= Solution
For the matrix's Euclidean <operator norm>, <orthogonal invariance of the Euclidean norm> gives
$$
\|Ax\|_2=\|\Sigma V^Tx\|_2\leq\sigma_1\|x\|_2.
$$
Taking $x=v_1$ gives equality, so $\|A\|_2=\sigma_1$. For the <Frobenius norm>, the <trace> identity and the <singular value decomposition> give
$$
\|A\|_F^2=\operatorname{tr}(A^TA)=\operatorname{tr}(V\Sigma^T\Sigma V^T)=\sum_j\sigma_j^2.
$$
Consequently
$$
\boxed{\|A\|_2=\sigma_1,\qquad\|A\|_F=\left(\sum_{j=1}^r\sigma_j^2\right)^{1/2}.}
$$
Here the Euclidean matrix norm means the <operator norm> induced by the vector <Euclidean norm>, rather than the entrywise Euclidean length, which is the <Frobenius norm>.
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