Mixing
What is definiteness of matrix $A+A^{*}$
$begingroup$
a)What is definiteness of matrix $A+A^{*}$ if matrix $A$ is defined
$$A= begin{bmatrix}
-1 & 1 & 3 \
1 & -3 & -2 \
-1 & 0 & -3 \
end{bmatrix} \ $$ b)Using conclusion from a), prove that real part of all matrix $A$ eigenvalues is negative. $$--------------------------------------$$ Is $A^{*}$ conjugate-transpose matrix? If it is, it should look like matrix A in a conjugate-transpose . Because matrix $A$ doesn't have complex elements, is then matrix $A+A^{*}=begin{bmatrix}
-1 & 1 & 3 \
1 & -3 & -2 \
-1 & 0 & -3 \
end{bmatrix} + begin{bmatrix}
-1 & 1 & -1 \
1 & -3 & 0 \
3 & -2 & -3 \
end{bmatrix} \[1ex] ?$
linear-algebra matrices
asked Jan 30 at 19:00
FiggaroFiggaro
355
$endgroup$
add a comment |
$begingroup$
a)What is definiteness of matrix $A+A^{*}$ if matrix $A$ is defined
$$A= begin{bmatrix}
-1 & 1 & 3 \
1 & -3 & -2 \
-1 & 0 & -3 \
end{bmatrix} \ $$ b)Using conclusion from a), prove that real part of all matrix $A$ eigenvalues is negative. $$--------------------------------------$$ Is $A^{*}$ conjugate-transpose matrix? If it is, it should look like matrix A in a conjugate-transpose . Because matrix $A$ doesn't have complex elements, is then matrix $A+A^{*}=begin{bmatrix}
-1 & 1 & 3 \
1 & -3 & -2 \
-1 & 0 & -3 \
end{bmatrix} + begin{bmatrix}
-1 & 1 & -1 \
1 & -3 & 0 \
3 & -2 & -3 \
end{bmatrix} \[1ex] ?$
linear-algebra matrices
asked Jan 30 at 19:00
FiggaroFiggaro
355
$endgroup$
$begingroup$
"Is $A^*$ conjugate-transpose matrix (of $A$)?" Usually, yes. What is the definition in your book/notes?
$endgroup$
– user587192
Jan 30 at 19:04
$begingroup$
Yes, it says exactly that. So, here I get $begin{bmatrix} -2 & 2 & 2 \ 2 & -6 & -2 \ 2 & -2 & -6 \ end{bmatrix}$ So I have a new matrix that is symmetric..with all diagonal elements lower than 0. I can use one more method, Sylvester's criterion to see if it is negative definite. But how can I prove b) ?
$endgroup$
– Figgaro
Jan 30 at 19:28
$begingroup$
Do you have difficulties finding all the eigenvalues?
$endgroup$
– user587192
Jan 30 at 19:30
$begingroup$
No, I can find them $4,sqrt{17} - 5, - 5 - sqrt{17}$. I wrote wrong, i have to use conclusion from a) and than to prove b), without accually calculationg eigenvalues..
$endgroup$
– Figgaro
Jan 30 at 19:41
add a comment |
$begingroup$
a)What is definiteness of matrix $A+A^{*}$ if matrix $A$ is defined
$$A= begin{bmatrix}
-1 & 1 & 3 \
1 & -3 & -2 \
-1 & 0 & -3 \
end{bmatrix} \ $$ b)Using conclusion from a), prove that real part of all matrix $A$ eigenvalues is negative. $$--------------------------------------$$ Is $A^{*}$ conjugate-transpose matrix? If it is, it should look like matrix A in a conjugate-transpose . Because matrix $A$ doesn't have complex elements, is then matrix $A+A^{*}=begin{bmatrix}
-1 & 1 & 3 \
1 & -3 & -2 \
-1 & 0 & -3 \
end{bmatrix} + begin{bmatrix}
-1 & 1 & -1 \
1 & -3 & 0 \
3 & -2 & -3 \
end{bmatrix} \[1ex] ?$
linear-algebra matrices
asked Jan 30 at 19:00
FiggaroFiggaro
355
$endgroup$
a)What is definiteness of matrix $A+A^{*}$ if matrix $A$ is defined
$$A= begin{bmatrix}
-1 & 1 & 3 \
1 & -3 & -2 \
-1 & 0 & -3 \
end{bmatrix} \ $$ b)Using conclusion from a), prove that real part of all matrix $A$ eigenvalues is negative. $$--------------------------------------$$ Is $A^{*}$ conjugate-transpose matrix? If it is, it should look like matrix A in a conjugate-transpose . Because matrix $A$ doesn't have complex elements, is then matrix $A+A^{*}=begin{bmatrix}
-1 & 1 & 3 \
1 & -3 & -2 \
-1 & 0 & -3 \
end{bmatrix} + begin{bmatrix}
-1 & 1 & -1 \
1 & -3 & 0 \
3 & -2 & -3 \
end{bmatrix} \[1ex] ?$
linear-algebra matrices
linear-algebra matrices
asked Jan 30 at 19:00
FiggaroFiggaro
355
asked Jan 30 at 19:00
FiggaroFiggaro
355
edited Jan 30 at 19:40
Figgaro
asked Jan 30 at 19:00
FiggaroFiggaro
355
asked Jan 30 at 19:00
FiggaroFiggaro
355
asked Jan 30 at 19:00
FiggaroFiggaro
355
355
$begingroup$
"Is $A^*$ conjugate-transpose matrix (of $A$)?" Usually, yes. What is the definition in your book/notes?
