Determinant problem

[geoffrey159]

Homework Statement

Given a matrix $A = (a_{ij})$, we define matrix $B = \begin{pmatrix} a_{11} & - a_{12} & a_{13} & \cdots \\ - a_{21} & a_{22} & -a_{23} & \cdots \\ a_{31} & - a_{32} & a_{33} & \cdots \\ \vdots & \vdots & \vdots & \vdots \end{pmatrix}$.
Another way to define $B$ is to say that $b_{ij} = a_{ij}$ if $i+j$ is even and $- a_{ij}$ otherwise.

The question is: show that $\det B = \det A$

I have my idea which I have written below, do you agree with my explanation? Thanks!

Homework Equations

The Attempt at a Solution

[/B]
By recursion over property ${\cal P}(n)$: the property is true for matrices of order $n$.
${\cal P}(2)$ is true, and we assume that that ${\cal P}(n-1)$ is true. Let us show that ${\cal P}(n)$ must be true.

Expanding the determinant according to the first line, I have:

$\det B = \sum_k b_{1k} \text{ cof}(b_{1k}) = \sum_{k \text{ even}} -a_{1k} \text{ cof}(b_{1k}) + \sum_{k \text{ odd}} a_{1k} \text{ cof}(b_{1k})$

But no matter the parity of $k$, if I reverse the sign of the $k-1$ first columns in the determinant hidden inside $\text{ cof}(b_{1k})$, which is of order $n-1$, and then use the recursion hypothesis, I have:

$\text{ cof}(b_{1k}) = (-1)^{k-1} \text{ cof}(a_{1k})$

Replacing the cofactors in the first formula, I have:

$\det B = \sum_{k \text{ even}} -a_{1k} (-\text{ cof}(a_{1k})) + \sum_{k \text{ odd}} a_{1k} \text{ cof}(a_{1k}) = \sum_k a_{1k} \text{ cof}(a_{1k}) = \det A$

This shows that ${\cal P}(n)$ is true, so it must be true for any integer $n$.

 

[fresh_42]

Homework Statement

Given a matrix $A = (a_{ij})$, we define matrix $B = \begin{pmatrix} a_{11} & - a_{12} & a_{13} & \cdots \\ - a_{21} & a_{22} & -a_{23} & \cdots \\ a_{31} & - a_{32} & a_{33} & \cdots \\ \vdots & \vdots & \vdots & \vdots \end{pmatrix}$.
Another way to define $B$ is to say that $b_{ij} = a_{ij}$ if $i+j$ is even and $- a_{ij}$ otherwise.

The question is: show that $\det B = \det A$

I have my idea which I have written below, do you agree with my explanation? Thanks!

Homework Equations

The Attempt at a Solution

[/B]
By recursion over property ${\cal P}(n)$: the property is true for matrices of order $n$.
${\cal P}(2)$ is true, and we assume that that ${\cal P}(n-1)$ is true. Let us show that ${\cal P}(n)$ must be true.

Expanding the determinant according to the first line, I have:

$\det B = \sum_k b_{1k} \text{ cof}(b_{1k}) = \sum_{k \text{ even}} -a_{1k} \text{ cof}(b_{1k}) + \sum_{k \text{ odd}} a_{1k} \text{ cof}(b_{1k})$

But no matter the parity of $k$, if I reverse the sign of the $k-1$ first columns in the determinant hidden inside $\text{ cof}(b_{1k})$, which is of order $n-1$, and then use the recursion hypothesis, I have:

$\text{ cof}(b_{1k}) = (-1)^{k-1} \text{ cof}(a_{1k})$

Replacing the cofactors in the first formula, I have:

$\det B = \sum_{k \text{ even}} -a_{1k} (-\text{ cof}(a_{1k})) + \sum_{k \text{ odd}} a_{1k} \text{ cof}(a_{1k}) = \sum_k a_{1k} \text{ cof}(a_{1k}) = \det A$

This shows that ${\cal P}(n)$ is true, so it must be true for any integer $n$.

I think the basic idea is correct, although it's a bit difficult to follow as the cofactors aren't very clear to see, resp. analyze. Do you apply the induction hypothesis on them?

Anyway, if you write the determinant as single formula, then the entire proof boils down to $|\,\mathcal{Sym}(n)\, : \,\mathcal{A}(n)\,|=2$.

 

[geoffrey159]

The problem with the determinants in the cofactors is that neighbor columns have the same sign and break the 'chessboard' structure.
That's why I reverse the sign of the $k-1$ first columns, in order to recover this structure, which allows me to apply ${\cal P}(n-1)$.

 

[fresh_42]

Have you considered to use a compact formula without recursion for the determinant?

 

[geoffrey159]

I have received an explanation in terms of permutations, but I didn't get it. That's why I tried to find another way. What do you mean ?

 

[fresh_42]

I have received an explanation in terms of permutations, but I didn't get it. That's why I tried to find another way. What do you mean ?

The determinant is the weighted sum over all diagonals, i.e. permutations: $\det A = \sum_{\sigma \in \mathcal{Sym}(n)} (-1)^{\operatorname{sgn}\sigma}a_{k\,\sigma(k)}$ Here you have only to analyze a single permutation $\sigma$. The question is: can there be an odd number of pairs $(k\,,\, \sigma(k))$ with $k + \sigma(k) \equiv 1 (2)\,$?

