conditions equivalent to singularity of matrix What conditions are equivalent to singularity of matrix $A\in \mathbb{R}^{n,n}$.
a. $\dim(ker A) \ge 0$
b. There is exist vector $b$ such that $Ax=b$ is contradictory.
c. $rank(A^T) < n$  
a. is true for each matrix, in other words $\dim$ can't be negative.
c. $rank(A^T) = rank(A)$.  Singularity means that some vector (row) is linearly dependent on other vectors.  Then, we may using elementary operations on rows, reset this row -> so it is true that $rank(A) < n$
Is it correct ?
When it comes to b. I suppose that it is true, however I can't prove it.  
 A: b. is equivalent to the fact that dimension of the kernel of the matrix is not null. By saying that there exists a vector b so that Ax=b is contradictory, it wants to say that there exists some vector b that belongs to the kernel of the matrix a. 
As that b would belong to the kernel of the matrix a, we could not generate it through the product Ax. For example, supose the next matrix $\left( \begin{array}{ccc}
1 & 3 \\
0 & 0  \\ \end{array} \right)$, whose range is obviusly spanned by $x^T=(1,0)$, and whose kernel is spanned by $x_k^T=(0,1)$. The n if we write the product $Ax=b$ in terms of the components of b, and take $b=x_k^T$, we see that:
$0=x_1+3x_2$
$k=0+0$
I am taking k instead of 1 to show that all the vectors from the kernel of A would not satisfy the equation. As you can see the second component of x leds to a contradiction as $k\neq 0,if k\neq0$, and k is not zero as we are assuming that vector b should be in the kernel of A.
Obviously yif the dimension of the kernel is null, there won't exist such a vector b.
A: a. Condition should be $dim \ker{A}>0$, as that implies that there is some column that is a linear combinación of the others. That would imply that the dimension of the range of A < n, implying that the rank of A is < n. Finally, that implies that the determinant of A is 0, so the Matrix is singular.
