This was the starred problem in our linear algebra class.
So I understand that if a $k$-by-$k$ matrix has rank $1$, it means that all rows are linearly dependant on lets say the first one. So if I have a matrix, I can write it as:
$$ \begin{pmatrix} a_1 & c_1*a_1 & c_2*a_1 & \cdots& c_k*a_1 \\ a_2 & c_1*a_2 & c_2*a_2 & \cdots & c_k*a_2 \\ a_3 & c_1*a_3 & c_2*a_3 & \cdots & c_k*a_3 \\ \vdots& \vdots& \vdots& \ddots & \vdots\\ a_k & c_1*a_k & c_2*a_k & \cdots& c_k*a_k \\ \end{pmatrix} $$
So if I write A+I:
$$ \begin{pmatrix} a_1+1 & c_1*a_1 & c_2*a_1 & \cdots& c_k*a_1 \\ a_2 & c_1*a_2+1 & c_2*a_2 & \cdots & c_k*a_2 \\ a_3 & c_1*a_3 & c_2*a_3+1 & \cdots & c_k*a_3 \\ \vdots& \vdots& \vdots& \ddots & \vdots\\ a_k & c_1*a_k & c_2*a_k & \cdots& c_k*a_k+1 \\ \end{pmatrix} $$
And $\det(A+I)=a_1+1$, since I can subtract $c_n$ ($n=1,2,3,\dots k$) times the first column from all the other columns and I am left with a upper triangular matrix.
So I would get $a_1+1=1+\text{tr}(A)$ and $a_1=\text{tr}(A)$, but this can't be right? I can't figure out where I am going wrong. Any help?
Of course I read the other one, but fine, no help from here.
