The norm of a matrix is defined as
\begin{equation}
\|A\| = \sup_{\|u\| = 1} \|Au\|
\end{equation}
Taking the singular value decomposition of the matrix $A$, we have
\begin{equation}
A = VD W^T
\end{equation}
where $V$ and $W$ are orthonormal and $D$ is a diagonal matrix. Since $V$ and $W$ are orthonormal, we have $\|V\| = 1$ and $\|W\| = 1$. Then $\|Av\| = \|D v\|$ for any vector $v$. Then we can maximize the norm of $Av$ by maximizing the norm of $Dv$.

By the definition of singular value decomposition, $D$ will have the singular values of $A$ on its main diagonal and will have zeros everywhere else. Let $\lambda_1, \ldots, \lambda_n$ denote these diagonal entries so that

\begin{equation}
D = \left(\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_n
\end{array}\right)
\end{equation}

Taking some $v = (v_1, v_2, \ldots, v_n)^T$, the product $Dv$ takes the form
\begin{equation}
Dv = \left(\begin{array}{c}
\lambda_1v_1 \\
\vdots \\
\lambda_nv_n
\end{array}\right)
\end{equation}
Maximizing the norm of this is the same as maximizing the norm squared. Then we are trying to maximize the sum
\begin{equation}
S = \sum_{i=1}^{n} \lambda_i^2v_i^2
\end{equation}
under the constraint that $v$ is a unit vector (i.e., $\sum_i v_i^2 = 1$). The maximum is attained by finding the largest $\lambda_i^2$ and setting its corresponding $v_i$ to $1$ and then setting each other $v_j$ to $0$. Then the maximum of $S$ (which is the norm squared) is the square of the absolutely largest eigenvalue of $A$. Taking the square root, we get the absolutely largest eigenvalue of $A$.