# Abstract Nonsense

## Halmos Chapter one Sections 10, 11, and 12: Subspaces, Calculus of Subspaces, and Dimension of Subspaces

1.

Problem: If $\mathcal{M}$ and $\mathcal{N}$ are finite-dimensional subspace with the same dimension $n$, and $\mathcal{M}\subseteq\mathcal{N}$, then $\mathcal{M}=\mathcal{N}$

Proof: Suppose not, then there is some $v\in\mathcal{N}-\mathcal{M}$. Then, if $\{x_1,\cdots,x_n\}$ is the finite basis for $\mathcal{M}$ we claim that $\{x_1,\cdots,x_n,v\}$ is l.i. in $\mathcal{N}$. To see this suppose that

$\alpha_1 x_1+\cdots+\alpha_n x_n+\alpha v=\bold{0}\quad (1)$

We may conclude that $\alpha=0$ otherwise

$\displaystyle v=\frac{-\alpha_1}{\alpha}x_1+\cdots+\frac{-\alpha_n}{\alpha} x_n\implies v\in\text{Span }\{x_1,\cdots,x_n\}=\mathcal{M}$

and thus $\alpha=0$.  Thus, $(1)$ may be rewritten as

$\alpha_1x_1+\cdots+\alpha_n x_n=\bold{0}$

and thus by the l.i. of $\{x_1,\cdots,x_n\}$ we see that this implies that $\alpha_1=\cdots=\alpha_n=0$ and so $\{x_1,\cdots,x_n,\alpha\}$ is a l.i. set in $\mathcal{N}$, but this contradicts that $\dim\mathcal{N}=n$. Thus, there is no $v\in\mathcal{N}-\mathcal{M}$ and so $\mathcal{M}=\mathcal{N}$

2.

Problem: If $\mathcal{M}$ and $\mathcal{N}$ are subspaces of a vector space $\mathcal{V}$ and if every vector in $\mathcal{V}$ belongs to either $\mathcal{N}$ or $\mathcal{M}$, then either $\mathcal{M}=\mathcal{V}$ or $\mathcal{N}=\mathcal{V}$

Proof: We prove the contrapositive, if $\mathcal{M},\mathcal{N}$ are subspaces of $\mathcal{V}$ such that neither equal $\mathcal{V}$ then $\mathcal{M}\cup\mathcal{N}\subsetneq\mathcal{V}$. To prove this, suppose not and $\mathcal{M}\cup\mathcal{N}=\mathcal{V}$. We first see by the condition that $\mathcal{M},\mathcal{N}\ne\mathcal{V}$ that there are $v\notin\mathcal{M}$ and $w\notin\mathcal{N}$. Now, since $\mathcal{M}\cup\mathcal{N}=\mathcal{V}$ we can easily see that $v\in\mathcal{N}$ and $w\in\mathcal{M}$, but where is $v+w$? If $v+w\in\mathcal{M}$ then by virtue of $\mathcal{M}$ being a subspace, so is $(v+w)+(-w)=v$ which contradicts our assumption that $v\notin\mathcal{M}$. Similarly, $v+w\in\mathcal{N}\implies (v+w)+(-v)=w\in\mathcal{N}$. Thus, $v+w\notin\mathcal{M}\cup\mathcal{N}=\mathcal{V}$ which is a contradiction. It follows that $\mathcal{N}\cup\mathcal{M}\ne\mathcal{V}$

3.

Problem: If $x,y$ and $z$ are vectors such that $x+y+z=\bold{0}$, then $x$ and $y$ span the same subspace as $y$ and $z$

Proof: We are attempting to show that $\text{Span }\{x,y\}=\text{Span }\{y,z\}$. To see this, let $v\in\text{Span }\{x,y\}$ then there exists $\alpha_1,\alpha_2\in F$ such that $\alpha_1x+\alpha_2y=v$, but notice that $x+y+z=0\implies -(y+z)=x$ and so this last equation may be restated as $-\alpha_1(z+y)+\alpha_2=-\alpha_1z+(\alpha_2-\alpha_1)y=v$. Thus, $v\in\text{Span }\{y,z\}$. Similarly, $x+y+z=0\implies z=-(x+y)$ and so $\alpha_1z+\alpha_2y=-\alpha_1 x+(\alpha_2-\alpha_2)y$ and so a similar reasoning may be applied. Of course, both of these sets equal $\text{Span }\{x,z\}$

Remark: One can clearly see how this extends to general linearly dependent sets.

