Two subspaces of $\Bbb R^n$

Question

If $A$ and $B$ are subspaces of $\mathbb R^n$. Is it possible to find a basis for $\mathbb R^n$ that contains a basis for $A$ and $B$?

It has been suggested to me that we define a basis for $A\cap B$ and then use that to define basises $A$ and $B$. I would like to understand why this approach is taken and how this is used to answer the above question. Please do not skip any details, I want to fully understand this method.

The main crux of my question is don't quite know how to answer this question. I also don't understand why the above was suggested.

I don't just want the answer as this is of minimal use to me. I want to understand how the answer was derived and why the particular path was chosen. I want be able apply the knowledge in similar cases

Answer 1

it is most certainly possible...and the approach suggested is the way to go. Basically you have a subspace \begin{equation} A+B=\{a+b:a \in A \text{ and } b \in B\}\end{equation} known as the sum of of $A$ and $B$; this is the smallest subspace containing $A$ and $B$. Also, the subspace $A \cap B$ is a subspace. So we can define, or find a basis for $A \cap B$ - let such a basis be $\alpha$. Now $\alpha$ is linearly independent, and any linearly independent set of vectors can be extended to be a basis - so we can extend $\alpha$ by adding a set of vectors $\beta$ so that $\alpha \cup \beta$ is a basis for $A$. Similarly we can extend $\alpha$ to $\alpha \cup \gamma$, to be a basis for $B$.

So then $\alpha \cup \beta \cup \gamma$ is a basis for $A+B$ - this is what we wanted right - now we must just extend it to be a basis for $\mathbb{R}^n$. So again since $\alpha \cup \beta \cup \gamma$ is linearly independent we can extend it to be a basis for $\mathbb{R}^n$, say by adding the set of vectors $\delta$. Then the basis $\alpha \cup \beta \cup \gamma \cup \delta$ is a basis for $\mathbb{R}^n$ that contains a basis for $A$ and $B$.

Why use this method - well it starts with defining exactly what you want - a basis for $A$ and $B$, and then extending it to be a basis for $\mathbb{R}^n$ - there is a theorem guaranteeing that any linearly independent set can be extended to be a basis for some vector space - so we can make use of that theorem if we first find a smaller basis. It is much more difficult/maybe not possible if you had to first find a general basis for $\mathbb{R}^n$ and then work back to find smaller bases for the subspaces. Please let me know if this is not clear to you...

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It was asked in the comments below that I clarify that $\alpha \cup \beta \cup \gamma$ is indeed a basis. ok: First, linear independence: by construction $\alpha \cup \beta$ (1) and $\alpha \cup \gamma$ (2) are linearly independent sets. Now, also by construction, span($\beta)$ is the complement of $A \cap B$ in $A$, and span($\gamma)$ is the complement of $A \cap B$ in $B$, so that none of the vectors in $\beta$ is in span$(\gamma)$. Now if you have a linearly independent set such as $\gamma$, and you add a vector $v$ NOT in the span of the set to form a new set $\gamma \cup \{v\}$, then $\gamma \cup \{v\}$ is linearly independent. That is why the vectors in $\beta \cup \gamma$ (3) are linearly independent. Combining the (1), (2) and (3) above, we have that $\alpha \cup \beta \cup \gamma$ is a linearly independent set.

For the second part we must just prove that every vector in A+B is in the span of $\alpha \cup \beta \cup \gamma$ - this one can just do by writing any vector $a+b \in A+B$ as: express $a$ as a linear combination of the vectors in $\alpha \cup \beta$ and $b$ as a linear combination of the vectors in $\alpha \cup \gamma$ and add the two expressions.

Answer 2

Let $\{v_1,\dots,v_k\}$ be a basis of $A\cap B$; you can find vectors $a_1,\dots,a_r\in A$ such that $\{v_1,\dots,v_k,a_1,\dots,a_r\}$ is a basis for $A$; similarly, there are $b_1,\dots,b_s\in B$ such that $\{v_1,\dots,v_k,b_1,\dots,b_s\}$ is a basis for $B$.

Your task is to prove that $$ \{v_1,\dots,v_k,a_1,\dots,a_r,b_1,\dots,b_s\} $$ is linearly independent. If it fails to be a basis for $\mathbb{R}^n$, just extend it to a basis.

Note that this has Grassmann's formula as a consequence: $$ \dim (A+B)=\dim A+\dim B-\dim(A\cap B) $$

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Here's how to tackle linear independence. Suppose $$ \gamma_1v_1+\dots+\gamma_kv_k+ \alpha_1a_1+\dots+\alpha_ra_r+ \beta_1b_1+\dots+\beta_sb_s=0. $$ Then we can consider $$ v=\gamma_1v_1+\dots+\gamma_kv_k+ \alpha_1a_1+\dots+\alpha_ra_r=-(\beta_1b_1+\dots+\beta_sb_s). $$ By hypothesis, $v\in A\cap B$, so we have $$ v=\delta_1v_1+\dots+\delta_kv_k $$ and so $$ (\gamma_1-\delta_1)v_1+\dots+(\gamma_k-\delta_k)v_k+ \alpha_1a_1+\dots+\alpha_ra_r=0 $$ which implies \begin{gather} \gamma_1-\delta_1=0,\dots,\gamma_k-\delta_k=0,\\ \alpha_1=0,\dots,\alpha_r=0 \end{gather} by linear independence of $\{v_1,\dots,v_k,a_1,\dots,a_r\}$.

Therefore $$ \gamma_1v_1+\dots+\gamma_kv_k=-(\beta_1b_1+\dots+\beta_sb_s) $$ and, by linear independence of $\{v_1,\dots,v_k,b_1,\dots,b_s\}$, we get \begin{gather} \gamma_1=0,\dots,\gamma_k=0\\ \beta_1=0,\dots,\beta_s=0. \end{gather}

Answer 3

Here's a general approach to certain classes of basis problems in finite-dimensional spaces (e.g., $R^n$). You work with two main facts.

i. If $S = \{ v_1, \ldots, v_k\}$ is a dependent set of vectors, then (a) at least one of the $v$s, say $v_i$, can be written as a linear combination of the others, and (b) The span of $S$ and the span of $S\ \{v_i\}$ (i.e., $S$ with $v_i$ removed) are the same. In short: you can always shrink a dependent set without reducing its span.

ii. If $S = \{ v_1, \ldots, v_k\}$ is independent, and $u \notin span(S)$, then $S' = \{ u, v_1, \ldots, v_k\}$ is also independent, and $span(S)$ is a proper subset of $span(S)$.

These two principles are what @Christiaan has used in his answer, more or less. I'd advise one more thing: draw a picture. Here, draw a Venn diagram of $A$, $B$, $A \cap B$, and where the basis elements you're seeking will have to fall. Then you can start work.

I also advise thinking about 2 cases in 3-space: (i) $A$ is a line and $B$ a plane (both through the origin). What's the intersection? What's the basis? What if the plane $B$ contains the line $A$? (ii) $A$ and $B$ are both planes, but they intersect in a line. What's the intersection? What's the basis? And what would happen if $A$ and $B$ were the same plane?

You can often do this sort of reasoning with examples for problems like this, because there are really only a few possibilities: for linear subspaces, the only intersections will be other linear subspaces of the same or smaller dimensions. You don't have to worry about two planes intersecting in a circle, etc.