|$ is the magnitude of $\mathbf{a}.$
test