Show that $|det(A_n)|=n^{n/2}$
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For k $ge2$ we recursively define $A_{2^k}$ as $begin{bmatrix} A_{2^{k-1}} & A_{2^{k-1}} \ A_{2^{k-1}} & -A_{2^{k-1}} end{bmatrix}$ and $A_1=[1]$
The problem is to show that $|det(A_n)|=n^{n/2}$
My attempt: we do an induction on $k$
$|det(A_2)|=2=2^{2/2}$. Induction hypothesis: $|det(A_{n})|=n^{n/2}$ and we want to show that $|det(A_{2n})|=(2n)^n$
using block matrix properties
$|det(A_{2n})|=|det(begin{bmatrix} A_{n} & A_{n} \ A_{n} & -A_{n} end{bmatrix})|=|det(-A)det(A+AA^{-1}A)|=|2^ndet(A_n)^2|=|2^nn^n|=(2n)^n$
Can somone confirm that there are no flaws in the reasoning please?
matrices
add a comment |
up vote
5
down vote
favorite
For k $ge2$ we recursively define $A_{2^k}$ as $begin{bmatrix} A_{2^{k-1}} & A_{2^{k-1}} \ A_{2^{k-1}} & -A_{2^{k-1}} end{bmatrix}$ and $A_1=[1]$
The problem is to show that $|det(A_n)|=n^{n/2}$
My attempt: we do an induction on $k$
$|det(A_2)|=2=2^{2/2}$. Induction hypothesis: $|det(A_{n})|=n^{n/2}$ and we want to show that $|det(A_{2n})|=(2n)^n$
using block matrix properties
$|det(A_{2n})|=|det(begin{bmatrix} A_{n} & A_{n} \ A_{n} & -A_{n} end{bmatrix})|=|det(-A)det(A+AA^{-1}A)|=|2^ndet(A_n)^2|=|2^nn^n|=(2n)^n$
Can somone confirm that there are no flaws in the reasoning please?
matrices
2
Looks ok. Alternatively you can prove (by induction on $k$) that $A_{2^k}A_{2^k}^T=2^k I$. As the determinants of the transposes are equal, the claim follows up to sign from this. The matrices you get are known as Hadamard matrices.
– Jyrki Lahtonen
Mar 2 at 17:57
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
For k $ge2$ we recursively define $A_{2^k}$ as $begin{bmatrix} A_{2^{k-1}} & A_{2^{k-1}} \ A_{2^{k-1}} & -A_{2^{k-1}} end{bmatrix}$ and $A_1=[1]$
The problem is to show that $|det(A_n)|=n^{n/2}$
My attempt: we do an induction on $k$
$|det(A_2)|=2=2^{2/2}$. Induction hypothesis: $|det(A_{n})|=n^{n/2}$ and we want to show that $|det(A_{2n})|=(2n)^n$
using block matrix properties
$|det(A_{2n})|=|det(begin{bmatrix} A_{n} & A_{n} \ A_{n} & -A_{n} end{bmatrix})|=|det(-A)det(A+AA^{-1}A)|=|2^ndet(A_n)^2|=|2^nn^n|=(2n)^n$
Can somone confirm that there are no flaws in the reasoning please?
matrices
For k $ge2$ we recursively define $A_{2^k}$ as $begin{bmatrix} A_{2^{k-1}} & A_{2^{k-1}} \ A_{2^{k-1}} & -A_{2^{k-1}} end{bmatrix}$ and $A_1=[1]$
The problem is to show that $|det(A_n)|=n^{n/2}$
My attempt: we do an induction on $k$
$|det(A_2)|=2=2^{2/2}$. Induction hypothesis: $|det(A_{n})|=n^{n/2}$ and we want to show that $|det(A_{2n})|=(2n)^n$
using block matrix properties
$|det(A_{2n})|=|det(begin{bmatrix} A_{n} & A_{n} \ A_{n} & -A_{n} end{bmatrix})|=|det(-A)det(A+AA^{-1}A)|=|2^ndet(A_n)^2|=|2^nn^n|=(2n)^n$
Can somone confirm that there are no flaws in the reasoning please?
matrices
matrices
asked Mar 2 at 17:28
John Cataldo
8581216
8581216
2
Looks ok. Alternatively you can prove (by induction on $k$) that $A_{2^k}A_{2^k}^T=2^k I$. As the determinants of the transposes are equal, the claim follows up to sign from this. The matrices you get are known as Hadamard matrices.
– Jyrki Lahtonen
Mar 2 at 17:57
add a comment |
2
Looks ok. Alternatively you can prove (by induction on $k$) that $A_{2^k}A_{2^k}^T=2^k I$. As the determinants of the transposes are equal, the claim follows up to sign from this. The matrices you get are known as Hadamard matrices.
