Question: Let $x$ be random variable, such that $E(x)=0,E(x^2)=1$ and $P(x^2\geq s^2)\geq\displaystyle\frac{C}{s^t}$, where $C>0,s\geq 1 , t>0$.
Let $m<n$ and $m,n$ are natural numbers very big. Let also $L\geq 1$ .
Consider 
$1-(1-\frac{n}{2}P(x^2\geq Ln))^m$.

Assume (\*) $\frac{n}{2}P(x^2\geq Ln)\leq \frac{2c\_0}{m}$, where $0 <c\_0 <0.6$;

using inequality $(1-y)^m\leq 1-\displaystyle\frac{my}{2}$, valid for $y\in [0, c\_0]$ , with natural $n$, we get

$$1-(1-\frac{n}{2}P(x^2\geq Ln))^m\geq\displaystyle\frac{nm}{4}P(x^2\geq Ln),$$

using assumption $P(x^2\geq s^2)\geq\displaystyle\frac{C}{s^t}$ with $s^2=Ln$ , we get
$$1-(1-\frac{n}{2}P(x^2\geq Ln))^m\geq \displaystyle\frac{nm}{4}P(x^2\geq Ln)\geq \frac{mnC'}{(Ln)^{\frac{t}{2}}}.$$

How to show, that if $t\geq 4$, then (\*) holds?

Answer: 
Technically it's *at most one*; for example, a collection consisting of a single non-zero vector is linearly independent, but if the vector space $V$ we're working in has dimension $>1$, there will be vectors that are not linear combinations of that vector (i.e. scalar multiples of it). **If** $n=\dim(V)$, **then** you are correct that it is actually exactly one.

If $v\_1,\ldots,v\_n$ are linearly independent, then by definition
$$\sum\_{i=1}^nc\_iv\_i=0\implies c\_i=0\text{ for all }i,$$
so if 
$$w=\sum\_{i=1}^nc\_iv\_i=\sum\_{i=1}^nd\_iv\_i$$
then $$0=w-w=\sum\_{i=1}^n(c\_i-d\_i)v\_i,$$
hence $c\_i-d\_i=0$ for all $i$, hence $c\_i=d\_i$ for all $i$. Thus, if the $v\_i$ are linearly independent, there is at most one way of writing a given vector as a linear combination of them.

Conversely, if there is at most one way of writing a given vector as a linear combination of the $v\_i$, that is if
$$\sum\_{i=1}^nc\_iv\_i=\sum\_{i=1}^nd\_iv\_i\implies c\_i=d\_i\text{ for all }i,$$
then if 
$$\sum\_{i=1}^nc\_iv\_i=0,$$
we have
$$\sum\_{i=1}^nc\_iv\_i=\sum\_{i=1}^n0v\_i\implies c\_i=0\text{ for all }i,$$
so the $v\_i$ are linearly independent.
Yes. Any vector that can be expressed as a linear combination of linearly independent vectors can be expressed in one and only one way.

To see this, suppose that $\mathbf{v}$ can be expressed as a linear combination of $\mathbf{v}\_1,\ldots,\mathbf{v}\_n$, with scalars $\alpha\_i$ and $\beta\_i$:
$$ \mathbf{v} = \alpha\_1\mathbf{v}\_1+\cdots + \alpha\_n\mathbf{v}\_n = \beta\_1\mathbf{v}\_1+\cdots + \beta\_n\mathbf{v}\_n.$$
Then:
$$\begin{align\*}
\mathbf{0} &= \mathbf{v}-\mathbf{v} \\
&=\bigl( \alpha\_1\mathbf{v}\_1+\cdots + \alpha\_n\mathbf{v}\_n\bigr)-\bigl( \beta\_1\mathbf{v}\_1+\cdots + \beta\_n\mathbf{v}\_n\bigr)\\
&= (\alpha\_1-\beta\_1)\mathbf{v}\_1 + \cdots + (\alpha\_n-\beta\_n)\mathbf{v}\_n.
\end{align\*}$$
Since $\mathbf{v}\_1,\ldots,\mathbf{v}\_n$ are linearly independent, this means $\alpha\_1-\beta\_1=\alpha\_2-\beta\_2=\cdots=\alpha\_n-\beta\_n = 0$, so $\alpha\_1=\beta\_1,\ldots,\alpha\_n=\beta\_n$. That is: the two expressions are actually identical.

One way to remember this is:

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> A set $S$ of vectors of $V$ spans $V$ if and only if every vector of $V$ can be written as a linear combination of vectors in $S$ in *at least one* way. A set $I$ of vectors of $V$ is linearly independent if and only if every vector of $V$ can be written as a linear combination of vectors in $I$ in *at most one* way. A set $B$ of vectors of $V$ is a basis if and only if every vector of $V$ can be written as a linear combination of vectors in $V$ in *exactly one* way.
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