4.

Problem: Suppose that $x$ and $y$ are vectors and $\mathcal{M}$ is a subspace in a vector space $\mathcal{V}$; let $\mathcal{H}$ be the subspace spanned by$\mathcal{M}$ and $x$, and let $\mathcal{K}$ be the subspace spanned by $\mathcal{M}$ and $y$. Prove that if $y\in\mathcal{H}$ but not in $\mathcal{M}$ then $x\in\mathcal{K}$

Proof: Let $\{x_1,\cdots,x_n\}$ be a basis for $\mathcal{M}$, then by $y\in\mathcal{H}$ we know there exists $\alpha_1,\cdots,\alpha_{n+1}$ such that $y=\alpha_1x+\cdots+\alpha_n x_n+\alpha_{n+1}x$. Now, if $\alpha_{n+1}=0$ we may rewrite this equation as $y=\alpha_1x_1+\cdots+\alpha_n x_n$ which implies that $y\in\text{Span }\{x_1,\cdots,x_n\}=\mathcal{M}$, which it assumed not to. Thus, $\alpha_{n+1}\ne 0$ and thus

$\displaystyle x=\frac{-\alpha_1}{\alpha_{n+1}}x_1+\cdots+\frac{-\alpha_{n}}{\alpha_{n+1}}x_n+\frac{1}{\alpha_{n+1}}y$

which states precisely that

$x\in\text{Span }\{x_1,\cdots,x_n,y\}=\mathcal{K}$

Remark: We’ve tacitly assumed something here which I’ll prove as a technical lemma, namely:

Lemma: Let $\mathcal{M}$ be a finite dimensional subspace of a vector space $\mathcal{V}$ with a basis $\{x_1,\cdots,x_n\}$ then

$\text{Span }\left(\mathcal{M}\cup\{v\}\right)=\text{Span }\{x_1,\cdots,x_n,v\}$

for any $v\in\mathcal{V}$

Proof: Recall that

$\displaystyle \text{Span }\left(\mathcal{M}\cup\{x\}\right)=\bigcap_{\underset{\mathcal{M}\cup\{v\}\subseteq\mathcal{S}}{\mathcal{S}\sqsubseteq\mathcal{V}}}\mathcal{S}$

(where $\sqsubseteq$ means ” is a subspace of”). Thus, noting that

$\text{Span }\left\{x_1,\cdots,x_n,v\right\}\sqsubseteq\mathcal{V}\text{ and }\mathcal{M}\cup\{v\}\subseteq\text{Span }\{x_1,\cdots,x_n,v\}$

we see from the definition that

$\text{Span }\left(\mathcal{M}\cup\{v\}\right)\subseteq\text{Span }\{x_1,\cdots,x_n,v\}$

conversely, to show the opposite inclusion holds it suffices to prove that

$\text{Span }\{x_1,\cdots,x_n,v\}\subseteq\mathcal{S}$

for a fixed but arbitrary subspace containing $\mathcal{M}\cup\{v\}$. But, this is immediate since any subspace containing $\mathcal{M}\cup\{v\}$ contains not only $\{x_1,\cdots,x_n,v\}$ but all of the possible linear combinations, and so the above inclusion holds. Thus,

$\displaystyle \text{Span }\{x_1,\cdots,x_n,v\}\subseteq\bigcap_{\underset{\mathcal{M}\cup\{v\}}{\mathcal{S}\sqsubseteq\mathcal{V}}}\mathcal{S}=\text{Span }\left(\mathcal{M}\cup\{v\}\right)$

from where the conclusion follows.

5.

