If a subset T of S is dependent, then S itself dependent

[Guest2]

1. Let $V$ be a vector space. Let $S \subset V$, and let $T \subset S$. Show that if $T$ is dependent then $S$ itself dependent.

If $T$ is dependent then there's finite set of distinct elements in $T$, say $y_{1}, \ldots, y_{n}$ and a corresponding set of scalars $a_1, \ldots, a_{n}$ such that $\sum_{i=1}^{n} a_iy_i = 0$. Since $T \subset S$, such $y_{1}, \ldots, y_{n}$ and $a_1, \ldots, a_{n}$ also belong to $S$, therefore $S$ itself dependent.

2. If one element in $S$ is scalar multiple of another, then $S$ is dependent.

Let the elements be $x_{1}$ and $x_2$, and the $c_1$ be the scalar. We have $x_{1} = c x_2 \implies x_1-cx_2 = 0$. Thus $S$ is dependent.

Are my attempts okay?

 

[Deveno]

1. Let $V$ be a vector space. Let $S \subset V$, and let $T \subset S$. Show that if $T$ is dependent then $S$ itself dependent.

If $T$ is dependent then there's finite set of distinct elements in $T$, say $y_{1}, \ldots, y_{n}$ and a corresponding set of scalars $a_1, \ldots, a_{n}$ such that $\sum_{i=1}^{n} a_iy_i = 0$. Since $T \subset S$, such $y_{1}, \ldots, y_{n}$ and $a_1, \ldots, a_{n}$ also belong to $S$, therefore $S$ itself dependent.

This is not quite true, we ALSO require that NOT ALL the $a_i$ are 0.

2. If one element in $S$ is scalar multiple of another, then $S$ is dependent.

Let the elements be $x_{1}$ and $x_2$, and the $c_1$ be the scalar. We have $x_{1} = c x_2 \implies x_1-cx_2 = 0$. Thus $S$ is dependent.

Are my attempts okay?

See my comment above. It's *very* close, which coefficient in your linear combination is non-zero?

 

[Guest2]

This is not quite true, we ALSO require that NOT ALL the $a_i$ are 0.
See my comment above. It's *very* close, which coefficient in your linear combination is non-zero?

Oh, I see I missed the same condition both times. We should have $c \ne 0$, since, as you said, all $a_i$ can't be zero; but also I think because if $c= 0$, then $x_1$ and $x_2$ wouldn't be distinct.

 

[Deveno]

Oh, I see I missed the same condition both times. We should have $c \ne 0$, since, as you said, all $a_i$ can't be zero; but also I think because if $c= 0$, then $x_1$ and $x_2$ wouldn't be distinct.

Oh, so close...

In your second example, if $c = 0$ we have a linearly dependent set, since $0x_1 = 0$, and the set consisting of the $0$-vector is always a linearly dependent set (in fact, any set CONTAINING the 0-vector is likewise linearly dependent).

But the non-zero coefficient we have in:

$x_1 - cx_2 = 0$

is the coefficient of $x_1$, which is $1$ and in a field, we can NEVER have (by definition) $1 = 0$.