$endgroup$
– user587192
Jan 30 at 19:04
$begingroup$
Yes, it says exactly that. So, here I get $begin{bmatrix} -2 & 2 & 2 \ 2 & -6 & -2 \ 2 & -2 & -6 \ end{bmatrix}$ So I have a new matrix that is symmetric..with all diagonal elements lower than 0. I can use one more method, Sylvester's criterion to see if it is negative definite. But how can I prove b) ?
$endgroup$
– Figgaro
Jan 30 at 19:28
$begingroup$
Do you have difficulties finding all the eigenvalues?
$endgroup$
– user587192
Jan 30 at 19:30
$begingroup$
No, I can find them $4,sqrt{17} - 5, - 5 - sqrt{17}$. I wrote wrong, i have to use conclusion from a) and than to prove b), without accually calculationg eigenvalues..
$endgroup$
– Figgaro
Jan 30 at 19:41
add a comment |
$begingroup$
"Is $A^*$ conjugate-transpose matrix (of $A$)?" Usually, yes. What is the definition in your book/notes?
$endgroup$
– user587192
Jan 30 at 19:04
$begingroup$
Yes, it says exactly that. So, here I get $begin{bmatrix} -2 & 2 & 2 \ 2 & -6 & -2 \ 2 & -2 & -6 \ end{bmatrix}$ So I have a new matrix that is symmetric..with all diagonal elements lower than 0. I can use one more method, Sylvester's criterion to see if it is negative definite. But how can I prove b) ?
$endgroup$
– Figgaro
Jan 30 at 19:28
$begingroup$
Do you have difficulties finding all the eigenvalues?
$endgroup$
– user587192
Jan 30 at 19:30
$begingroup$
No, I can find them $4,sqrt{17} - 5, - 5 - sqrt{17}$. I wrote wrong, i have to use conclusion from a) and than to prove b), without accually calculationg eigenvalues..
$endgroup$
– Figgaro
Jan 30 at 19:41
$begingroup$
"Is $A^*$ conjugate-transpose matrix (of $A$)?" Usually, yes. What is the definition in your book/notes?
$endgroup$
– user587192
Jan 30 at 19:04
$begingroup$
"Is $A^*$ conjugate-transpose matrix (of $A$)?" Usually, yes. What is the definition in your book/notes?
$endgroup$
– user587192
Jan 30 at 19:04
$begingroup$
Yes, it says exactly that. So, here I get $begin{bmatrix} -2 & 2 & 2 \ 2 & -6 & -2 \ 2 & -2 & -6 \ end{bmatrix}$ So I have a new matrix that is symmetric..with all diagonal elements lower than 0. I can use one more method, Sylvester's criterion to see if it is negative definite. But how can I prove b) ?
$endgroup$
– Figgaro
Jan 30 at 19:28
$begingroup$
Yes, it says exactly that. So, here I get $begin{bmatrix} -2 & 2 & 2 \ 2 & -6 & -2 \ 2 & -2 & -6 \ end{bmatrix}$ So I have a new matrix that is symmetric..with all diagonal elements lower than 0. I can use one more method, Sylvester's criterion to see if it is negative definite. But how can I prove b) ?
$endgroup$
– Figgaro
Jan 30 at 19:28
$begingroup$
Do you have difficulties finding all the eigenvalues?
$endgroup$
– user587192
Jan 30 at 19:30
$begingroup$
Do you have difficulties finding all the eigenvalues?
$endgroup$
– user587192
Jan 30 at 19:30
$begingroup$
No, I can find them $4,sqrt{17} - 5, - 5 - sqrt{17}$. I wrote wrong, i have to use conclusion from a) and than to prove b), without accually calculationg eigenvalues..
$endgroup$
– Figgaro
Jan 30 at 19:41
$begingroup$
No, I can find them $4,sqrt{17} - 5, - 5 - sqrt{17}$. I wrote wrong, i have to use conclusion from a) and than to prove b), without accually calculationg eigenvalues..
$endgroup$
– Figgaro
Jan 30 at 19:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
a) We can calculate the eigenvalues of $A+A^*$ or check the principle minors to find it is negative definite.
