Mixing
If $A + A^t = 2I$ then $det(A) geq 1$
$begingroup$
Let $A$ be a $n times n$ real matrix such that $$ A + A^t = 2I, $$ where $I$ is the $n times n $ identity matrix.
Prove that $det(A) geq 1$.
It is obvious that tr$(A) = n$. Furthermore, we also have that $$A - 2I = -A^t, $$ so we get that $$det(A-2I) = (-1)^n cdot det(A).$$
Now, we let $lambda_1, lambda_2, cdots, lambda_n in mathbb{C}$ be the eigenvalues of $A$, so we get that $$(lambda_1 - 2)(lambda_2 - 2) cdots (lambda_n - 2) = (-1)^nlambda_1lambda_2 cdots lambda_n, $$ but I couldn't derive anything about the product $lambda_1lambda_2 cdots lambda_n = det(A)$ (the only known thing is $lambda_1 + cdots + lambda_n = n$).
Also, I tried the same approach for $2 times 2$ and $3 times 3 $ matrices, but it didn't lead to anything (especially since the eigenvalues can be complex numbers).
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 24 '18 at 18:23
CosminCosmin
1,4811527
$endgroup$
add a comment |
$begingroup$
Let $A$ be a $n times n$ real matrix such that $$ A + A^t = 2I, $$ where $I$ is the $n times n $ identity matrix.
Prove that $det(A) geq 1$.
It is obvious that tr$(A) = n$. Furthermore, we also have that $$A - 2I = -A^t, $$ so we get that $$det(A-2I) = (-1)^n cdot det(A).$$
Now, we let $lambda_1, lambda_2, cdots, lambda_n in mathbb{C}$ be the eigenvalues of $A$, so we get that $$(lambda_1 - 2)(lambda_2 - 2) cdots (lambda_n - 2) = (-1)^nlambda_1lambda_2 cdots lambda_n, $$ but I couldn't derive anything about the product $lambda_1lambda_2 cdots lambda_n = det(A)$ (the only known thing is $lambda_1 + cdots + lambda_n = n$).
Also, I tried the same approach for $2 times 2$ and $3 times 3 $ matrices, but it didn't lead to anything (especially since the eigenvalues can be complex numbers).
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 24 '18 at 18:23
CosminCosmin
1,4811527
$endgroup$
add a comment |
$begingroup$
Let $A$ be a $n times n$ real matrix such that $$ A + A^t = 2I, $$ where $I$ is the $n times n $ identity matrix.
Prove that $det(A) geq 1$.
It is obvious that tr$(A) = n$. Furthermore, we also have that $$A - 2I = -A^t, $$ so we get that $$det(A-2I) = (-1)^n cdot det(A).$$
Now, we let $lambda_1, lambda_2, cdots, lambda_n in mathbb{C}$ be the eigenvalues of $A$, so we get that $$(lambda_1 - 2)(lambda_2 - 2) cdots (lambda_n - 2) = (-1)^nlambda_1lambda_2 cdots lambda_n, $$ but I couldn't derive anything about the product $lambda_1lambda_2 cdots lambda_n = det(A)$ (the only known thing is $lambda_1 + cdots + lambda_n = n$).
Also, I tried the same approach for $2 times 2$ and $3 times 3 $ matrices, but it didn't lead to anything (especially since the eigenvalues can be complex numbers).
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 24 '18 at 18:23
CosminCosmin
1,4811527
$endgroup$
Let $A$ be a $n times n$ real matrix such that $$ A + A^t = 2I, $$ where $I$ is the $n times n $ identity matrix.
Prove that $det(A) geq 1$.
It is obvious that tr$(A) = n$. Furthermore, we also have that $$A - 2I = -A^t, $$ so we get that $$det(A-2I) = (-1)^n cdot det(A).$$
Now, we let $lambda_1, lambda_2, cdots, lambda_n in mathbb{C}$ be the eigenvalues of $A$, so we get that $$(lambda_1 - 2)(lambda_2 - 2) cdots (lambda_n - 2) = (-1)^nlambda_1lambda_2 cdots lambda_n, $$ but I couldn't derive anything about the product $lambda_1lambda_2 cdots lambda_n = det(A)$ (the only known thing is $lambda_1 + cdots + lambda_n = n$).
