My linear book had an exercise that demonstrated that a nilpotent matrix A has det(A)=0
$A^k=0$ is the nilpotent condition. $det(A^k)=(det(A))^k$ and since $det(A^k)=0 \Leftrightarrow det(A)=0$.
My questions:
Is the revers relationship also true, i.e. if a determinant=0 is then the matrix nilpotent?
Does a nilpotent matrix A decreases the degree of dimensions ,by at least one , with every application on vector x i.e. space to plane to line to point?
All of your questions can be answered with the theory of Jordan Normal Form:
Every square matrix over an algebraically closed field (e.g. $\Bbb C$) is conjugate to a block-diagonal matrix where each block is of the form: \begin{bmatrix} \lambda & 1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda & 1 & 0 & \cdots & 0 \\ 0 & 0 & \lambda & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \\ 0 & 0 & 0 & 0 & \cdots & \lambda \end{bmatrix} i.e. $\lambda$ on the diagonal, and $1$ on the superdiagonal.
The total number of a specific $\lambda$ on the diagonal of the big matrix is the same as the multiplicity of $(x-\lambda)$ in the characteristic polynomial of the matrix.
Since each block is upper triangular, its powers have $\lambda^n$ on the diagonal; in particular:
Now the determinant is the product of all the eigenvalues, so:
Then one can now clearly see that $\det = 0$ does not imply nilpotent, simply by having a matrix with a $0$-block and a nonzero-block, such as the one suggested in the comment, i.e. $\begin{pmatrix}1&0\\0&0\end{pmatrix}$, which consists of a $1$-block of size $1$ and a $0$-block of size $1$.
As a sidenote, one can see that every nilpotent matrix is conjugate to a matrix with $0$ everywhere except $1$ on some superdiagonal entries.
Now if you observe a $0$-block $A$, and if you consider powers of $A$, it's basically shifting the diagonal upward until it disappears, i.e. $$\begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{pmatrix} \mapsto \begin{pmatrix}0&0&1&0\\0&0&0&1\\0&0&0&0\\0&0&0&0\end{pmatrix} \mapsto \begin{pmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix} \mapsto \begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$$
It so happens that these intermediate forms are all in Reduced Row Echelon Form, so you can read off the rank quite easily: in each block the rank decreases by $1$ each time, until it becomes $0$.
So if $A$ consists of a $0$-block of size 5 and a $0$-block of size 3 and nothing else, then the ranks of the powers of $A$ are $4+2$, $3+1$, $2+0$, $1+0$, and $0+0$.
From this example we can see that generally the rank of $A^n$ keeps decreasing until it reaches $0$, if $A$ is nilpotent.