Problem: Suppose that $\mathcal{L},\mathcal{M}$, and $\mathcal{N}$ are subspaces of a vector space $\mathcal{V}$.

a) Show that the equation

$\mathcal{L}\cap\left(\mathcal{M}+\mathcal{N}\right)=\left(\mathcal{L}\cap\mathcal{M}\right)+\left(\mathcal{L}\cap\mathcal{N}\right)$

isn’t necessarily true.

b) Prove that

$\mathcal{L}\cap\left(\mathcal{M}+\left(\mathcal{L}\cap\mathcal{N}\right)\right)=\left(\mathcal{L}\cap\mathcal{M}\right)+\left(\mathcal{L}\cap\mathcal{N}\right)$

Proof:

a) Consider $\mathbb{R}^3$ as a vector space and consider the three subspaces of $\mathbb{R}^3$ over $\mathbb{R}$

$\mathcal{L}=\left\{(\alpha,0,0):\alpha\in\mathbb{R}\right\}$

$\mathcal{M}=\left\{(\beta,\beta,0):\beta\in\mathbb{R}\right\}$

and

$\mathcal{N}=\left\{(0,\gamma,0):\gamma\in\mathbb{R}\right\}$

Notice that

$\mathcal{L}\cap\mathcal{M}=\mathcal{L}\cap\mathcal{N}=\{(0,0,0)\}$

But, since $(1,1,0)\in\mathcal{M}$ and $(0,-1,0)\in\mathcal{N}$ we see that $(1,1+-1,0)=(1,0,0)\in\mathcal{M}+\mathcal{N}$ and so $\mathcal{L}\cap\left(\mathcal{M}+\mathcal{N}\right)$ is not trivial.

b) Let $v\in\mathcal{L}\cap\left(\mathcal{M}+\left(\mathcal{L}\cap\mathcal{N}\right)\right)$ then $v\in\mathcal{L}$ and $v=x+y$ for some $x\in\mathcal{M}$ and some $y\in\mathcal{L}\cap\mathcal{N}$. But, since $y,v\in\mathcal{L}$ and $x=v-y$ we see that $x\in\mathcal{L}$. Thus, $x\in\mathcal{L}\cap\mathcal{M}$ and so $v=x+y$ where $x\in\mathcal{M}\cap\mathcal{L}$ and $y\in\mathcal{L}\cap\mathcal{N}$ and so $v\in\left(\mathcal{L}\cap\mathcal{M}\right)+\left(\mathcal{L}\cap\mathcal{N}\right)$

Conversely, let $v\in\left(\mathcal{L}\cap\mathcal{M}\right)+\left(\mathcal{L}\cap\mathcal{M}\right)$, then $v=x+y$ where $x\in\mathcal{L}\cap\mathcal{M}$ and $y\in\mathcal{L}\cap\mathcal{N}$. But, since $v$ is the sum of two elements of $\mathcal{L}$, which is a subspace we immediately see that $v\in\mathcal{L}$. But, with the same credence we can can see that $x\in\mathcal{M}$ so that $v=x+y$ where $x\in\mathcal{M}$ and $y\in\mathcal{L}\cap\mathcal{N}$, thus $v\in\mathcal{M}+\left(\mathcal{L}\cap\mathcal{N}\right)$. Putting this together we get that $v\in\mathcal{L}\cap\left(\mathcal{M}+\left(\mathcal{L}\cap\mathcal{N}\right)\right)$.

The conclusion follows.

6.

Problem:

a) Can it happen that a non-trivial subspace of a vector space $\mathcal{V}$ has a unique complement?

b) If $\mathcal{M}$ is an $m$-dimensional subspace of an $n$-dimensional vector space, then every complement $\mathcal{N}$ of $\mathcal{M}$ has dimension $n-m$

Proof:

a) Yes,there are lots of examples ;). In fact, if

b) By assumption we have that $\mathcal{M}$ and $\mathcal{N}$ are subspaces of $\mathcal{V}$ and so we know there exists bases $\{x_1,\cdots,x_m\}$ and $\{y_1,\cdots,y_k\}$ for $\mathcal{M}$ and $\mathcal{N}$ respectively. We will show that $\{x_1,\cdots,x_m,y_1,\cdots,y_k\}$ is a basis for $\mathcal{V}$. To see this suppose that

