Find the determinant $Δ_n$
$begingroup$
Find the determinant $Delta_n$
$A_n = begin{bmatrix} 0 & 1 & 0 &dots &dots&0\ -1 & 0 &1 & 0&&vdots\ 0&-1 & 0 &1 &ddots&vdots\ vdots& & & &ddots&0\ vdots& & ddots &ddots &ddots&1\ 0 & dots & dots &0&-1&0 end{bmatrix} in M_n(mathbb{K})$
After doing some tests I conclude $Delta_n$ is $0$ if $n$ is odd and $1$ if $n$ is even. How can I prove it formally? Any hint?
linear-algebra matrices determinant
asked Jan 9 at 21:21
Jhon KnowsJhon Knows
83
$endgroup$
add a comment |
$begingroup$
Find the determinant $Delta_n$
$A_n = begin{bmatrix} 0 & 1 & 0 &dots &dots&0\ -1 & 0 &1 & 0&&vdots\ 0&-1 & 0 &1 &ddots&vdots\ vdots& & & &ddots&0\ vdots& & ddots &ddots &ddots&1\ 0 & dots & dots &0&-1&0 end{bmatrix} in M_n(mathbb{K})$
After doing some tests I conclude $Delta_n$ is $0$ if $n$ is odd and $1$ if $n$ is even. How can I prove it formally? Any hint?
linear-algebra matrices determinant
asked Jan 9 at 21:21
Jhon KnowsJhon Knows
83
$endgroup$
1
$begingroup$
Do the Laplace expansion along the first row and column to get $Delta_n=Delta_{n-2}$. Then $Delta_1$ and $Delta_2$ are obvious.
$endgroup$
– A.Γ.
Jan 9 at 21:40
$begingroup$
An alternate approach for the case of odd $n$: $A$ is skew-symmetric. So, if $n$ is odd, we have $$ det(A) = det(A^T) = det(-A) = (-1)^n det(A) = -det(A) $$ That is, we have $det(A) = -det(A)$ which means that $det(A) = 0$.
$endgroup$
– Omnomnomnom
Jan 9 at 22:43
$begingroup$
You could also use more general results about tridiagonal Toeplitz matrices
$endgroup$
– Omnomnomnom
Jan 9 at 22:46
add a comment |
$begingroup$
Find the determinant $Delta_n$
$A_n = begin{bmatrix} 0 & 1 & 0 &dots &dots&0\ -1 & 0 &1 & 0&&vdots\ 0&-1 & 0 &1 &ddots&vdots\ vdots& & & &ddots&0\ vdots& & ddots &ddots &ddots&1\ 0 & dots & dots &0&-1&0 end{bmatrix} in M_n(mathbb{K})$
After doing some tests I conclude $Delta_n$ is $0$ if $n$ is odd and $1$ if $n$ is even. How can I prove it formally? Any hint?
linear-algebra matrices determinant
asked Jan 9 at 21:21
Jhon KnowsJhon Knows
83
$endgroup$
Find the determinant $Delta_n$
$A_n = begin{bmatrix} 0 & 1 & 0 &dots &dots&0\ -1 & 0 &1 & 0&&vdots\ 0&-1 & 0 &1 &ddots&vdots\ vdots& & & &ddots&0\ vdots& & ddots &ddots &ddots&1\ 0 & dots & dots &0&-1&0 end{bmatrix} in M_n(mathbb{K})$
After doing some tests I conclude $Delta_n$ is $0$ if $n$ is odd and $1$ if $n$ is even. How can I prove it formally? Any hint?
linear-algebra matrices determinant
linear-algebra matrices determinant
asked Jan 9 at 21:21
Jhon KnowsJhon Knows
83
asked Jan 9 at 21:21
Jhon KnowsJhon Knows
83
edited Jan 9 at 21:33
Jhon Knows
asked Jan 9 at 21:21
Jhon KnowsJhon Knows
83
asked Jan 9 at 21:21
Jhon KnowsJhon Knows
83
asked Jan 9 at 21:21
Jhon KnowsJhon Knows
83
83
1
$begingroup$
Do the Laplace expansion along the first row and column to get $Delta_n=Delta_{n-2}$. Then $Delta_1$ and $Delta_2$ are obvious.
$endgroup$
– A.Γ.
Jan 9 at 21:40
$begingroup$
An alternate approach for the case of odd $n$: $A$ is skew-symmetric. So, if $n$ is odd, we have $$ det(A) = det(A^T) = det(-A) = (-1)^n det(A) = -det(A) $$ That is, we have $det(A) = -det(A)$ which means that $det(A) = 0$.
$endgroup$
– Omnomnomnom
Jan 9 at 22:43
$begingroup$
You could also use more general results about tridiagonal Toeplitz matrices
$endgroup$
– Omnomnomnom
Jan 9 at 22:46
add a comment |
1
$begingroup$
Do the Laplace expansion along the first row and column to get $Delta_n=Delta_{n-2}$. Then $Delta_1$ and $Delta_2$ are obvious.
$endgroup$
– A.Γ.
Jan 9 at 21:40
$begingroup$
An alternate approach for the case of odd $n$: $A$ is skew-symmetric. So, if $n$ is odd, we have $$ det(A) = det(A^T) = det(-A) = (-1)^n det(A) = -det(A) $$ That is, we have $det(A) = -det(A)$ which means that $det(A) = 0$.
