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I want to show that the determinant of the following matrix, $f(t)$, has a degree which is less or equal than $1$.

$$\left[ \begin{matrix} c_1-t & a-t & a-t & ... & a-t \\ b-t & c_2-t & a-t & ... & a-t \\ b-t & b-t & c_3-t & ... & a-t \\ \vdots & \vdots & \vdots &\ddots &\vdots \\ b-t & b-t & b-t & ... & c_n-t \\ \end{matrix} \right]$$

where $a,b,c_1,c_2, ... , c_n \in \mathbb{C} $

I've tried using induction on $n$ but this has led me nowhere. I'd appreciate some help.

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    $\begingroup$ Maybe start row-reducing first. Notice that row one minus row two will give you lots of zeros in the first row. $\endgroup$ Oct 9, 2016 at 17:45
  • $\begingroup$ Ok thank you. I think I have solved it now. If anyone wishes to see the solution just tell me to post it. $\endgroup$
    – user92596
    Oct 9, 2016 at 18:04

1 Answer 1

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The concrete way: Subtract row $1$ from all other rows. This will eliminate $t$ from all other rows. The claim now follows from symbolic determinant expansion along row $1$.

The general way: You could represent your Matrix as $$\begin{align} M &= S - tuv^\top & u^\top = v^\top &= \begin{bmatrix}1&1&\cdots&1\end{bmatrix} \\ S &= ((s_{ij})) & s_{ij} &= \begin{cases} a & \text{if $i<j$} \\ b & \text{if $i>j$} \\ c_i & \text{if $i=j$} \end{cases} \end{align}$$ and use the matrix determinant lemma which gives $$\det(M) = \det(S - tuv^\top) = \det(S) - t v^\top \operatorname{adj}(S)\,u$$ which proves your claim because $S,u,v$ are independent from $t$. Note that this works regardless of the concrete values in $S,u,v$. It suffices that $t$ brings at most a rank-one update to the matrix.

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