Also, I tried the same approach for $2 times 2$ and $3 times 3 $ matrices, but it didn't lead to anything (especially since the eigenvalues can be complex numbers).
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 24 '18 at 18:23
CosminCosmin
1,4811527
asked Dec 24 '18 at 18:23
CosminCosmin
1,4811527
asked Dec 24 '18 at 18:23
CosminCosmin
1,4811527
asked Dec 24 '18 at 18:23
CosminCosmin
1,4811527
asked Dec 24 '18 at 18:23
CosminCosmin
1,4811527
1,4811527
add a comment |
add a comment |
2 Answers
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$begingroup$
Note that $A - I$ is a real skew-symmetric matrix, so all of its eigenvalues are pure imaginary (or zero) and appear in complex conjugate pairs. Therefore the eigenvalues of $A$ are either $1$ or of the form $1 + c i, 1-c i$ for some $cinmathbb{R}$. Note that $(1+ci)(1-ci) = 1+c^2 geq 1$. Therefore
$$det(A) = prod (1 + c^2) geq 1$$
answered Dec 24 '18 at 18:43
ODFODF
1,486510
$endgroup$
$begingroup$
Thanks! That's exactly what I just though about now, once I tried to write out the matrices and compute the sum.
$endgroup$
– Cosmin
Dec 24 '18 at 18:44
add a comment |
$begingroup$
Let $A = I+B.$ One can immediately notice that $B$ is skew-symmetric or $B = -B^T.$ This means the eigenvalues of $A$ are precisely $1+lambda_j$, where $lambda_j$ are the eigenvalues of $B.$ Now, it's well-known that the eigenvalues of a skew-symmetric matrix comes in pairs $pm ilambda_j.$ This means that:
$$det A = prod(1+lambda^2_j)geq 1.$$
You are encouraged to fill in the details separating the cases where $n$ is odd or even.
answered Dec 24 '18 at 18:44
dezdichadodezdichado
6,4591929
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
Note that $A - I$ is a real skew-symmetric matrix, so all of its eigenvalues are pure imaginary (or zero) and appear in complex conjugate pairs. Therefore the eigenvalues of $A$ are either $1$ or of the form $1 + c i, 1-c i$ for some $cinmathbb{R}$. Note that $(1+ci)(1-ci) = 1+c^2 geq 1$. Therefore
$$det(A) = prod (1 + c^2) geq 1$$
answered Dec 24 '18 at 18:43
ODFODF
1,486510
$endgroup$
$begingroup$
Thanks! That's exactly what I just though about now, once I tried to write out the matrices and compute the sum.
$endgroup$
– Cosmin
Dec 24 '18 at 18:44
add a comment |
$begingroup$
Note that $A - I$ is a real skew-symmetric matrix, so all of its eigenvalues are pure imaginary (or zero) and appear in complex conjugate pairs. Therefore the eigenvalues of $A$ are either $1$ or of the form $1 + c i, 1-c i$ for some $cinmathbb{R}$. Note that $(1+ci)(1-ci) = 1+c^2 geq 1$. Therefore
$$det(A) = prod (1 + c^2) geq 1$$
answered Dec 24 '18 at 18:43
ODFODF
1,486510
$endgroup$
$begingroup$
Thanks! That's exactly what I just though about now, once I tried to write out the matrices and compute the sum.
$endgroup$
– Cosmin
Dec 24 '18 at 18:44
add a comment |
$begingroup$
Note that $A - I$ is a real skew-symmetric matrix, so all of its eigenvalues are pure imaginary (or zero) and appear in complex conjugate pairs. Therefore the eigenvalues of $A$ are either $1$ or of the form $1 + c i, 1-c i$ for some $cinmathbb{R}$. Note that $(1+ci)(1-ci) = 1+c^2 geq 1$. Therefore
$$det(A) = prod (1 + c^2) geq 1$$
answered Dec 24 '18 at 18:43
ODFODF
1,486510
$endgroup$
Note that $A - I$ is a real skew-symmetric matrix, so all of its eigenvalues are pure imaginary (or zero) and appear in complex conjugate pairs. Therefore the eigenvalues of $A$ are either $1$ or of the form $1 + c i, 1-c i$ for some $cinmathbb{R}$. Note that $(1+ci)(1-ci) = 1+c^2 geq 1$. Therefore
$$det(A) = prod (1 + c^2) geq 1$$
answered Dec 24 '18 at 18:43
ODFODF
1,486510
answered Dec 24 '18 at 18:43
ODFODF
1,486510
answered Dec 24 '18 at 18:43
ODFODF
1,486510
answered Dec 24 '18 at 18:43
ODFODF
1,486510
1,486510
$begingroup$
Thanks! That's exactly what I just though about now, once I tried to write out the matrices and compute the sum.
