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2) Determine whether or not each given subset of $\mathbb R^4$ is a subspace of $\mathbb R^4$?

a) $(a - b, a, 1 - b, a + b)$ where $a,b$ are real numbers

b) $(c - 2b, 3a + c, b - a, 2c - b)$ where $a,b,c$ real numbers

c) $(a, c, b, a^2)$ where $a,b,c$ real numbers

d) $\operatorname{span} \{(1, 1, 0, 1)\} \cup \{1, 0, 1, 1\}$

What do I look for in these? I'm having trouble justifying

3) Define $ H =\{ (a_1 a_2 a_3 a_4) \in M_{2\times2} : 3a_1 + 7a_2 − 2a_3 − a_4 = 0\}$. Show $H$ is $a$ subspace of $M_{2\times2}$.

Not really sure how to show this

4) Suppose $A$ is a $3\times2$ matrix and $B$ is a $2\times3$ matrix. Show that $AB$ cannot be invertible. Might $BA$ be invertible?

I know $AB$ is not invertible, because if $A$ is an $m \times n$ matrix and $B$ is an $n \times m$ matrix where $n < m$, then $AB$ is not invertible. But, how do I argue that $BA$ might be invertible? Thanks guys, any help is appreciated <3

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For your first two questions, if the subset of a vector space contains the $0$ vector and is closed under addition and scalar multiplication, it's a subspace. – dmk May 16 '14 at 14:32
Thanks so would 1) c) not be a subspace because it is not closed under vector addition? And d) because it doesn't contain the zero vector? – Kyle May 16 '14 at 14:48
With c, right: In general, $a^2 + b^2 \neq (a + b)^2$. With d, is it possible to write $0$ as a linear combination of $\langle 1, 1, 0, 1 \rangle$ and $\langle 1, 0, 1, 1 \rangle$? – dmk May 16 '14 at 14:51
I was reading the last question (4) backwards. Anyway, the question sounds like it's asking for an existence proof, so perhaps an example would suffice? Have you tried dropping some easy (i.e. small) numbers into such matrices and multiplying them together? – dmk May 16 '14 at 14:58
for d) I thought it wasn't possible because it looks like you could never make components 2 and 3 nonzero, is this not right? For B it is basically asking show that AB is never invertible under these conditions, but BA might be – Kyle May 16 '14 at 15:02
up vote 0 down vote accepted

Kyle, I'll expand a bit on the last question I asked. In a way, though, you might have noticed that the question gave the answer. When I wrote, "A vector space $V = \text{span}(v)$," I gave away the fact that it was a vector space — which means it contains the $0$ vector. In any case, even if you didn't notice that (I didn't notice it when I wrote it!), you probably knew or assumed that something spanned by one vector does not contain only one vector; that would have been a boring question. So where do those other vectors come from? From the scalar multiples of the spanning vector. Since we're dealing with vectors over the real numbers, those scalars are all the real numbers (e.g., $0$) as well.

I just noticed, actually, that d is written in a rather strange way:

$$\text{span}\{(1,1,0,1) \}\cup \{1,0,1,1 \}$$

This would most likely be read as the union of the span of a vector $(1,1,0,1)$ with another vector $(1,0,1,1)$ — note we are not looking at the span of the second vector. (The English teacher in me says that the braces around the second group of numbers means it's a set of some sort, but that interpretation, while more precise, makes no sense in the context of $\mathbb{R}^4$ — or in any other context, as far as I know.) Usually when I've seen questions like this, they're written like so:

$$\text{span}\{(1,1,0,1), (1,0,1,1) \}$$

This means it's the span of two vectors. (It would mean the same thing as $\text{span}\{(1,1,0,1) \} \cup \text{span} \{ (1,0,1,1) \}$, but that's much wordier.) A vector $v$ in this subset would be of the form $v = a\cdot (1,1,0,1) + b \cdot (1,0,1,1)$, where $a,b$ are real numbers. Would such a subset form a subspace?

A vector $u$ in the earlier subset would be of the form $u = c\cdot (1,1,0,1) + d \cdot (1,0,1,1)$, where $c\in \mathbb{R}$ and $d \in \{0,1\}$. A subset containing vectors of this form would not be closed under addition or scalar multiplication (why?) and so would not be a subspace.


As for the last question: In my experience, questions that want you to "show that" when something is always true, it also happens to be true in this particular case (yes, that's supposed to sound stupid) give you a problem on a relatively small scale so that you can work through the nuts and bolts of it. In other words, they probably don't want you to quote a theorem :). mirgee handles it the way I would have, but if you haven't seen linear transformations yet, let

$$AB = \left[ \begin{matrix} a & b \\ c & d \\ e & f \\ \end{matrix} \right] \left[ \begin{matrix} g & h & i \\ j & k & l \\ \end{matrix} \right] $$

What's the determinant of $AB$? We hope, as we approach the bottom of the second page of arithmetic, that it's $0$! Yes, it's obnoxious. It's mindless, which means you have to be careful. But alas, that may be want they want.

As for $BA$, if you've ever heard of least squares, which is used in linear regression (finding a line that best approximates a set of data), one method employs multiplying a $2\times n$ matrix by an $n \times 2$ matrix precisely because it is (often) invertible.

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For your last question, consider the fact that we have $\dim B A=\dim B$ if $A$ is "onto". This is a simple consequence of the fact that since $A(\mathbb R^2) \subset \subset \mathbb R^3$, $B(A(\mathbb R^2)) \subset \subset B(\mathbb R^3)$, so $\dim BA \leq \dim B(\mathbb R^3)= \dim B$. If $A$ is "onto" there is obviously an equality.

In this case, the dimension of either $A$ and $B$ is at most $2$. For $AB$ to be invertible, $\dim AB = 3$, which is impossible, as $\dim AB \leq \min \{\dim A, \dim B\}$. But for $BA$ to be invertible, $\dim BA = 2$, which is possible. Just find any two such matrices with maximum possible rank...

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