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

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:

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

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:

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

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?).

• I think $f$ restricted to $U_1\cap U_2$ is an isomorphism from $U_1\cap U_2$ to $f(U_1)\cap f(U_2)$, so that the two spaces have the same dimension? Commented Feb 24, 2020 at 15:53

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.

Hope this helps.

• You've proved that $g$ is an isomorphism of sets (i.e., a bijective function), but not that it's an isomorphism of vector spaces. (The latter's not hard at all, and follows from properties of $f$, but it still needs to be proved.) Commented Feb 24, 2020 at 16:10
• I was thinking that a bijective linear map is automatically an isomorphism. Sorry for being unclear. Commented Feb 24, 2020 at 16:43
• Oof that's a very nice proof. I hope I can get to a standard where I can write such nice proofs. Commented Feb 24, 2020 at 17:49

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.

• Oh shit that’s a nice proof. Nice nice Commented Feb 25, 2020 at 2:12