– Jyrki Lahtonen
Mar 2 at 17:57
2
2
Looks ok. Alternatively you can prove (by induction on $k$) that $A_{2^k}A_{2^k}^T=2^k I$. As the determinants of the transposes are equal, the claim follows up to sign from this. The matrices you get are known as Hadamard matrices.
– Jyrki Lahtonen
Mar 2 at 17:57
Looks ok. Alternatively you can prove (by induction on $k$) that $A_{2^k}A_{2^k}^T=2^k I$. As the determinants of the transposes are equal, the claim follows up to sign from this. The matrices you get are known as Hadamard matrices.
– Jyrki Lahtonen
Mar 2 at 17:57
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Your proof looks good.
Alternately, for the induction-step, notice that
$$
begin{vmatrix}
A_n & A_n \
A_n & -A_n
end{vmatrix}
=
begin{vmatrix}
A_n & A_n \
0 & -2A_n
end{vmatrix},
$$
since
$$
begin{bmatrix}
A_n & A_n \
A_n & -A_n
end{bmatrix}
sim
begin{bmatrix}
A_n & A_n \
0 & -2A_n
end{bmatrix}
$$
(the symbol $`sim'$ denotes row-equivalent).
Thus,
$$|A_{2n}| = |A_n|cdot |-2A_n| = n^{n/2}(-2)^n n^{n/2} = 2^n n^n = (2n)^{2n/2}$$.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Your proof looks good.
Alternately, for the induction-step, notice that
$$
begin{vmatrix}
A_n & A_n \
A_n & -A_n
end{vmatrix}
=
begin{vmatrix}
A_n & A_n \
0 & -2A_n
end{vmatrix},
$$
since
$$
begin{bmatrix}
A_n & A_n \
A_n & -A_n
end{bmatrix}
sim
begin{bmatrix}
A_n & A_n \
0 & -2A_n
end{bmatrix}
$$
(the symbol $`sim'$ denotes row-equivalent).
Thus,
$$|A_{2n}| = |A_n|cdot |-2A_n| = n^{n/2}(-2)^n n^{n/2} = 2^n n^n = (2n)^{2n/2}$$.
add a comment |
up vote
4
down vote
accepted
Your proof looks good.
Alternately, for the induction-step, notice that
$$
begin{vmatrix}
A_n & A_n \
A_n & -A_n
end{vmatrix}
=
begin{vmatrix}
A_n & A_n \
0 & -2A_n
end{vmatrix},
$$
since
$$
begin{bmatrix}
A_n & A_n \
A_n & -A_n
end{bmatrix}
sim
begin{bmatrix}
A_n & A_n \
0 & -2A_n
end{bmatrix}
$$
(the symbol $`sim'$ denotes row-equivalent).
Thus,
$$|A_{2n}| = |A_n|cdot |-2A_n| = n^{n/2}(-2)^n n^{n/2} = 2^n n^n = (2n)^{2n/2}$$.
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Your proof looks good.
Alternately, for the induction-step, notice that
$$
begin{vmatrix}
A_n & A_n \
A_n & -A_n
end{vmatrix}
=
begin{vmatrix}
A_n & A_n \
0 & -2A_n
end{vmatrix},
$$
since
$$
begin{bmatrix}
A_n & A_n \
A_n & -A_n
end{bmatrix}
sim
begin{bmatrix}
A_n & A_n \
0 & -2A_n
end{bmatrix}
$$
(the symbol $`sim'$ denotes row-equivalent).
Thus,
$$|A_{2n}| = |A_n|cdot |-2A_n| = n^{n/2}(-2)^n n^{n/2} = 2^n n^n = (2n)^{2n/2}$$.
Your proof looks good.
Alternately, for the induction-step, notice that
$$
begin{vmatrix}
A_n & A_n \
A_n & -A_n
end{vmatrix}
=
begin{vmatrix}
A_n & A_n \
0 & -2A_n
end{vmatrix},
$$
since
$$
begin{bmatrix}
A_n & A_n \
A_n & -A_n
end{bmatrix}
sim
begin{bmatrix}
A_n & A_n \
0 & -2A_n
end{bmatrix}
$$
(the symbol $`sim'$ denotes row-equivalent).
Thus,
$$|A_{2n}| = |A_n|cdot |-2A_n| = n^{n/2}(-2)^n n^{n/2} = 2^n n^n = (2n)^{2n/2}$$.
edited Nov 21 at 19:06
answered Mar 2 at 17:40
Pietro Paparella
1,305514
1,305514
add a comment |
add a comment |
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Looks ok. Alternatively you can prove (by induction on $k$) that $A_{2^k}A_{2^k}^T=2^k I$. As the determinants of the transposes are equal, the claim follows up to sign from this. The matrices you get are known as Hadamard matrices.
– Jyrki Lahtonen
Mar 2 at 17:57