Proof that greatest entry of a unit vector is $\leq 1$ and $\geq \frac{1}{\sqrt{n}}$

Question

I have a real orthogonal matrix so the column vectors form an orthogonal system and thus the vectors have length one.

I now want to show that for an arbitrary column vector $v_k \in \mathbb{R^n}$ the absolute value of the greatest entry $|v_{k_i}|$

i) is less or equal than one

ii) is greater or equal than $\frac{1}{\sqrt{n}}$. So

$$\dfrac{1}{\sqrt{n}} \leq \;\max\{|v_{k_1}|,|v_{k_2}|,...,|v_{k_n}|\} = |v_{k_i}|\; \leq 1.$$

i) This is intuitive but I struggle to come up with a proof. Can I assume that for an orthogonal matrix, the vectors are always orthonormal regarding the standard inner product or do other inner products are possible, too? If not I would prove it using an indirect proof, else I don't know.

ii) I have no idea how to approach this, any hints are welcome.

Thank you

Answer 1

Every column has norm $1$. So, the sum of the squares of its entries is equal to $1$ and therefore no entry can have absolute value greater than $1$.

But if all of them had absolute value smaller than $\frac1{\sqrt n}$, then the sum of the squares of the entries would be smaller than$$\overbrace{\frac1n+\frac1n+\cdots+\frac1n}^{n\text{ times}}=1.$$

Answer 2

Let $v \in \mathbb R^n$ be a unit vector. That is, $\sum |v_i|^2=1$. Let $k$ be the index of the coordinate with largest absolute value. Then,

$$|v_k|^2 \leq \sum |v_i|^2 = 1 \Rightarrow |v_k| \leq 1$$ and $$|v_k|^2 = \frac{1}{n} \cdot \sum |v_k|^2 \geq \frac{1}{n} \cdot \sum |v_i|^2=\frac{1}{n} \Rightarrow |v_k| \geq \frac{1}{\sqrt n}$$