Am I using quotient spaces correctly in this linear algebra proof?

[Eclair_de_XII]

Homework Statement

Let $X$ be a vector space and $Y$ a subspace with $\dim Y = \dim X$. Prove that $Y=X$.

Relevant Equations

Quotient space of X mod Y:

$X/Y=\{x_1,x_2\in X:\,x_1-x_2\in Y\}$

Equivalence class of x w.r.t. quotient space:

$\{x\}_Y=\{x_0\in X:\,x_0-x\in Y\}\subset X/Y$

Theorem:

$\dim Y + \dim {X/Y} = \dim X$

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Assume that $X/Y$ is defined. Since $\dim Y = \dim X$, it follows that $\dim {X/Y}=0$ and that $X/Y=\{0\}$.

Suppose that $Y$ is a proper subspace of $X$. Then there is an $x\in X$ such that $x\notin Y$.

Let us consider the equivalence class:

$\{x\}_Y=\{x_0\in X:\,x_0-x\in Y\}$

This is a subset of $X/Y=\{0\}$, which means that because $\{x\}_Y$ is a vector space, $\{0\}\subset \{x\}_Y$. Hence, $\{x\}_Y=\{0\}$.

This implies that $0-x\in Y$ and that $x\in Y$, since $Y$ is closed under scalar multiplication. This contradicts the fact that $x\notin Y$.

Hence, there cannot be an $x\in X$ that is not in $Y$. Thus, $X=Y$.

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I am a bit worried about the existence of $X/Y$. Is it defined for every vector space? I am also worried that I am misinterpreting the meaning of quotient spaces and equivalence classes.

 

[fresh_42]

It's a bit complicated but correct. You have a subspace $Y\subseteq X$. Then you can choose a basis $\{x_1,\ldots,x_m,x_{m+1},\ldots,x_n\}$ where $m=\dim Y$ and $n=\dim X$. With $m=n$ you get $X=Y$. But of course, this is proven along the lines of quotient spaces. As often, the essential point is: what are you allowed to use?

I am a bit worried about the existence of $X/Y.$ Is it defined for every vector space?

For every vector space $X$ and every subspace $Y$. It only requires that $Y$ is a linear space.

I am also worried that I am misinterpreting the meaning of quotient spaces and equivalence classes.

No. You were right. As mentioned above: the easy way with bases is proven by the fact - and now comes the real reason why your proof works: every short exact sequence $Y\rightarrowtail X \twoheadrightarrow X/Y$ in the category of vector spaces splits, i.e. there is a monomorphism $X/Y \rightarrowtail X$ such that the composition of these homomorphisms is the identity on $X/Y$.

This is not true in the category of groups, but it is for vector spaces. It simply means that $x_{m+1}+Y,\ldots,x_n+Y$ is a basis of $X/Y$ with the basis as described above.

 

[Eclair_de_XII]

As often, the essential point is: what are you allowed to use?

Everything before page twenty-five of Linear Algebra and its Applications by Peter Lax. And this includes that theorem I cited. And I did learn about linear independence, span, bases, sums of vector spaces, theorems pertaining to those sums, quotient spaces, and equivalence classes that partition those spaces. Oh, and that theorem I cited.

As mentioned above: the easy way with bases is proven by the fact

Does that proof go something like:

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Let $S_X=x_1,...,x_n$ be a basis for $X$ and $S_Y=y_1,..,y_n$ be a basis for $Y$.

Suppose $S_Y$ does not span $X$. Then there is $x_i$ that cannot be written as a linear combination of the vectors in $S_Y$.

Hence, the sequence $x_i,y_1,...,y_n$ is linearly independent.

Suppose it spans $X$. Then it is a basis of $X$ of length $n+1$. This contradicts the fact that every basis of $X$ must have exactly $n$ elements.

If it does not, continue checking the vectors of $S_X$ until we have a basis for $X$ of length at least $n+2$ and at most $2n>n$.

Either way, this is a contradiction. $S_Y$ must span $X$ as a result.

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