Vector Spaces: Finding a basis and Dimension

I could really use some step-by-step help on these two problems please. Thank You in advance.

1.) Let $V = \{{\bf{A|A}}$ is an $n \times n$ matrix, $n$ fixed, det$({\bf{A}}) = 0$ }. Is $V$, with the usual addition and

scalar multiplication, a vector space? Give reason. If yes, find the dimension and a basis for $V$.

2.) Let $V = \{f(x)|f(x) = (ax + b)e^{-x},\; a,b\; \in\; \mathbb{R}\}$. Is $V$, with the usual addition and scalar

multiplication, a vector space? Give reason. If yes, find the dimension and basis for $V$.

-

(1) Is the set $\mathbf{V}=\{A\mid A\text{ is an }n\times n\text{ matrix and }\det(A)=0\}$ a vector space, under the usual addition and scalar multiplication of matrices?
Since the set of all $n\times n$ matrices is a vector space, the question is really whether this is a subspace (all the axioms of a vector space will necessarily hold, except perhaps for the existence of a zero vector, the existence of inverses, and the "hidden" axioms that the set must be closed under vector addition and scalar multiplication: the sum of two vectors in $\mathbf{V}$ must lie in $\mathbf{V}$, and every scalar multiple of a vector in $\mathbf{V}$ lies in $\mathbf{V}). Scalar multiplication is easy: if$\alpha\in\mathbb{R}$and$A$is any$n\times n$matrix, then we know that$\det(\alpha A) = \alpha^n \det(A)$. So$\det(\alpha A)=0$if and only if$\alpha = 0$or$\det(A)=0$. So, if$A\in\mathbf{V}$, then$\alpha A\in\mathbf{V}$for all scalars$\alpha$. What about vector addition? We would need to show that if$A$and$B$both have nonzero determinant, then so does$A+B$. But this is not the case, as AgustÃ­ Roig points out. It should be easy to come up with a similar example for any$n\gt 0$. So$\mathbf{V}$is not closed under addition (exhibit an explicit pair of matrices, both in$\mathbf{V}$, but whose sum is not in$\mathbf{V}$; sometimes the sum of two matrices in$\mathbf{V}$is in$\mathbf{V}$, the point is that it doesn't always lie in$\mathbf{V}$, so give an example!). (2). Again, since all real valued functions form a vector space, the only issue is whether the set$\mathbf{V} = \{ f(x)\mid f(x) = (ax+b)e^{-x},\ a,b\in\mathbb{R}\}$is closed under sums and scalar multiplication. Suppose$f(x),g(x)\in\mathbf{V}$. Will$f(x)+g(x)$lie in$\mathbf{V}as well? Write \begin{align*} f(x) &= (ax+b)e^{-x}\\ g(x) &= (cx+d)e^{-x}\\ f(x)+g(x) &= \Bigl( (ax+b)e^{-x}\Bigr) + \Bigl( (cx+d)e^{-x}\Bigr)\\ &= \Bigl( (ax+b) + (cx+d)\Bigr) e^{-x}\\ &= \Bigl( (a+c)x + (b+d)\Bigr) e^{-x}. \end{align*} So settingA=a+c$and$B=b+d$, which are real numbers because all of$a,b,c,d$are real numbers, then we see that we can write$f(x)+g(x)$in the form$(Ax+B)e^{-x}$. So if each of$f(x)$and$g(x)$are in$\mathbf{V}$, then$f(x)+g(x)\in\mathbf{V}$. Now you need to show that if$f(x)\in\mathbf{V}$and$\alpha\in\mathbb{R}$, then$\alpha f(x)\in\mathbf{V}$. I'll leave that to you to do. What about the dimension of$\mathbf{V}$? If you want to describe an element of$\mathbf{V}$, you really only need to specify two things: the value of$a$and the value of$b$. that suggests that the dimension will be$2$. Can you find a set of$2$linearly independent functions, both in$\mathbf{V}$, that span$\mathbf{V}$? - The first one is not a vector space as sum of two singular matrix may be nonsingular. For example $$\begin{pmatrix} 1 & 0 \\\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\\ 0 & 1 \end{pmatrix}$$ The second one is a vector space of dimension 2 as$xe^{-x}$and$e^{-x}$are linearly independent continuas functions. If$axe^{-x} + be^{-x} = 0$for$a,b \in \mathbb{R}$, Then$ax +b = 0$as a continuas function on$\mathbb{R}$. Putting$x = 0,1$we have$b = 0$and$a +b = 0$. Hence$a = b = 0$. - Actually, what you need in the first example is to show that the sum of two singular matrices may be non singular. – Raskolnikov Mar 4 '11 at 13:28 @Christopher: The exact same way: find two singular matrices whose sum is *non*singular. – Arturo Magidin Mar 4 '11 at 14:18 @Christopher:I am extremely sorry for not reading the first part of the question carefully. Agusti Roig is correct. I have first canceled$e^{-x}$as it is always a non zero continuous function. Then I conclude that$ax +b$is a everywhere zero continuas function. – Anjan Gupta Mar 4 '11 at 14:37 I downvoted your otherwise fine answer because the very first sentence is not the proof of the statement the OP was looking for, something that Raskolnikov also noticed. If you edit your answer, I'll gladly remove my downvote. – Billare Mar 5 '11 at 1:00 I think that Anjan really wanted to say that the sum of two singular matrices may be non-singular. For instance: $$\begin{pmatrix} 1 & 0 \\\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\\ 0 & 1 \end{pmatrix}$$ -  @Christopher: No, that statement is irrelevant to problem (1). That statement gives that it is possible for two matrices not in$\mathbf{V}$whose sum is in$\mathbf{V}$. But that is irrelevant to whether$\mathbf{V}$is a vector space or not. (Analogy:$\frac{1}{2}+\frac{1}{2}$shows that it is possible for two non-integers to add up to an integer. But it does not tell us anything about whether the sum of two integers will be an integer or not.) The example$I_n + (-I_n) = 0\$ is useless for this problem. – Arturo Magidin Mar 4 '11 at 14:20