If $a_{ij}=\max(i,j)$, calculate the determinant of $A$

Question

If $A$ is an $n \times n$ real matrix and

$$a_{ij}=\max(i,j)$$

for $i,j = 1,2,\dots,n$, calculate the determinant of $A$.

So, we know that

$$A=\begin{pmatrix} 1 & 2 & 3 & \dots & n\\ 2 & 2 & 3 & \dots & n\\ 3 & 3 & 3 & \dots & n\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ n& n & n & \dots & n \end{pmatrix}$$

but what do I do after?

Answer 1

Let $d_n$ be the determinant of the $n\times n$ matrix

We can also write it as a recurrence

By expanding on the last row (or column) we observe that all but the minors of last two columns have linear dependent columns, so we have:

$d_n=-\frac{n^2}{n-1}d_{n-1}+nd_{n-1}=-\frac{n}{n-1}d_{n-1}$

Coupled with $d_1=1$ we get $d_n=(-1)^{n-1}n$

Answer 2

You can also do $L_i-L_{i+1}$ which would give you a triangular matrix and the determinant is only the product of the coefficients of the diagonal

Answer 3

Use row operation to transform it to $$ \begin{pmatrix} 1 & 2 & 3 & ... &n \\ 1 & 0 & 0 & ... &0 \\ 3 & 3 & 3 & ... & n\\ ...&... &... &... &... \\ n& n & n & n & n \end{pmatrix}, $$ then to (determinant is multiplied by $-1$) $$ \begin{pmatrix} 1 & 0 & 0 & ... &0 \\ 0 & 2 & 3 & ... &n \\ 0 & 3 & 3 & ... & n\\ ...&... &... &... &... \\ 0& n & n & n & n \end{pmatrix}, $$ and repeat the process.

Thus the determinant should be $(-1)^{n-1}n$.

Hope this helps.

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It turns out this is just a reformulation of @G.Sassatelli's comment. Maybe I should delete it?

Answer 4

Hint: By induction on $n$. Subtract row $n-1$ from $n$, row $n-2$ from $n-3$ and so on. Then expand by cofactors along the last column.

Answer 5

Apply row operations $R_{j} \leftarrow R_{j} - R_{1}, \, \forall j \in \left \{2, ...n\right \}$ on $A$ to get the following matrix:

$A_1 = \begin{pmatrix} 1& 2 & 3 & ... & n-1 & n\\ 1& 0& 0 & ... & 0& 0\\ 2& 1& 0 & ... & 0& 0\\ ...& ...& ... & ... & ...& ...\\ n-2& n-3 & n-4 & ...& 0& 0\\ n-1& n-2& n-3& ...& 1& 0 \end{pmatrix}$

Now, apply the following row operations in the order of $j = 2,3,...,n-1$,

$R_{k} \leftarrow R_{k} - (k-1)R_{j}, \, \forall k \in \left \{j+1, j+2, ..., n\right \}$

to get the following matrix:

$A_2 = \begin{pmatrix} 1& 2 & 3 & ... & n-1 & n\\ 1& 0& 0 & ... & 0& 0\\ 0& 1& 0 & ... & 0& 0\\ ...& ...& ... & ... & ...& ...\\ 0& 0 & 0 & ...& 0& 0\\ 0& 0& 0& ...& 1& 0 \end{pmatrix}$

Now apply the following row operations $R_1 \leftarrow (j-1)R_j, \, \forall j \in {2,3,...,n}$ to get the following matrix:

$A_3 = \begin{pmatrix} 0& 0 & 0 & ... & 0 & n\\ 1& 0& 0 & ... & 0& 0\\ 0& 1& 0 & ... & 0& 0\\ ...& ...& ... & ... & ...& ...\\ 0& 0 & 0 & ...& 0& 0\\ 0& 0& 0& ...& 1& 0 \end{pmatrix}$

Note that $A_3$ has a value of $n$ at $(1,n)$th cell, all $1$s on the first lower off-diagonal and all other elements are $0$s. Now, compute the determinant along the first row to get:

$|A_3| = (-1)^{n+1}n$

Since, all the above row operations do not change the value of the determinant, we obtain,

$|A| = |A_1| = |A_2| = |A_3| = (-1)^{n+1}$

Answer 6

Hint to a more artsy proof:

Whenever $\mathcal{A}$ is a logical statement, we shall write $\left[\mathcal{A}\right]$ for the integer $\begin{cases} 1, & \text{ if } \mathcal{A} \text{ is true}; \\ 0, & \text{ if } \mathcal{A} \text{ is false} \end{cases}$. (This is called the Iverson bracket notation.)

Let $B$ be the $n\times n$-matrix whose $\left(i,j\right)$-th entry is $n \left[j=n\right] - \left[i \leq j < n\right]$. Here is how $B$ looks like: $$ B = \left( \begin{array}{ccccc} -1 & -1 & \cdots & -1 & n \\ 0 & -1 & \cdots & -1 & n \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & -1 & n \\ 0 & 0 & \cdots & 0 & n \end{array} \right) . $$ The matrix $B$ is upper-triangular, and so $\det B = n \left(-1\right)^{n-1}$.

Let $C$ be the $n \times n$-matrix whose $\left(i,j\right)$-th entry is $\left[i \geq j\right]$. Here is how $C$ looks like: $$ C = \left(\begin{array}{ccccc} 1 & 0 & \cdots & 0 & 0 \\ 1 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & 1 & \cdots & 1 & 0 \\ 1 & 1 & \cdots & 1 & 1 \end{array} \right) . $$ The matrix $C$ is lower-triangular, and so $\det C = 1$.

Now, check that $A = BC$, and conclude.