Proof verification: $\dim(U_1 \cap U_2) = \dim(f(U_1) \cap f(U_2))$

Question

Okay, so the following actually isn't a theorem asserted by the book I'm using. It's actually an example given to illustrate the importance of isomorphisms. It was a specific example with specific numbers but I want to try and prove that it holds in general. So, here's the assertion:

Let $f:V \to W$ be an isomorphism. Let $U_1$ and $U_2$ be subspaces of V. Then, the following holds:

$$\dim(U_1 \cap U_2) = \dim(f(U_1) \cap f(U_2))$$

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Proof Attempt:

Let $(v_1,v_2,....,v_n)$ be a basis for $U_1 \cap U_2$. Thus, that n-tuple is linearly independent and spans the intersection of the two subspaces.

By a previously proven result, if $(v_1,v_2,....,v_n)$ is a linearly independent list of vectors, then $\beta = (f(v_1),f(v_2),....,f(v_n))$ is a linearly independent list of vectors. To show that this list forms a basis for $f(U_1) \cap f(U_2)$, we just need to show that it generates that set.

Consider the following linear combination:

$\sum_{k=1}^{n} \alpha_k f(v_k)$

Due to the linearity of f, we can rewrite the above as follows:

$f[\sum_{k=1}^{n} \alpha_k v_k]$.

$L(\beta) \subset f(U_1) \cap f(U_2)$. So, we have to prove containment in the other direction. Let $w \in f(U_1) \cap f(U_2)$. Then, $w \in f(U_1)$ and $w \in f(U_2)$.

By the surjectivity of f, there exists a vector $u_1 \in U_1$ such that $f(u_1) = w$ and there exists a vector $u_2 \in U_2$ such that $f(u_2) = w$. By the injectivity of f, it's clear that $u_1 = u_2$.

That also means that $u_1 \in U_1 \cap U_2$ so it can be written as a linear combination of the vectors $(v_1,v_2,...,v_n)$. Therefore, $f(u_1)$ can be written as a linear combination of the vectors in $\beta$ and this proves that $f(U_1) \cap f(U_2) \subset L(\beta)$. This proves that $\beta$ generates $f(U_1) \cap f(U_2)$.

So, $\beta$ forms a basis for $f(U_1) \cap f(U_2)$ and that basis has length $n$. Similarly, the basis for $U_1 \cap U_2$ also has length n. Thus, we conclude that:

$\dim(U_1 \cap U_2) = \dim(f(U_1) \cap f(U_2))$.

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Once again, I need to know if my argument above is correct or not. I'm also pretty sure that there's a short and slick proof of this but I certainly can't see it (maybe with the use of set identities?).

Answer 1

Yes, I think your proof is correct. And in general you don't need the basis to prove this.

Claim:

$$g:U_1\cap U_2\rightarrow f(U_1)\cap f(U_2)\\x\mapsto f(x)$$ is an isomorphism, if $f:V\rightarrow W$ is an injective linear map.

Proof:

Suppose $g(x)=f(x)=0$ then by the injectivity of $f$, $x=0$, so $g$ is injective.

Take any $y\in f(U_1)\cap f(U_2)$. By the definition of $f(U_1)$ and $f(U_2)$ there exist $x_1\in U_1$ and $x_2\in U_2$ such that $y=f(x_1)=f(x_2)$. Since $f$ is injective, we see $x_1=x_2\in U_1\cap U_2$. So $y=g(x_1)$. Thus $g$ is surjective.

Hence $g$ is a bijective linear transformation.

By set-theory, we know $g$ has a unique inverse function $g^{-1}:f(U_1)\cap f(U_2)\rightarrow U_1\cap U_2$. Then, for any $x,y\in f(U_1)\cap f(U_2)$ and $f\in F$, where $F$ is the base field, we have $$ g(g^{-1}(fx+y))=fx+y=fg(g^{-1}(x))+g(g^{-1}(y))=g(fg^{-1}(x)+g^{-1}(y)). $$ Since $g$ is injective, this shows that $g^{-1}$ is linear. Therefore $g$ is a linear isomorphism.

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Hope this helps.

Answer 2

You may want to consider a geometric interpretation of the given result. Let us consider the planes $P_{xy}$ and $P_{xz}$ within $\textbf{R}^{3}$. Consequently, we have that $P_{xy}\cap P_{xz} = L_{x}$, where $L_{x}$ is the $x$-axis.

Since $f$ is an isomorphism, it is non-singular and, therefore, it takes basis onto basis. In other words, the sets $f(P_{xy})$ and $f(P_{xz})$ are also planes within $\textbf{R}^{3}$. Moreover, the proposed result states that $f(P_{xy})\cap f(P_{xz})$ is a line as well.

Here it is an alternative approach to solve the problem.

Let $\mathcal{B} = \{u_{1},u_{2},\ldots,u_{k}\}$ be a basis for $U_{1}\cap U_{2}$. Then we can extend it to a basis for $U_{1}$ as well as to a basis for $U_{2}$. Precisely, $\mathcal{B}_{1} = \{u_{1},u_{2},\ldots,u_{k},a_{k+1},\ldots,a_{m}\}$ and $\mathcal{B}_{2} = \{u_{1},u_{2},\ldots,u_{k},b_{k+1},\ldots,b_{n}\}$. Once $f$ is an isomorphism, it takes basis onto basis, from whence we conclude that \begin{align*} & f(\mathcal{B}_{1}) = \{f(u_{1}),f(u_{2}),\ldots,f(u_{k}),f(a_{k+1}),\ldots,f(a_{m})\}\quad\text{is a basis for $f(U_{1})$}\\\\ & f(\mathcal{B}_{2}) = \{f(u_{1}),f(u_{2}),\ldots,f(u_{k}),f(b_{k+1}),\ldots,f(b_{n})\}\quad\text{is a basis for $f(U_{2})$} \end{align*}

If $w\in f(U_{1})\cap f(U_{2})$, then $w\in f(U_{1})$ and $w\in f(U_{2})$. Hence we have that \begin{align*} w = & \sum_{i=1}^{k}\alpha_{i}f(u_{i}) + \sum_{i=k+1}^{m}\alpha_{i}f(a_{i}) = \sum_{i=1}^{k}\beta_{i}f(u_{i}) + \sum_{i=k+1}^{n}\beta_{i}f(b_{i}) \Longrightarrow\\\\ & \sum_{i=1}^{k}(\alpha_{i} - \beta_{i})f(u_{i}) + \sum_{i=k+1}^{m}\alpha_{i}f(a_{i}) - \sum_{i=k+1}^{n}\beta_{i}f(b_{i}) = 0 \Longrightarrow\\\\ & \alpha_{i} = \beta_{j} = 0\,\,\text{for}\,\,k+1\leq i \leq m\,\,\text{and}\,\,k+1\leq j \leq n. \end{align*}

In other words, $f(U_{1})\cap f(U_{2}) = \text{Span}\{f(u_{1}),f(u_{2}),\ldots,f(u_{k})\}$.

Since the set $\{f(u_{1}),f(u_{2}),\ldots,f(u_{k})\}$ is linear independent, we conclude that $\dim(U_{1}\cap U_{2}) = \dim(f(U_{1})\cap f(U_{2}))$, as desired.