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This answer makes sense to me except for the last part. Could anyone provide a simple example for me?

For example I'm thinking c0=1,c1=2,c2=2,and c3=3

So the polynomial formed would be f(t) = (t-1)(t-3) = t^2-4t+3

But how exactly does f(ci) = 0 for those? wouldn't it not work for the non-distinct numbers? Thanks!

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    You want $f(t)=(t-c_0)(t-c_1)(t-c_3)$. Note that only $i=q$ is excluded, not $i=p$ – 2017-02-26
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    If $c_p = c_q$ for some $p \neq q$ then you can take $f$ to be the polynomial $f(t) = \prod_{i \neq q} (t - c_i) = (t - c_0)(t - c_1) \dots (t - c_{q - 1})(t - c_{q + 1})\dots(t - c_n)$ and then $f$ will be a polynomial of degree $n$ (because you multiply only $n$ linear polynomials, not $n + 1$!) and $T(f) = 0$. – 2017-02-26

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Well, for your concrete example of $c_i$ we can take $$f(t) = (t - 1)(t - 2)(t - 3) = (t^2 - 3t + 2)(t - 3) = t^3 - 6t^2 - 7t - 6. $$

Then

$$ T(f) = \begin{pmatrix} f(1) \\ f(2) \\ f(2) \\ f(3) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}. $$

The polynomial $f$ is a polynomial of degree $3$ which is non-zero but $T(f) = 0$.