$\displaystyle \sum_{i\leqslant n}\alpha_i x_i+\sum_{i\leqslant k}\beta_i y_i=\bold{0}$

then

$\displaystyle \sum_{i\leqslant n}\alpha_i x_i=\sum_{i\leqslant k}(-\beta_i)y_i$

but the LHS is an element of $\mathcal{M}$ and the RHS an element of $\mathcal{N}$ and since $\mathcal{M}\cap\mathcal{N}=\{\bold{0}\}$ it follows that

$\displaystyle \sum_{i\leqslant n}\alpha_i x_i=\sum_{i\leqslant k}(-\beta_i)y_i=\bold{0}$

and by the l.i. of $\{x_1,\cdots,x_m\}$ and $\{y_1,\cdots,y_k\}$ this implies that

$\alpha_1=\cdots=\alpha_m=\beta_1=\cdots=\beta_k=0$

from where l.i. follows. Now, the fact that

$\text{Span }\{x_1,\cdots,x_m,y_1,\cdots,y_k\}=\mathcal{V}$

is fairly clear. By assumption for every $v\in\mathcal{V}$ there exists $x\in\mathcal{M}$ and $y\in\mathcal{N}$ such that

$\displaystyle v=x+y=\sum_{i\leqslant m}\alpha_i x_i+\sum_{i\leqslant k}\beta_i y_,i,\quad \alpha_1,\cdots,\alpha_m,\beta_1,\cdots,\beta_k\in F$

since

$\text{Span }\{x_1,\cdots,x_m\}=\mathcal{M}\text{ and }\text{Span }\{y_1,\cdots,y_k\}=\mathcal{N}$

It follows that $\{x_1,\cdots,x_m,\cdots,y_1,\cdots,y_k\}$ is a basis for $\mathcal{V}$ and so

$\text{card }\{x_1,\cdots,x_m,y_1,\cdots,y_k\}=m+k=\dim_F\left(\mathcal{V}\right)=n$

from where it follows that

$\dim_F\left(\mathcal{N}\right)=k=n-m$

7.

Problem:

a) Show that if both $\mathcal{M},\mathcal{N}$ are three-dimensional subspaces of a five dimensional vecto space, then $\mathcal{M}$ and $\mathcal{N}$ are disjoint.

b) If $\mathcal{M}$ and $\mathcal{N}$ are finite dimensional subspaces of a vector space, then

$\dim\left(\mathcal{M}+\mathcal{N}\right)=\dim\left(\mathcal{M}\right)+\dim\left(\mathcal{N}\right)-\dim\left(\mathcal{M}\cap\mathcal{N}\right)$

Proof:

a)

Lemma: Let $\mathcal{M},\mathcal{N}$ be subspaces of some vector space $\mathcal{V}$. Then, if $\{x_1,\cdots,x_m\}$ and $\{y_1,\cdots,y_n\}$ are respective bases then

$\mathcal{M}+\mathcal{N}=\text{Span }\left(\mathcal{M}\cup\mathcal{N}\right)=\text{Span }\{x_1,\cdots,x_m,y_1,\cdots,y_n\}$

Proof: We first prove that

$\text{Span }\left(\mathcal{M}\cup\mathcal{N}\right)=\text{Span }\left\{x_1,\cdots,x_m,y_1,\cdots,y_n\right\}$

To do this we first remember that

$\mathcal{M}+\mathcal{N}\overset{\text{de}\text{f.}}{=}\left\{x+y:x\in\mathcal{M}\text{ and }y\in\mathcal{N}\right\}$

and since evidently $\text{Span }\left\{x_1,\cdots,x_m,y_1,\cdots,y_n\right\}$ a subspace of $\mathcal{V}$ which contains $\mathcal{M}\cup\mathcal{N}$ it is immediate that

$\text{Span }\left(\mathcal{M}\cup\mathcal{N}\right)\subseteq\text{Span }\{x_1,\cdot,x_m,y_1,\cdots,y_n\}$

Conversely, let $v\in\text{Span }\{x_1,\cdots,x_m,y_1,\cdots,y_n\}$. To show that $\text{Span }\left(\mathcal{M}\cup\mathcal{N}\right)$ it suffices to show that $v\in\mathcal{S}$ for a fixed but arbitrary subspace containing $\mathcal{M}\cup\mathcal{N}$. But, clearly $\mathcal{M}\cup\mathcal{N}\subseteq\mathcal{S}$ implies that $\{x_1,\cdots,x_m,y_1,\cdots,y_m\}\subseteq\mathcal{S}$, but since $\mathcal{S}$ is a subspace we see that any linear combination of elements of $\{x_1,\cdots,x_m,y_1,\cdots,y_n\}$ is contained in $S$, in particular by definition $v$ is a linear combination of such vectors and so $v\in\mathcal{S}$. The conclusion follows.

