I have a $n \times n$ matrix $\mathbf{M}$ defined as, \begin{equation}(\mathbf{M})_{ij} = \begin{cases} {2i \choose j} \quad 1 \le j \le 2i, 1 \le i \le n \\ 0 \quad \mathrm{otherwise} \end{cases} \end{equation}
I want to prove that this matrix is invertible i.e. the determinant is non-zero. I tried various various by induction, and using some tricks in this question on Pascal matrices, but I was not able to reach a resolution.
What would be a good way to approach this?
There are a few approaches to solve this problem, but I'll give one fairly simple computational approach, and one proof by literature. There's probably a slick combinatorial way too, but I don't have the time to look into it right now. Hopefully someone else will be able to see one!
For the computational approach, recall we can compute the determinant by row reducing to an upper triangular matrix, then looking at the product of the diagonal entries.
But it's easy to see that we can just... do that. We can kill off the entries below the diagonal, by repeatedly subtracting off rows. Let's do a big enough example to showcase the pattern, say the $4 \times 4$ case.
$$ \begin{pmatrix} 2 & 1 & 0 & 0 \\ 4 & 6 & 4 & 1 \\ 6 & 15 & 20 & 15 \\ 8 & 28 & 56 & 70 \end{pmatrix} $$
First, we kill off column $1$ by subtracting multiples of row $1$.
$$ \begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 4 & 4 & 1 \\ 0 & 12 & 20 & 15 \\ 0 & 24 & 56 & 70 \end{pmatrix} $$
Next, we kill off column $2$ by subtracting multiples of row $2$.
$$ \begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 4 & 4 & 1 \\ 0 & 0 & 8 & 12 \\ 0 & 0 & 32 & 64 \end{pmatrix} $$
And column $3$, by subtracting multiples of row $3$.
$$ \begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 4 & 4 & 1 \\ 0 & 0 & 8 & 12 \\ 0 & 0 & 0 & 16 \end{pmatrix} $$
This is upper triangular, with nonzero entries on the diagonal. So its determinant (which equals the determinant of our original matrix, since we never had to swap or rescale rows) is nonzero (indeed, a power of $2$, as observed in the comments).
Since you say you are only looking for an approach, I'll stop here. There's a lot of very interesting questions left1, and I don't actually know the answer to all of them! Hopefully you'll find some joy in figuring the answers out for yourself ^_^.
1: Can we show by induction this always works? Will we ever have to rescale a row? Or will integer combinations prior rows always suffice? Why do we get powers of $2$ on the diagonal? Will we always?
I should say a brief word about why this approach does always work, though. In the interest of not spoiling any of the questions I outlined above, I'll cite a result which tells you directly that these matrices are always invertible. In Scott Kersey's Invertibility of Submatrices of the Pascal Matrix and Birkhoff Interpolation (available here), it is shown that a matrix formed by picking rows and columns of the infinite pascal matrix is invertible if and only if it has no $0$s on the diagonal. Of course, it's easy to see that your matrices will always satisfy this condition (indeed, the diagonal entries of your matrix are $\binom{2j}{j}$), so your matrices are invertible.
Kersey's paper is quite legible, and is absolutely worth reading if you're interested in these sorts of things!
I hope this helps ^_^
Let $M_{ij}={2i\choose j}$ and let ${\mathbf M}_n$ be the $ (n+1) \times (n+1)$ matrix with entries $M_{ij}$ for $i=0,\cdots, n$ and $j=0,\cdots, n$.
Notice that the matrix ${\mathbf M}_n$ includes the $0^{th}$ row and column, but removing these does not effect the determinant, as noted in comments.
We wish to show that: $$2^{n+1}{\rm Det}({\mathbf M}_{n})={\rm Det}({\mathbf M}_{n+1}),$$ as we may then conclude by induction that ${\rm Det}({\mathbf M}_{n})=2^{n(n+1)/2}$, as conjectured in comments.
