1. ## Subspaces

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

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

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

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

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

This gives that $\displaystyle W_1+W_2=W_1$.

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

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

$\displaystyle 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) $$\displaystyle \Rightarrow \ W_1 \cap W_2=W_2 (this is implied because they have the same dimension over the same vector space so they must be the same subspace). 2. Originally Posted by Showcase_22 \displaystyle dim \ W_1 \geq \ dim \ W_2 \Rightarrow dim \ W_1-dim \ W_2 \geq 0 \Rightarrow \ dim \ W_1-dim \ W_2 \geq k \geq 0 The value of k represents the dimension of a basis for the vectors in \displaystyle W_1 that are not in \displaystyle W_2. Hence \displaystyle W_2 \subset W_1 (Was the proof above enough to justify this step?). No! Therefore \displaystyle \forall \ u \in W_1 and \displaystyle \forall v \in W_2, \displaystyle u+v \in W_1+W_2 \subset W_1. Conversely, \displaystyle \underline{0} \in W_2 so \displaystyle W_1 \subset W_1+W_2. This gives that \displaystyle W_1+W_2=W_1. \displaystyle dim(W_1 +W_2)=dim(W_1)+dim(W_2)-dim(W_1 \cap W_2) but we know that \displaystyle dim(W_1+W_2)=dim(W_1) \displaystyle dim(W_1)=dim(W_1)+dim(W_2)-dim(W_1 \cap W_2) \Rightarrow dim(W_1 \cap W_2)=dim(W_2)$$\displaystyle \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 $\displaystyle W_1 \subseteq W_2$ or $\displaystyle W_2 \subseteq W_1,$ then we're done. otherwise, $\displaystyle W_1 \cap W_2$ would be a proper subspace of both $\displaystyle W_1$ and $\displaystyle W_2.$ thus: $\displaystyle \dim W_1 \geq 1 + \dim W_1 \cap W_2$ and $\displaystyle \dim W_2 \geq 1 + \dim W_1 \cap W_2.$

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

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

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

where the second "mayor o equal" mean subspace

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

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

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

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

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

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