Successive minima

Remarks

The infimum is attained, and $0 < \lambda_1 \le \lambda_2 \le \dots \le \lambda_n < \infty$. Example: for $\Lambda = \Z^2$ and $K = [-1, 1] \times [-2, 2]$, one has $\lambda K = [-\lambda, \lambda] \times [-2\lambda, 2\lambda]$. The vector $(0, 1)$ lies in $\lambda K$ exactly when $\lambda \ge 1/2$, and no nonzero integer vector lies in $\lambda K$ for $\lambda < 1/2$, so $\lambda_1 = 1/2$. A vector independent of $(0, 1)$ has first coordinate of absolute value at least $1$, so it needs $\lambda \ge 1$, and $(1, 0)$ enters at $\lambda = 1$; so $\lambda_2 = 1$. Non-example: the strip $K = \R \times [-1, 1]$ is convex and symmetric but not compact. Then $(1, 0) \in \lambda K$ for every $\lambda > 0$, so the infimum defining $\lambda_1$ is $0$ and the minima lose their meaning. When $K$ is the closed unit ball, $\lambda_k$ is the smallest $r$ such that $\Lambda$ has $k$ linearly independent vectors of length at most $r$.