Show the identity $\ker A\ne 0 \iff \det A = 0$ I don't understand the following line from the proof of eigenvalues and charasteristic polynomial:
$$\ker A\ne 0 \iff \det A = 0$$
Thank you
 A: Suppose $k \in \text{ker}(A), k \neq \vec{0}$. Then because $Ak=0$ , we consider the rows of the matrix A,$A_i$, and note that as vectors in $\mathbb{R}^n$, they are linearly dependent, because we have that $A_1k_1 + ... + A_nk_n=0$ for at least some non-negative integers $k_i$ (these are the entries of $k$). So $A_1 = \frac{-A_2k_2 - ... - A_nk_n}{k_1}$,assuming $k_1 \neq 0$ without loss of generality. In the matrix $A$, we perform the following row-change operation : from the first row, subtract $k_2$ times the second row, followed by $k_3$ times the third row, and so on so forth. Note that we'll eventually get the entire row as a zero row. Note that subtracting multiples of rows doesn't change the determinant,and the determinant of a matrix with a zero row is zero, so $det(A)=0$ . We'll be using this argument later, so note it down if you need.
For the other way ($det A = 0 \implies ker A \neq 0)$, we use the contrapositive way to prove this i.e. we assume that $ker A = 0$ and show that $det A \neq 0$. This will prove it.
Suppose $ker A = 0$. That means, by the rank nullity theorem (You surely must know this theorem) that the rank of $A$ is $n$, where $A$ has dimensions $n*n$. So A has linearly independent rows. Now, we triangularize the matrix in  a manner that does not change it's determinant: figure out for yourself how we make the a matrix of the form 
$$
\begin{pmatrix}
a\ b\ c \\
d\ e\ f \\
g\ h\ i \\
\end{pmatrix}
$$
into a matrix of the form:
$$
\begin{pmatrix}
j\ 0\ 0 \\
k\ l\ 0 \\
m\ n\ p \\
\end{pmatrix}
$$
where $a ... i$ are any real numbers, and $j ... p$ are obtained after triangularization. ( Just use row subtraction, and also generalize for $n*n$ matrices.)
I now claim that $j , l $ and $p$, which are the diagonal entries of the matrix, are not zero. Suppose, for example, $p=0$. Then the three rows of the matrix are all vectors in $R^2$, so they are linearly dependent (because they are three vectors in a space of dimension two,namely $\mathbb{R}^2$),which is a contradiction. If $l=0$, then the the first and second vectors are linearly dependent since they are two vectors in $\mathbb{R}$. Of course, if j=0, then the whole first row is zero, so the rows are linearly dependent,again a contradiction.
I'll leave you to generalize this to $n*n$ matrices. For our final step, note that the determinant of a triangular matrix is just the product of the diagonal entries! But none of the diagonal entries are zero, hence their product is non-zero, hence $det(A) \neq 0$!
Finally, we are done. This is a matter of lines if you have the necessary background, but a proof with elementary roots always appeals more to me, and it's better for people who need that kind of approach i.e. work from definition and use standard tricks. Please reply if any doubts.
