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 $V$.
 A: 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 $a = b = 0$.
A: Okay, this got a bit mangled. 
(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 Agustí 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 setting $A=a+c$ 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 $\mathbf{V}$?
A: I think that Anjan really wanted to say that the sum of two singular matrices may be non-singular. For instance:
$$
\begin{pmatrix}
1  &  0  \\\
0  &  0
\end{pmatrix}
+
\begin{pmatrix}
0  &  0  \\\
0  &  1
\end{pmatrix}
=
\begin{pmatrix}
1  &  0  \\\
0  &  1
\end{pmatrix}
$$
