1. ## Subspaces

Show that if $dim \ W_1 \geq \ dim \ W_2$ and $dim(W_1+W_2)=1+dim(W_1 \cap W_2)$, then $W_1+W_2=W_1$ and $W_1 \cap W_2=W_2$.
Proving $W_1+W_2=W_1$
$dim \ W_1 \geq \ dim \ W_2 \Rightarrow dim \ W_1-dim \ W_2 \geq 0 \Rightarrow \ dim \ W_1-dim \ W_2 \geq k \geq 0$

The value of k represents the dimension of a basis for the vectors in $W_1$ that are not in $W_2$.

Hence $W_2 \subset W_1$ (Was the proof above enough to justify this step?).

Therefore $\forall \ u \in W_1$ and $\forall v \in W_2$, $u+v \in W_1+W_2 \subset W_1$.

Conversely, $\underline{0} \in W_2$ so $W_1 \subset W_1+W_2$.

This gives that $W_1+W_2=W_1$.

Proving $W_1 \cap W_2=W_1$
$dim(W_1 +W_2)=dim(W_1)+dim(W_2)-dim(W_1 \cap W_2)$

but we know that $dim(W_1+W_2)=dim(W_1)$

$dim(W_1)=dim(W_1)+dim(W_2)-dim(W_1 \cap W_2) \Rightarrow dim(W_1 \cap W_2)=dim(W_2)$ $\Rightarrow \ W_1 \cap W_2=W_2$

(this is implied because they have the same dimension over the same vector space so they must be the same subspace).

2. Originally Posted by Showcase_22

$dim \ W_1 \geq \ dim \ W_2 \Rightarrow dim \ W_1-dim \ W_2 \geq 0 \Rightarrow \ dim \ W_1-dim \ W_2 \geq k \geq 0$

The value of k represents the dimension of a basis for the vectors in $W_1$ that are not in $W_2$.

Hence $W_2 \subset W_1$ (Was the proof above enough to justify this step?). No!

Therefore $\forall \ u \in W_1$ and $\forall v \in W_2$, $u+v \in W_1+W_2 \subset W_1$.

Conversely, $\underline{0} \in W_2$ so $W_1 \subset W_1+W_2$.

This gives that $W_1+W_2=W_1$.

$dim(W_1 +W_2)=dim(W_1)+dim(W_2)-dim(W_1 \cap W_2)$

but we know that $dim(W_1+W_2)=dim(W_1)$

$dim(W_1)=dim(W_1)+dim(W_2)-dim(W_1 \cap W_2) \Rightarrow dim(W_1 \cap W_2)=dim(W_2)$ $\Rightarrow \ W_1 \cap W_2=W_2$

(this is implied because they have the same dimension over the same vector space so they must be the same subspace).
if $W_1 \subseteq W_2$ or $W_2 \subseteq W_1,$ then we're done. otherwise, $W_1 \cap W_2$ would be a proper subspace of both $W_1$ and $W_2.$ thus: $\dim W_1 \geq 1 + \dim W_1 \cap W_2$ and $\dim W_2 \geq 1 + \dim W_1 \cap W_2.$

therefore: $\dim (W_1+W_2)=\dim W_1 + \dim W_2 - \dim W_1 \cap W_2 \geq 2 + \dim W_1 \cap W_2,$ contradicting our assumption that $\dim(W_1 + W_2)=1 + \dim W_1 \cap W_2.$ Q.E.D.

3. $W_1$ and $W_2$ are in the same vectorial space?

a question: If $V_1$ and $V_2$ are in the same vectorial space, I can tell: $
\dim \left( {V_1 } \right) \leqslant \dim \left( {V_2 } \right) \Rightarrow V_1 \leqslant V_2
$
?

where the second "mayor o equal" mean subspace

4. Originally Posted by Nacho
$W_1$ and $W_2$ are in the same vectorial space?
of course! otherwise $W_1 + W_2$ would be undefined!

a question: If $V_1$ and $V_2$ are in the same vectorial space, I can tell: $
\dim \left( {V_1 } \right) \leqslant \dim \left( {V_2 } \right) \Rightarrow V_1 \leqslant V_2
$
?

where the second "mayor o equal" mean subspace
not necessarily!

5. Originally Posted by Nacho
$W_1$ and $W_2$ are in the same vectorial space?

a question: If $V_1$ and $V_2$ are in the same vectorial space, I can tell: $
\dim \left( {V_1 } \right) \leqslant \dim \left( {V_2 } \right) \Rightarrow V_1 \leqslant V_2
$
?

where the second "mayor o equal" mean subspace
The command is \subseteq