Edit: Corrected errors.

 

[geoffrey159]

I think I get it, but it is a little difficult:

$\begin{align*} \det B &= \sum_{\sigma\in S_n} \epsilon(\sigma) b_{\sigma(1)1}\cdots b_{\sigma(n)n}\\ &= \sum_{\sigma\in S_n} \epsilon(\sigma) a_{\sigma(1)1}\cdots a_{\sigma(n)n} (-1)^{\#S_\sigma} \end{align*}$

where set $S_\sigma = \{\text{couples }(k,\sigma(k)) \text{ such that } k+\sigma(k) \text{ is odd}, 1\le k \le n \}$

But for any permuatition $\sigma$ of $\{1,...,n\}$: $\sum_k (k + \sigma(k)) = \underbrace{n(n+1)}_\text{even}$

So $\#S_\sigma$ must be even and $\det B = \sum_{\sigma\in S_n} \epsilon(\sigma) a_{\sigma(1)1}\cdots a_{\sigma(n)n} = \det A$


EDITED: exhausted :-)

 

[fresh_42]

It must be true for every single diagonal, for otherwise it might not compensate in the sum. Furthermore doesn't the signature of the permutation change, so it has to be true for every single permutation: $\begin{bmatrix}1&2&3&\ldots &n\\k_1&k_2&k_3&\ldots &k_n\end{bmatrix}$. Now we change the sign of the diagonal in total by $$\prod_{i=1}^n (-1)^{i+k_i}=\prod_{i=1}^n (-1)^{i+\sigma(i)}$$.

 

[geoffrey159]

Thank you for your help !

 

[Ray Vickson]

Homework Statement

Given a matrix $A = (a_{ij})$, we define matrix $B = \begin{pmatrix} a_{11} & - a_{12} & a_{13} & \cdots \\ - a_{21} & a_{22} & -a_{23} & \cdots \\ a_{31} & - a_{32} & a_{33} & \cdots \\ \vdots & \vdots & \vdots & \vdots \end{pmatrix}$.
Another way to define $B$ is to say that $b_{ij} = a_{ij}$ if $i+j$ is even and $- a_{ij}$ otherwise.

The question is: show that $\det B = \det A$

I have my idea which I have written below, do you agree with my explanation? Thanks!

Homework Equations

The Attempt at a Solution

[/B]
By recursion over property ${\cal P}(n)$: the property is true for matrices of order $n$.
${\cal P}(2)$ is true, and we assume that that ${\cal P}(n-1)$ is true. Let us show that ${\cal P}(n)$ must be true.

Expanding the determinant according to the first line, I have:

$\det B = \sum_k b_{1k} \text{ cof}(b_{1k}) = \sum_{k \text{ even}} -a_{1k} \text{ cof}(b_{1k}) + \sum_{k \text{ odd}} a_{1k} \text{ cof}(b_{1k})$

But no matter the parity of $k$, if I reverse the sign of the $k-1$ first columns in the determinant hidden inside $\text{ cof}(b_{1k})$, which is of order $n-1$, and then use the recursion hypothesis, I have:

$\text{ cof}(b_{1k}) = (-1)^{k-1} \text{ cof}(a_{1k})$

Replacing the cofactors in the first formula, I have:

$\det B = \sum_{k \text{ even}} -a_{1k} (-\text{ cof}(a_{1k})) + \sum_{k \text{ odd}} a_{1k} \text{ cof}(a_{1k}) = \sum_k a_{1k} \text{ cof}(a_{1k}) = \det A$

This shows that ${\cal P}(n)$ is true, so it must be true for any integer $n$.

Another way is to change the signs of every second column (so all columns look the same), then multiply the resulting matrix by
$$\pmatrix{1 & 0 & 0 & \ldots & 0 \\
0 & -1 & 0 & \ldots & 0 \\
0 & 0 & 1 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & \pm 1}
$$
The first move changes the sign of the determinant of $B$ in a predictable way, and the second move changes it by another predictable way and gets you back to $\det(A)$.

 

[StoneTemplePython]

piggybacking here, use Ray's matrix

$\mathbf S:= \pmatrix{1 & 0 & 0 & \ldots & 0 \\ 0 & -1 & 0 & \ldots & 0 \\ 0 & 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & \pm 1}$

verify the involution $\mathbf {SS} = \mathbf I \to \mathbf S = \mathbf S^{-1}$

recognize that the two matrices $\mathbf A$ and $\mathbf B$ are similar and hence have same determinant, characteristic polynomial, etc.

i.e. confirm
$\mathbf B = \mathbf S \mathbf A \mathbf S = \mathbf S \mathbf A \mathbf S^{-1}$

that's it.

 

[Ray Vickson]

piggybacking here, use Ray's matrix

$\mathbf S:= \pmatrix{1 & 0 & 0 & \ldots & 0 \\ 0 & -1 & 0 & \ldots & 0 \\ 0 & 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & \pm 1}$

verify the involution $\mathbf {SS} = \mathbf I \to \mathbf S = \mathbf S^{-1}$

recognize that the two matrices $\mathbf A$ and $\mathbf B$ are similar and hence have same determinant, characteristic polynomial, etc.

i.e. confirm
$\mathbf B = \mathbf S \mathbf A \mathbf S = \mathbf S \mathbf A \mathbf S^{-1}$

that's it.

Much nicer!