# Linear Algebra: Vector Subspaces problem

I am currently stuck on number 16:

So theorem 4.5 says:

If $W$ is a non empty subset of a vector space $V$, then $W$ is a subspace of $V$ if and only if the following closure conditions hold:

1. If $\mathbf{u}$ and $\mathbf{v}$ are in (subspace) $W$ then $\mathbf{u} + \mathbf{v}$ is in $W$.
2. If $\mathbf{u}$ is in $W$ and $c$ is any scalar, then $c\mathbf{u}$ is in $W$.

Questions:
1) What is defined as a closure?

2) I tried doing 16 by saying that A is the 2x2 matrix $\left(\begin{array}{cc}1&0\\0&0\end{array}\right)$, and that B is the 2x2 matrix $\left(\begin{array}{cc}0&0\\0&1\end{array}\right)$. Both of these are singular. So if you add the two together you get $\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$ which is nonsingular. Thus we can conclude that it is not closed under addition. So no.1 is not satisfied.

3) How do I know if condition 2 is satisfied? If I multiply some scalar, such as 2, by matrix A from 2), I would get $\left(\begin{array}{cc}2&0\\0&0\end{array}\right)$. Is this considered to be part of the subspace?

-
Are you stuck on 16, or on 15? You discuss 15. – Arturo Magidin Jun 23 '11 at 20:49
stuck on 16 ... my choice of matrices was arbitrary – chaver Jun 23 '11 at 20:55
@Virtuoso: But 16 is not about singular vs. nonsingular, it's about whether the matrix equals its square or not. Your $A$ and $B$, as well as your $A+B$, all satisfy being equal to their own squares. Also, keep in mind the two uses of "arbitrary"; you may have picked your $A$ and $B$ in an "arbitrary manner", but they are specific matrices in your set, not "arbitrary matrices". A single example does not suffice to show closure always is satisfied, though a single example does suffice to show it is not always satisfied. A single counter example is enough, but not a single example. – Arturo Magidin Jun 23 '11 at 21:03
So wait a second... so I have to prove that W is not a subspace of a vector space by showing that the condition placed in the problem (A^2=A) does not meet the two criteria that define a subspace (Theorem 4.5)... Is this the correct way to think about it? – chaver Jun 23 '11 at 21:08
@Virtuoso: You have ten different subsets $W$ in the extract you quoted; if you want to talk about only one of them, you should have just copied the definition of that one (it gets worse when you mention one problem, 16, and then start talking about a different one, 15; it's very confusing). You are not putting conditions on the "problem", you are putting conditions on the elements of the set. To show $W$ is not a subspace, you have to show that it fails to meet at least one of the conditions in Theorem 4.5: (cont) – Arturo Magidin Jun 23 '11 at 21:17

1. "Closure" in this context means that if you take things inside the set, and you perform the required operation (adding them, or scalar multiplication in this case), then the result will not fall "outside" the set, it will stay "inside"; it all happens inside a "closed environment", so to speak. It's not a something, but it's a property that the set may (or may fail to) have.

2. I think you mean 15, not 16. That is an example to show that in the case of $n=2$, the set in question is not "closed under vector sums" (your first "closure condition"), so that shows 15 is not a subspace. You may want to modify it slightly to show this works for any $n\gt 1$, not just $n=2$; (and for bonus points, figure out why, in the case of $n=1$, the set of singular matrices is a subspace.

3. If your set consists of all singular $2\times 2$ matrices, then the matrix $$\left(\begin{array}{cc} 2 & 0\\ 0 & 0 \end{array}\right)$$ is in the set if and only if it is a singular $2\times 2$ matrix. Is it a singular $2\times 2$ matrix? It's not whether it is or it is not "considered part of the subspace", it's whether it is or is not in the set. That is an objective condition (remember, the bouncer at the door only lets people in the set if they are members, and it's not a matter of "consideration"; either the object is in the set, or not).

(As it happens, the set of singular $n\times n$ matrices is closed under scalar multiplication; that is, the second of your "closure conditions" does hold; if you know that a square matrix is singular if and only if its determinant is $0$, you can prove this using some simple properties of the determinant).

You say you are stuck on 16, the set of all $n\times n$ matrices $A$ such that $A^2=A$.

A very easy matrix with this property is the identity matrix. What happens if you multiply the identity matrix by a scalar? Will it still have the property? Why or why not? If not, give a specific example. What if you add the identity m atrix to itself; will the resulting matrix will be equal to its own square?

-
so what is "inside the set" described as? if I use matrix A, and I add it to itself... is this outiside the set since its not equal to A ? (or 2=/=1 ?) Who is the set defined by? is it just arbitrary like I just explained with matrix A ? – chaver Jun 23 '11 at 21:11
The "set" in problem 16 is "All $n\times n$ matrices that are equal to their own square". If $A$ is a matrix, then it is "inside the set" if and only if (i) it is $n\times n$; and (ii) $A^2$ is equal to $A$. "Inside the set" is not an absolute term, it depends on what set you are considering. – Arturo Magidin Jun 23 '11 at 21:14
Ok, so say I add the identity matrix to itself, and I get [2,0;0,2], and here I am testing first condition, do I say that it does not pass the first condition because my result squared is not equal to [2,0;0,2] ? in this case it also does not pass the second condition since if you multiply scalar 2 (for example) to the identity matrix you get the same thing?... just to be clear, you can say that W is defined to be all n x n matrices that have A^2 = A.. correct? – chaver Jun 24 '11 at 2:08
@Virtuoso: In this case, you are looking at the $n=2$ case only. Your $W$ is the collection/set of all $2\times 2$ matrices which, when multiplied by themselves, equal themselves ($A^2=A$ is true). The identity is in this set; but, as you note, if you add the identity to itself, it's not in $W$ (because it does not satisfy the membership requirement). So $W$ fails to satisfy the first closure condition. And likewise, taking the identity, which is in $W$, and multiplying it by $2$ gives you a matrix which is not in $W$, so $W$ also fails to satisfy the second closure condition. – Arturo Magidin Jun 24 '11 at 3:39
@Virtuoso: That's essentially what you said, just cast in more standard, unambiguous mathematical language. Now, you should also verify that the corresponding set fails to be a subspace no matter what $n$ is (you just did it for $n=2$, it should be done for any $n$, not just $n=2$). – Arturo Magidin Jun 24 '11 at 3:40