The usual recurrence relation for Pascal's triangle, iterated twice, tells us that $$M_{(i+1)j}= M_{ij}+2M_{i(j-1)}+M_{i(j-2)}.\qquad[1]$$
Take the final column of ${\mathbf M}_n$, double it, and add the column to the left to it. Then take the second to last column, double it and add the column to the left of it. Continue doing this to each column in turn, working from right to left. When you get to the leftmost column, just double it. Call the resulting matrix $\mathbf{A}$.
Let $\vec{R}_i$ denote the row: $$(M_{i1} M_{i2}\cdots M_{i(n+1)})$$
Then from $[1]$ we know that $2M_{i(j-1)}+M_{i(j-2)}=M_{(i+1)j}- M_{ij}$, so $${\mathbf A}=\left( \begin{array}{ccc} -&\vec{R}_1-\vec{0}&-\\ -&\vec{R}_2-\vec{R}_1&-\\ -&\vdots&-\\ -&\vec{R}_{n+1}-\vec{R}_n&- \end{array} \right)$$
As we doubled $n+1$ columns, in the process of constructing ${\mathbf A}$ from ${\mathbf M}_n$, we know: $${\rm Det}({\mathbf A})=2^{n+1}{\rm Det}({\mathbf M}_{n}).$$
Now add the first row of ${\mathbf A}$ to the second, then add the new second row to the third and so on, to obtain:$$ \left( \begin{array}{ccc} -&\vec{R}_1&-\\ -&\vec{R}_2&-\\ -&\vdots&-\\ -&\vec{R}_{n+1}&- \end{array} \right)$$
Finally add a row of zeroes at the top, and a column of ones on the left, and we have ${\mathbf M}_{n+1}$.
We conclude:$$2^{n+1}{\rm Det}({\mathbf M}_{n})={\rm Det}({\mathbf A})={\rm Det}({\mathbf M}_{n+1}),$$ as required.
Note, the exact same argument shows that if we take rows of Pascal's triangle which are multiples of $k$, instead of even ones, the determinant of the resulting matrix is: $k^{n(n+1)/2}$.
Finally, here is an illustration of the argument for $n=2$:
$${\mathbf M}_2=\left(\begin{array}{ccc}1&0&0\\ 1&2&1\\ 1&4&6 \end{array}\right)$$
Doubling columns and adding column on left (right to left):$$ \left(\begin{array}{ccc}1&0&0\\ 1&2&1\\ 1&4&6 \end{array}\right) \to \left(\begin{array}{ccc}1&0&0\\ 1&2&4\\ 1&4&16 \end{array}\right) \to \left(\begin{array}{ccc} 1&1&0\\ 1&5&4\\ 1&9&16 \end{array}\right) \to \left(\begin{array}{ccc} 2&1&0\\ 2&5&4\\ 2&9&16 \end{array}\right) $$
Thus determinant doubled 3 times.
Adding each row to row below (top to bottom): $$ \left(\begin{array}{ccc} 2&1&0\\ 2&5&4\\ 2&9&16 \end{array}\right) \to \left(\begin{array}{ccc} 2&1&0\\ 4&6&4\\ 2&9&16 \end{array}\right) \to \left(\begin{array}{ccc} 2&1&0\\ 4&6&4\\ 6&15&20 \end{array}\right) $$ Adding row of zeroes to top and column of ones to left: $$ \left(\begin{array}{ccc} 2&1&0\\ 4&6&4\\ 6&15&20 \end{array}\right) \to \left(\begin{array}{ccc} 0&0&0\\ 2&1&0\\ 4&6&4\\ 6&15&20 \end{array}\right) \to \left(\begin{array}{cccc} 1&0&0&0\\ 1&2&1&0\\ 1&4&6&4\\ 1&6&15&20 \end{array}\right)={\mathbf M}_3 $$ So ${\rm Det}({\mathbf M}_3)=8{\rm Det}({\mathbf M}_2)$.
\mathbbfor individual items, but for large structures, like the set of integers, the reals, etc. You can use just $M,$ or $\mathbf M$ (\mathbf M) for a single matrix. – Thomas Andrews Jul 06 '21 at 00:04