$endgroup$
– Omnomnomnom
Jan 9 at 22:43
$begingroup$
You could also use more general results about tridiagonal Toeplitz matrices
$endgroup$
– Omnomnomnom
Jan 9 at 22:46
1
1
$begingroup$
Do the Laplace expansion along the first row and column to get $Delta_n=Delta_{n-2}$. Then $Delta_1$ and $Delta_2$ are obvious.
$endgroup$
– A.Γ.
Jan 9 at 21:40
$begingroup$
Do the Laplace expansion along the first row and column to get $Delta_n=Delta_{n-2}$. Then $Delta_1$ and $Delta_2$ are obvious.
$endgroup$
– A.Γ.
Jan 9 at 21:40
$begingroup$
An alternate approach for the case of odd $n$: $A$ is skew-symmetric. So, if $n$ is odd, we have $$ det(A) = det(A^T) = det(-A) = (-1)^n det(A) = -det(A) $$ That is, we have $det(A) = -det(A)$ which means that $det(A) = 0$.
$endgroup$
– Omnomnomnom
Jan 9 at 22:43
$begingroup$
An alternate approach for the case of odd $n$: $A$ is skew-symmetric. So, if $n$ is odd, we have $$ det(A) = det(A^T) = det(-A) = (-1)^n det(A) = -det(A) $$ That is, we have $det(A) = -det(A)$ which means that $det(A) = 0$.
$endgroup$
– Omnomnomnom
Jan 9 at 22:43
$begingroup$
You could also use more general results about tridiagonal Toeplitz matrices
$endgroup$
– Omnomnomnom
Jan 9 at 22:46
$begingroup$
You could also use more general results about tridiagonal Toeplitz matrices
$endgroup$
– Omnomnomnom
Jan 9 at 22:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
Expanding by the last row, prove that
$$Delta_n=begin{vmatrix}0&1&0&dots&0&0&0 \ -1&0&1&dots&0&0&0 \ 0&-1& 0 & cdots&0&0&0\
vdots&&& vdots &&&vdots \0&0&0&cdots&0&1&0 \0&0&0&dots&-1&0&0 \0&0&0&dots&0&-1&1end{vmatrix}=Delta_{n-2} ;text{ (expanding by the last column)}$$
answered Jan 9 at 22:33
BernardBernard
119k740113
$endgroup$
add a comment |
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$begingroup$
Hint:
Expanding by the last row, prove that
$$Delta_n=begin{vmatrix}0&1&0&dots&0&0&0 \ -1&0&1&dots&0&0&0 \ 0&-1& 0 & cdots&0&0&0\
vdots&&& vdots &&&vdots \0&0&0&cdots&0&1&0 \0&0&0&dots&-1&0&0 \0&0&0&dots&0&-1&1end{vmatrix}=Delta_{n-2} ;text{ (expanding by the last column)}$$
answered Jan 9 at 22:33
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
Hint:
Expanding by the last row, prove that
$$Delta_n=begin{vmatrix}0&1&0&dots&0&0&0 \ -1&0&1&dots&0&0&0 \ 0&-1& 0 & cdots&0&0&0\
vdots&&& vdots &&&vdots \0&0&0&cdots&0&1&0 \0&0&0&dots&-1&0&0 \0&0&0&dots&0&-1&1end{vmatrix}=Delta_{n-2} ;text{ (expanding by the last column)}$$
answered Jan 9 at 22:33
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
Hint:
Expanding by the last row, prove that
$$Delta_n=begin{vmatrix}0&1&0&dots&0&0&0 \ -1&0&1&dots&0&0&0 \ 0&-1& 0 & cdots&0&0&0\
vdots&&& vdots &&&vdots \0&0&0&cdots&0&1&0 \0&0&0&dots&-1&0&0 \0&0&0&dots&0&-1&1end{vmatrix}=Delta_{n-2} ;text{ (expanding by the last column)}$$
answered Jan 9 at 22:33
BernardBernard
119k740113
$endgroup$
Hint:
Expanding by the last row, prove that
$$Delta_n=begin{vmatrix}0&1&0&dots&0&0&0 \ -1&0&1&dots&0&0&0 \ 0&-1& 0 & cdots&0&0&0\
vdots&&& vdots &&&vdots \0&0&0&cdots&0&1&0 \0&0&0&dots&-1&0&0 \0&0&0&dots&0&-1&1end{vmatrix}=Delta_{n-2} ;text{ (expanding by the last column)}$$
answered Jan 9 at 22:33
BernardBernard
119k740113
edited Jan 10 at 18:18
answered Jan 9 at 22:33
BernardBernard
119k740113
answered Jan 9 at 22:33
BernardBernard
119k740113
answered Jan 9 at 22:33
BernardBernard
119k740113
119k740113
add a comment |
add a comment |
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$begingroup$
Do the Laplace expansion along the first row and column to get $Delta_n=Delta_{n-2}$. Then $Delta_1$ and $Delta_2$ are obvious.
$endgroup$
– A.Γ.
Jan 9 at 21:40
$begingroup$
An alternate approach for the case of odd $n$: $A$ is skew-symmetric. So, if $n$ is odd, we have $$ det(A) = det(A^T) = det(-A) = (-1)^n det(A) = -det(A) $$ That is, we have $det(A) = -det(A)$ which means that $det(A) = 0$.
$endgroup$
– Omnomnomnom
Jan 9 at 22:43
$begingroup$
You could also use more general results about tridiagonal Toeplitz matrices
$endgroup$
– Omnomnomnom
Jan 9 at 22:46