That means that for any $xne 0$ we have $x^* (A+A^*)x <0$.
b) Since $(A+A^*)$ is negative definite, we have for every eigenvector $v$ of $A$ with eigenvalue $lambda$:
$$v^*(A+A^*)v = v^*Av +(Av)^*v = v^*lambda v + (lambda v)^*v = lambda v^2 + lambda^* v^2 = 2Re(lambda) v^2 < 0$$
Therefore $Re(lambda)<0$.
answered Jan 31 at 0:55
Klaas van AarsenKlaas van Aarsen
4,3421822
$endgroup$
add a comment |
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1 Answer
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$begingroup$
a) We can calculate the eigenvalues of $A+A^*$ or check the principle minors to find it is negative definite.
That means that for any $xne 0$ we have $x^* (A+A^*)x <0$.
b) Since $(A+A^*)$ is negative definite, we have for every eigenvector $v$ of $A$ with eigenvalue $lambda$:
$$v^*(A+A^*)v = v^*Av +(Av)^*v = v^*lambda v + (lambda v)^*v = lambda v^2 + lambda^* v^2 = 2Re(lambda) v^2 < 0$$
Therefore $Re(lambda)<0$.
answered Jan 31 at 0:55
Klaas van AarsenKlaas van Aarsen
4,3421822
$endgroup$
add a comment |
$begingroup$
a) We can calculate the eigenvalues of $A+A^*$ or check the principle minors to find it is negative definite.
That means that for any $xne 0$ we have $x^* (A+A^*)x <0$.
b) Since $(A+A^*)$ is negative definite, we have for every eigenvector $v$ of $A$ with eigenvalue $lambda$:
$$v^*(A+A^*)v = v^*Av +(Av)^*v = v^*lambda v + (lambda v)^*v = lambda v^2 + lambda^* v^2 = 2Re(lambda) v^2 < 0$$
Therefore $Re(lambda)<0$.
answered Jan 31 at 0:55
Klaas van AarsenKlaas van Aarsen
4,3421822
$endgroup$
add a comment |
$begingroup$
a) We can calculate the eigenvalues of $A+A^*$ or check the principle minors to find it is negative definite.
That means that for any $xne 0$ we have $x^* (A+A^*)x <0$.
b) Since $(A+A^*)$ is negative definite, we have for every eigenvector $v$ of $A$ with eigenvalue $lambda$:
$$v^*(A+A^*)v = v^*Av +(Av)^*v = v^*lambda v + (lambda v)^*v = lambda v^2 + lambda^* v^2 = 2Re(lambda) v^2 < 0$$
Therefore $Re(lambda)<0$.
answered Jan 31 at 0:55
Klaas van AarsenKlaas van Aarsen
4,3421822
$endgroup$
a) We can calculate the eigenvalues of $A+A^*$ or check the principle minors to find it is negative definite.
That means that for any $xne 0$ we have $x^* (A+A^*)x <0$.
b) Since $(A+A^*)$ is negative definite, we have for every eigenvector $v$ of $A$ with eigenvalue $lambda$:
$$v^*(A+A^*)v = v^*Av +(Av)^*v = v^*lambda v + (lambda v)^*v = lambda v^2 + lambda^* v^2 = 2Re(lambda) v^2 < 0$$
Therefore $Re(lambda)<0$.
answered Jan 31 at 0:55
Klaas van AarsenKlaas van Aarsen
4,3421822
answered Jan 31 at 0:55
Klaas van AarsenKlaas van Aarsen
4,3421822
answered Jan 31 at 0:55
Klaas van AarsenKlaas van Aarsen
4,3421822
answered Jan 31 at 0:55
Klaas van AarsenKlaas van Aarsen
4,3421822
4,3421822
add a comment |
add a comment |
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$begingroup$
"Is $A^*$ conjugate-transpose matrix (of $A$)?" Usually, yes. What is the definition in your book/notes?
$endgroup$
– user587192
Jan 30 at 19:04
$begingroup$
Yes, it says exactly that. So, here I get $begin{bmatrix} -2 & 2 & 2 \ 2 & -6 & -2 \ 2 & -2 & -6 \ end{bmatrix}$ So I have a new matrix that is symmetric..with all diagonal elements lower than 0. I can use one more method, Sylvester's criterion to see if it is negative definite. But how can I prove b) ?
$endgroup$
– Figgaro
Jan 30 at 19:28
$begingroup$
Do you have difficulties finding all the eigenvalues?
$endgroup$
– user587192
Jan 30 at 19:30
$begingroup$
No, I can find them $4,sqrt{17} - 5, - 5 - sqrt{17}$. I wrote wrong, i have to use conclusion from a) and than to prove b), without accually calculationg eigenvalues..
$endgroup$
– Figgaro
Jan 30 at 19:41