We now prove that

$\mathcal{M}+\mathcal{N}=\text{Span }\left(\mathcal{M}\cup\mathcal{N}\right)$

To see this, we first begin by remembering a definition, namely:

$\displaystyle \text{Span }\left(\mathcal{M}\cup\mathcal{N}\right)\overset{\text{de}\text{f.}}{=}\bigcap_{\underset{\mathcal{M}\cup\mathcal{N}\subseteq\mathcal{S}}{\mathcal{S}\sqsubseteq\mathcal{V}}}\mathcal{S}$

To see this let $v\in\mathcal{M}+\mathcal{N}$, to show that $v\in\text{Span }\left(\mathcal{M}\cup\mathcal{N}\right)$ it suffices to show that $v\in\mathcal{S}$ for a fixed but arbitrary subspace containing $\mathcal{M}\cup\mathcal{N}$, but this is immediate since $v=x+y,\text{ }x,y\in\mathcal{M}\cup\mathcal{N}$ and so $x,y\in\mathcal{S}$, but since it’s a subspace $x+y=v\in\mathcal{S}$

Conversely, we merely note that $\mathcal{M}+\mathcal{N}$ is a subspace which contains $\mathcal{M}+\mathcal{N}$ and so by definition

$\text{Span }\left(\mathcal{M}\cup\mathcal{N}\right)\subseteq\mathcal{M}+\mathcal{N}$

and so the conclusion follows. $\blacksquare$

Lemma: Let $\mathcal{M},\mathcal{N}$ be two subspace of $\mathcal{V}$ (over $F$) with

$\dim_F\left(\mathcal{M}\right)=m\text{ and }\dim_F\left(\mathcal{N}\right)=n$

then if

$\mathcal{M}\cap\mathcal{N}=\{\bold{0}\}$

then $\mathcal{M}+\mathcal{N}$ is a subspace and

$\dim_F\left(\mathcal{M}+\mathcal{N}\right)=m+n$

Proof: The fact that it’s a subspace is elementary, and clear thus it suffices to prove the second claim. To do this let $\{x_1,\cdots,x_m\}$ and $\{y_1,\cdots,y_n\}$ be bases for $\mathcal{M}$ and $\mathcal{N}$ respectively, we claim that $\{x_1,\cdots,x_m,y_1,\cdots,y_n\}$ is a base for $\mathcal{M}+\mathcal{N}$. To do this suppose that

$\displaystyle \sum_{j\leqslant m}\alpha_j x_j+\sum_{j\leqslant n}\beta_j y_j=\bold{0}$

for some $\alpha_1,\cdots,\alpha_m,\beta_1,\cdots,\beta_n\in F$, then

$\displaystyle \sum_{j\leqslant m}\alpha_j x_j=\sum_{j\leqslant n}\left(-\beta_j\right)y_j$

but the LHS is an element of $\mathcal{M}$ and the RHS an element of $\mathcal{N}$ and so by assumption

$\displaystyle \sum_{j\leqslant m}\alpha_j x_j=\sum_{j\leqslant n}\left(-\beta_j\right)y_j=\bold{0}$

and so by the l.i. of $\{x_1,\cdots,x_m\}$ and $\{y_1,\cdots,y_n\}$ we see that

$\alpha_1=\cdots=\alpha_m=\beta_1=\cdots=\beta_n=\bold{0}$

and so $\{x_1,\cdots,x_m,y_1,\cdots,y_n\}$ is l.i. Next, we note from the above that

$\mathcal{M}+\mathcal{N}=\text{Span }\{x_1,\cdots,x_m,y_1,\cdots,y_n\}$

and thus it follows that $\{x_1,\cdots,x_m,y_1,\cdots,y_n\}$ is a basis for $\mathcal{M}+\mathcal{N}$ and so in particular

