Why are these determinants $0$?

Question

These $3$ matrices below have determinants of $0$. Increasing each element by $1$ still results in a determinant of $0$:

\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix} \begin{bmatrix}2&3&4\\5&6&7\\8&9&10\end{bmatrix} \begin{bmatrix}3&4&5\\6&7&8\\9&10&11\end{bmatrix}

Also, if I square each element of the first matrix the determinant isn't $0$, it's $-216$, but if I continue this pattern to a $4 \times 4$ matrix the determinant then is $0$:

\begin{bmatrix}1&4&9\\16&25&36\\49&64&81\end{bmatrix} \begin{bmatrix}1&4&9&16\\25&36&49&64\\81&100&121&144\\169&196&225&256\end{bmatrix}

This pattern seems to work for matrices with dimensions $2$ higher than what the elements are raised to:

\begin{bmatrix}1&8&27&64&125\\216&343&512&729&1000\\1331&1728&2197&2744&3375\\4096&4913&5832&6859&8000\\9261&10648&12167&13824&15625\end{bmatrix} \begin{bmatrix}1&16&81&256&625&1296\\2401&4096&6561&10000&14641&20736\\28561&38416&50625&65536&83521&104976\\130321&160000&194481&234256&279841&331776\\390625&456976&531441&614656&707281&810000\\923521&1048576&1185921&1336336&1500625&1679616\end{bmatrix}

Can someone please explain why this is?

Answer 1

Let $A,B,C,D$ be rows of the matrices. For first three you find: $C-2B+A=0$. For the "squared" matrices $D-3C+3B-A=0$. Can you take it from here?

Answer 2

Your matrices before you start raising the entries to powers have the particular property that there are constants $a, b, c$ such that the element at position $i,j$ has the value $a+bi+cj$.

(For your particular $n\times n$ matrices you have $a=-n$, $b=n$, $c=1$, but these precise values are not important).

A typical column of one of the matrices therefore looks like $$ \begin{pmatrix} k \\ k+b \\ k+2b \\ \vdots \\ k+(n-1)b \end{pmatrix} $$ where $k$ is different for each column while $b$ is always the same.

When you raise the elements to the $(n-2)$th power, you get columns of the form $$ \begin{pmatrix} k^{n-2} \\ (k+b)^{n-2} \\ (k+2b)^{n-2} \\ \vdots \\ (k+(n-1)b)^{n-2} \end{pmatrix} $$ What is important here is that each entry in the column is a polynomial in $k$ of degree $n-2$, and the same $n$ polynomials generate all the columns of the matrix.

Since polynomials of degree at most $n-2$ constitute a real vector space of dimension $n-1$, there must be a linear relation between these polynomials, and this linear relation becomes a linear relation between the rows of the element-by-element exponentiated matrix. Therefore the matrix is singular and its determinant is $0$.

If you raise to the $(n-1)$th power, however, the relevant vector space of polynomials has dimension $n$, and there is no longer any reason to expect that the $n$ polynomials that make up a column are linearly dependent. This removes the reason why the matrix would be singular, and in general it isn't.

Answer 3

What @user is getting at in their answer is the fact that the determinant is an alternating multilinear function of the rows or columns of a matrix. That is to say, we can think of determinant as function like this: $$ \det \left( \begin{bmatrix} a_{11} \\ a_{21} \\ a_{31} \end{bmatrix}, \begin{bmatrix} a_{12} \\ a_{22} \\ a_{23} \end{bmatrix}, \begin{bmatrix} a_{13} \\ a_{23} \\ a_{33} \end{bmatrix} \right) = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{bmatrix} $$ And then if $v_1, v_2, v_3, v_4$ are vectors, and $\lambda$ is number (scalar) we have $$ \det(v_1 + v_4, v_2, v_3) = \det (v_1, v_2, v_3) + \det(v_4, v_2, v_3) \\ \det(\lambda v_1, v_2, v_3) = \lambda \det(v_1, v_2, v_3) $$ We can do this splitting in any argument of the function. This is what I mean when I said determinant is "multilinear." It preserves the linear operations of + and scalar multiplication. Also, we have $$ \det(v_1, v_1, v_2) = \det(v_3, v_2, v_3) = \det(v_1, v_2, v_2) = 0 $$ This is what I mean when I say the determinant is "alternating." If any two rows are the same, then $\det = 0$. Combining these two properties, if any column is a linear combination of the other columns, then the determinant will be zero.

In all of your matrices, one column can be written as a linear combination of the others, so the determinant is zero.

Answer 4

The forward difference operator $\delta:p(x)\mapsto p(x)-p(x+1)$ maps any polynomial with degree $d\geq 1$ into a polynomial with degree $d-1$. In particular, if we apply $\delta^n$ to a polynomial $p(x)$ with degree $d<n$ we get the constant $0$: $$ \sum_{k=0}^{n}\binom{n}{k}(-1)^k p(x+k) = 0. \tag{1} $$ Let us consider $(1)$ with $n=3$ and $p(x)=x^2$. It gives $$ x^2-3(x+1)^2+3(x+2)^2-(x+3)^2 = 0 \tag{2} $$ hence $(1,-3,3,-1)^T$ belongs to the kernel of the linear map associated to the $4\times 4$ matrix $M$ such that $M_{i,j}=(1+4i+j)^2$. In particular $\det M=0$. Similarly, if $M$ is the $6\times 6$ matrix such that $M_{i,j}=(1+6i+j)^4$ we have $(1,-5,10,-10,5,-1)^T\in\ker M$ and $\det M=0$.

This is just a generalization of the fact that $$ \det\left[\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}+x\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}\right]=0 $$ since the middle column is the average of the adjacent columns.

Answer 5

Let

$$\mathrm M := \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9\end{bmatrix}$$

whose determinant is $0$ and whose adjugate is the following symmetric matrix

$$\mbox{adj} (\mathrm M) = \begin{bmatrix} -3 & 6 & -3\\ 6 & -12 & 6\\-3 & 6 & -3\end{bmatrix}$$

whose rows and columns add to zero. Using the matrix determinant lemma, for all $\gamma \in \mathbb R$

$$\det \left( \mathrm M + \gamma \, 1_3 1_3^\top \right) = \det (\mathrm M) + \gamma 1_3^\top \mbox{adj} (\mathrm M) \, 1_3 = 0 + \gamma \cdot 0 = 0$$