Suppose $A$ is nilpotent, and $B = c_0I_n + c_1A + \cdots + c_{m-1}A^{m-1}$, show $\det(B) = 0$ iff $c_0 = 0$ Let $A$ be an $n\times n$ matrix, with real entries. If $A$ is nilpotent, so $A^m = 0$ for some $m \ge 1$, $B = c_0I_n + c_1A + \cdots + c_{m-1}A^{m-1}$, where $c_i \in R$ for each $i$, show that $\det(B) = 0$ if and only if $c_0 = 0$.
My attempt: 
So, $c_0 = 0$ implies $\text{det}(B) = 0$ is obvious: if $c_0 = 0$, then $\det(B) = \det((c_1 + c_2A + \cdots + c_{m-1}A^{m-2})A) = \det(c_1 + c_2A + \cdots + c_{m-1}A^{m-2})\cdot \det(A) = 0,$ since $\det(A) = 0$ if $A$ is nilpotent.
I'm not sure how to tackle the reverse direction. If $\text{det}(B) = 0$, we know: $B$ is not invertible, we know that $\det(c_0I_n + c_1A + \cdots + c_{m-1}A^{m-1}) = 0$. But I don't know where to go from here! A hint in the right direction would be appreciated.
 A: If $N$ is nilpotent of order $k$, then $I-N$ is invertible because
$$
    (I-N)(I+N+N^{2}+\cdots+N^{k-1})=I-N^{k}=I.
$$
Now suppose $A$ is nilpotent and further suppose $c_0 \ne 0$. Then
$$
         c_0 I + c_1 A + c_2 A^{2}+\cdots + c_{m-1}A^{m-1} = c_0(I-N),
$$
where $N$ is nilpotent. Hence, the above is invertible if $c_0 \ne 0$. However, the above is definitely not invertible if $c_0=0$ because, in that case, the above is nilpotent.
A: If the vector space is $R^n$ extend it to $C^n$  . We still have $A^m=0$ because any  $v \in C^n$ is equal to $x +i y$ where $x,y \in R^n$, and $A^m z=A^m x +i A^m y=0$.A crucial point is that $$ A \text { has no non-zero eigenvalues}$$ because $A z =k z$ with $k\ne 0$ and $z\ne 0$ implies $A^m z =k^m z \ne 0$.  Consider the polynomial $ p(v)=\sum_{j=0}^{j=m-1} c_j v^j$ for $v \in C$ with  $c_0\ne 0$ in the case where $ d=\deg p >1$. We have $p(v)= c_0 \prod_{j=1}^{j=d}(v-v_j)$ for all $v \in C $ for some non-zero $ v_1,\ldots,v_d \in C$...$$\text{ Assume  } \det (B)=\det( p(A))=0.$$ Then $c_0^{-1} p(A) z =0$ for some non-zero vector $z$, so let $d_0$ be the least $k$ with $1\le k \le d$ such that $\prod_{j=1}^{j=k}(A-v_j) z =0.$ We have $d_0>1$ ,else $A z=v_1 z$ (which makes  $v_1$ a non-$0$ eigenvalue of $ A$.)  But if $d_0>1$ , put $w=\prod_{j=1}^{j=d_o-1}(A-v_j) z$.  Then we have $w\ne 0$ and $(A-v_{d_0}) w =0$ (because all the terms $A-v_j$ commute with each other) which makes $v_{d_0}$ a non-zero eigenvalue of $A$.$$\text{ So we have a contradiction from assuming  } \det (B)=0.$$ I leave the case $\deg (p) <2$ up to you. Remark: I am using the usual notation $A-v=A-vI$ when $A$ is a linear operator and $v$ is a scalar.(So that the expressions $p(A)$ and $A-v_j$ make sense.) After seeing another answer I am reminded that I always seem to find the hardest proof.
A: Do the following:
Consider $B^n$ the n-th power of B, with $n\geq m$ 
$B^n = (c_0 I + c_1 A + \cdots + c_{m-1} A^{m-1}) \cdot \ldots \cdot (c_0 I + c_1 A + \cdots + c_{m-1} A^{m-1})$
$B^n = c_0^n I^n + \sum_{2n \geq i\geq n} \alpha_i A^i  = c_0^n I$
With $\sum_{2n \geq i\geq n} \alpha_i A^i$ I mean that $B^n$ is equal to $c_0^n I$ plus some powers of $A$ multiplied by an scalar but the exponent of those powers of $A$ is grater than $m$ therefore $A^i=0$ and $B^n = c_0^n I$.
Now use what you know $0 = det(B)^n = det(B^n) = det(c_0^n I) = c_0^k det(I) = c_0^k$ therefore $c_0 = 0$.
