Linear Algebra questions on terminology and concepts If my reduced row echelon of the matrix looks like
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 &0\\
0 & 1 & 0 & 0 &0 &0\\
0 & 0 & 0 & 0 &0 & 0\\
0 & 0 & 0 & 0 &0 & 0\\
0 & 0 & 0 & 0 &0 &0\end{pmatrix} 
then using the linear algebra terminology I would say that the dimension or the degrees of freedom is two correct? 
What is meant by $Ker(M) \neq 0$? I don't understand this and would need an example. Also if the $Ker(M) = 0$ is the matrix $M$ invertible? Does the converse of this statement hold? Please help I want to get this material to pass my linear algebra class
 A: You could say that the dimension or the degrees of freedom "is two" for this matrix, but you'd be being pretty vague. Here are some more precise things you could say:
The dimension of the column space is two, i.e. there are two linearly independent vectors which span the space spanned by the columns. Similarly for the row space.
The nullspace or kernel has dimension $3$. The kernel is the space that your matrix sends to $0$. You can see that it has dimension three in this case because
$$\left( \begin{matrix} 0\\0\\1\\0\\0\end{matrix}\right),\left( \begin{matrix} 0\\0\\0\\1\\0\end{matrix}\right),\left( \begin{matrix} 0\\0\\0\\0\\1\end{matrix}\right) $$
are all send to $0$, while all nonzero combinations of the other two basis vectors are not.
The image has dimension $2$. This is the dimension of the space your matrix maps onto. It's got the same dimension as the column space, because each column is the image of a basis vector.
A central fact in linear algebra is the rank-nullity theorem. This says that for any linear transformation $T$ from $V$ to $U$ the dimension of the kernel and the dimension of the image sum to the dimension of the domain, $V$. So if $V=U$, or more generally if they have the same dimension, then if $\ker T=0$ the rank-nullity theorem tells us the image of $T$ has the full dimension of $U$, which does mean that $T$ is invertible. 
However, if $U$ and $V$ have different dimensions, $T$ can have no kernel but $0$ and still not be invertible: consider the map from $\mathbb{R}^2\to\Bbb{R}^3$ that sends $\left(\begin{matrix}1\\0\end{matrix}\right)$ to $\left(\begin{matrix}1\\0\\0\end{matrix}\right)$ and
$\left(\begin{matrix}0\\1\end{matrix}\right)$ to 
$\left(\begin{matrix}0\\1\\0\end{matrix}\right)$. This isn't invertible, because for instance nothing maps to $\left(\begin{matrix}0\\0\\1\end{matrix}\right)$.
A: Let me address a few concepts here, my apologies that this answer got so long. It is not too clear what exactly your question is (since you seem to be confused on the terminology a bit) so hopefully I'll be able to address most of your issues.
First let me begin with the concept of a basis. A basis is a set of vectors in a vector space. A basis is defined as a set which is both linearly independent and spanning. Every element in the vector space can be written in a unique linear combination of a set of basis vectors.
Next, the concept of dimension in linear algebra is very specific. Formally, dimension is a number associated with a vector space. If $V$ is a vector space (if you are uncomfortable with arbitrary vector spaces,  then just replace all my $V$s with $\mathbb{R}^n$) then the dimension of $V$ is the number of vectors in a basis of $V$. It turns out that this definition is well defined since all basis of a vector space have the same size.
Very informally, dimension addresses the size (or "degree of freedom" if you will) of the vector space. In $\mathbb{R}^2$ we have the standard basis 
$$\left\{\begin{pmatrix}1 \\ 0 \end{pmatrix},\ \begin{pmatrix}0 \\ 1\end{pmatrix}\right\}$$
which spans the space. Any vector in $\mathbb{R}^2$ can be written uniquely as
$$a\begin{pmatrix}1 \\ 0 \end{pmatrix} + b\begin{pmatrix}0 \\ 1 \end{pmatrix}$$
for scalars $a,\ b\in\mathbb{R}$, so I guess you can say that the "degree of freedom" is $2$ since you may vary the two scalar coefficients $a$ and $b$ (be careful with this "degree of freedom" idea though since it is very ambiguous, it is useful as an intuitive aid but make sure you know the proper definition of dimension). Similarly, the dimension of $\mathbb{R}^3$ would be $3$ and in general $\dim\left(\mathbb{R}^n\right) = n$.
Finally, let me address your actual question. From the above, you can see that dimension is a term reserved for a vector space. It makes no real sense to ask about the dimension of a matrix. If you are referring to the height and width of the matrix, then I would simply talk about the size of the matrix. That being said, there are several very special and very important vector spaces which are associated with each matrix. The three you may have been exposed to are the rowspace, columnspace and the nullspace (or kernel). It turns out that the dimension of the columnspace and the rowspace are always the same and we give it the name rank. The rank of a matrix is the dimension of its rowspace and columnspace.
The dimension of the nullspace is called the nullity of the matrix. If you look at your matrix in reduced row echelon form and count the number of pivot entries (leading $1$s) then that gives you the rank of the matrix. The remaining non-pivot columns each correspond to a variable which is free to be adjusted. We call these free variables and each free variable gives one "degree of freedom" to the nullspace; more formally, the nullity of the matrix is the number of free variables. The matrix you posted has $2$ pivot columns (columns one and two) and $4$ non-pivot columns (columns three to six). That means it has rank $2$ and nullity $4$. I'm not sure if you know how to find the nullspace of the matrix by inspecting its reduced row echelon form, I will assume you do and move on. 
When we say the kernel is zero (or trivial), we really mean that the dimension of the kernel is $0$, i.e. it contains only the zero vector. From the above, this can happen if and only if each column of the matrix is a pivot column. If the matrix happens to be square (invertible is a term normally reserved only for square matrices, it's a bit tricky to talk about rectangular inverses), then a trivial kernel implies that the matrix is row equivalent to the identity matrix which of course means that it is invertible. The converse also holds. In fact, a square matrix has trivial kernel if and only if it is row equivalent to the identity matrix if and only if it is invertible. This is only three conditions of a long chain of equivalencies sometimes termed the invertible matrix theorem which you will eventually (or may have already) be exposed to.