$endgroup$
– Cosmin
Dec 24 '18 at 18:44
add a comment |
$begingroup$
Thanks! That's exactly what I just though about now, once I tried to write out the matrices and compute the sum.
$endgroup$
– Cosmin
Dec 24 '18 at 18:44
$begingroup$
Thanks! That's exactly what I just though about now, once I tried to write out the matrices and compute the sum.
$endgroup$
– Cosmin
Dec 24 '18 at 18:44
$begingroup$
Thanks! That's exactly what I just though about now, once I tried to write out the matrices and compute the sum.
$endgroup$
– Cosmin
Dec 24 '18 at 18:44
add a comment |
$begingroup$
Let $A = I+B.$ One can immediately notice that $B$ is skew-symmetric or $B = -B^T.$ This means the eigenvalues of $A$ are precisely $1+lambda_j$, where $lambda_j$ are the eigenvalues of $B.$ Now, it's well-known that the eigenvalues of a skew-symmetric matrix comes in pairs $pm ilambda_j.$ This means that:
$$det A = prod(1+lambda^2_j)geq 1.$$
You are encouraged to fill in the details separating the cases where $n$ is odd or even.
answered Dec 24 '18 at 18:44
dezdichadodezdichado
6,4591929
$endgroup$
add a comment |
$begingroup$
Let $A = I+B.$ One can immediately notice that $B$ is skew-symmetric or $B = -B^T.$ This means the eigenvalues of $A$ are precisely $1+lambda_j$, where $lambda_j$ are the eigenvalues of $B.$ Now, it's well-known that the eigenvalues of a skew-symmetric matrix comes in pairs $pm ilambda_j.$ This means that:
$$det A = prod(1+lambda^2_j)geq 1.$$
You are encouraged to fill in the details separating the cases where $n$ is odd or even.
answered Dec 24 '18 at 18:44
dezdichadodezdichado
6,4591929
$endgroup$
add a comment |
$begingroup$
Let $A = I+B.$ One can immediately notice that $B$ is skew-symmetric or $B = -B^T.$ This means the eigenvalues of $A$ are precisely $1+lambda_j$, where $lambda_j$ are the eigenvalues of $B.$ Now, it's well-known that the eigenvalues of a skew-symmetric matrix comes in pairs $pm ilambda_j.$ This means that:
$$det A = prod(1+lambda^2_j)geq 1.$$
You are encouraged to fill in the details separating the cases where $n$ is odd or even.
answered Dec 24 '18 at 18:44
dezdichadodezdichado
6,4591929
$endgroup$
Let $A = I+B.$ One can immediately notice that $B$ is skew-symmetric or $B = -B^T.$ This means the eigenvalues of $A$ are precisely $1+lambda_j$, where $lambda_j$ are the eigenvalues of $B.$ Now, it's well-known that the eigenvalues of a skew-symmetric matrix comes in pairs $pm ilambda_j.$ This means that:
$$det A = prod(1+lambda^2_j)geq 1.$$
You are encouraged to fill in the details separating the cases where $n$ is odd or even.
answered Dec 24 '18 at 18:44
dezdichadodezdichado
6,4591929
answered Dec 24 '18 at 18:44
dezdichadodezdichado
6,4591929
answered Dec 24 '18 at 18:44
dezdichadodezdichado
6,4591929
answered Dec 24 '18 at 18:44
dezdichadodezdichado
6,4591929
6,4591929
add a comment |
add a comment |
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