$\dim_F\left(\mathcal{M}+\mathcal{N}\right)=\text{card }\{x_1,\cdots,x_m,y_1+\cdots,y_n\}=m+n$

thus the lemma is proven. $\blacksquare$

Corollary: Given an $\ell$-dimensional vector space $\mathcal{V}$ and $m,n$-dimensional vector spaces $\mathcal{M}$ and $\mathcal{N}$ such that $m+n>\ell$ then $\mathcal{M}\cap\mathcal{N}\supsetneq\{\bold{0}\}$

b) Let $\{x_1,\cdots,x_m\}$ be a basis for $\mathcal{M}\cap\mathcal{N}$, then we may extend it to a basis $\{x_1,\cdots,x_m,y_1,\cdots,y_k\}$ (where $x_i\ne y_j,\text{ }\forall i,j$)  for $\mathcal{M}$ and a basis $\{x_1,\cdots,x_m,z_1,\cdots,z_\ell\}$( same condition) for $\mathcal{N}$. We claim that $\{x_1,\cdots,x_m,y_1,\cdots,y_k,z_1,\cdots,z_\ell\}$ is a basis for $\mathcal{M}+\mathcal{N}$. It is evident that

$\text{Span }\{x_1,\cdots,x_m,y_1,\cdots,y_k,z_1,\cdots,z_\ell\}=\mathcal{M}+\mathcal{N}$

and so it suffices to prove l.i. Suppose that

$\displaystyle \sum_{j\leqslant m}\alpha_j x_j+\sum_{j\leqslant k}\beta_j y_j+\sum_{j\leqslant \ell}\gamma_j z_j=\bold{0}$

where not all the coefficients are zero. Now, clearly we must have that $\beta_j\ne 0$ and $\gamma_i\ne 0$ for some $i,j$ since $\{x_1,\cdots,x_m,y_1,\cdots,y_k\}$ and $\{x_1,\cdots,x_m,z_1,\cdots,z_\ell\}$  are l.i. Thus

$\displaystyle \sum_{j\leqslant m}\alpha_j x_m+\sum_{j\leqslant \ell}\gamma_j z_j=\sum_{j\leqslant k}\beta_j y_j$

and so there is some non-zero (considering what we just said) vector

$v\in\text{Span }\{y_1,\cdots,y_k\}\cap\text{Span }\{x_1,\cdots,x_m,z_1,\cdots,z_\ell\}$

But, this implies that $v\in \mathcal{M}\cap\mathcal{N}$ and so $v\in\text{Span }\{y_1,\cdots,y_k\}\cap\text{Span }\{x_1,\cdots,x_m\}$ and since the two sets are disjoint it follows that $v=\bold{0}$ which is a contradiction. It follows that $\{x_1,\cdots,x_m,y_1,\cdots,y_k,z_1,\cdots,z_\ell\}$ is a basis  for $\mathcal{M}+\mathcal{N}$. Note, that in particular this implies that $y_j\ne z_i,\text{ }\forall i,j$. So

$\text{card }\{x_1,\cdots,x_m,y_1,\cdots,y_k\}+\text{card }\{x_1,\cdots,x_m,z_1,\cdots,z_\ell\}=m+k+m+\ell$

(the last part using the disjointness of the sets), but this is equal to

$m+k+\ell+m=\text{card }\{x_1,\cdots,x_m,y_1,\cdots,y_k,z_1,\cdots,z_\ell\}+\text{card }\{x_1,\cdots,x_m\}$

But, noticing that $m=\dim\left(\mathcal{M}\cap\mathcal{N}\right)$, $m+k=\dim\mathcal{M}$, $m+\ell=\dim\mathcal{N}$, and $m+k+\ell=\dim\left(\mathcal{M}+\mathcal{N}\right)$ we may put this all together to get

$\dim\mathcal{M}+\dim\mathcal{N}=\dim\left(\mathcal{M}+\mathcal{N}\right)+\dim\left(\mathcal{M}\cap\mathcal{N}\right)$

as desired.