Proving an inner product
$begingroup$
Let $V$ be the vector space of all $2$ by $2$ matrices. Define $langle A, Brangle = mathrm{tr}(A^TDB)$, where $D = begin{pmatrix}7&1\ 1&1end{pmatrix}$. Prove that $<A, B>$ defines an inner product on $V$.
So I do see that the inner product is symmetric and bilinear as $D$ is equal to its transpose. But I struggling to prove the idea of positive definiteness as that would require that $D$ have positive eigenvalues.
But after calculating the eigenvalues, I get that they are $4+sqrt{10}$ and $4-sqrt{10}$. Since the second eigenvalue is negative, I am not sure how this inner product is positive definite.
Any help would be highly appreciated!
linear-algebra matrices inner-product-space trace
edited Dec 17 '18 at 2:35
Arturo Magidin
265k34590918
asked Dec 17 '18 at 1:03
sktsasussktsasus
1,015418
$endgroup$
add a comment |
$begingroup$
Let $V$ be the vector space of all $2$ by $2$ matrices. Define $langle A, Brangle = mathrm{tr}(A^TDB)$, where $D = begin{pmatrix}7&1\ 1&1end{pmatrix}$. Prove that $<A, B>$ defines an inner product on $V$.
So I do see that the inner product is symmetric and bilinear as $D$ is equal to its transpose. But I struggling to prove the idea of positive definiteness as that would require that $D$ have positive eigenvalues.
But after calculating the eigenvalues, I get that they are $4+sqrt{10}$ and $4-sqrt{10}$. Since the second eigenvalue is negative, I am not sure how this inner product is positive definite.
Any help would be highly appreciated!
linear-algebra matrices inner-product-space trace
edited Dec 17 '18 at 2:35
Arturo Magidin
265k34590918
asked Dec 17 '18 at 1:03
sktsasussktsasus
1,015418
$endgroup$
3
$begingroup$
The second isn't negative. $sqrt{10}<sqrt{16}=4$.
$endgroup$
– jgon
Dec 17 '18 at 1:08
$begingroup$
Oh dear how embarrassing...Thanks for letting me know.
$endgroup$
– sktsasus
Dec 17 '18 at 1:12
1
$begingroup$
No worries, I've certainly made similar mistakes before.
$endgroup$
– jgon
Dec 17 '18 at 1:14
add a comment |
$begingroup$
Let $V$ be the vector space of all $2$ by $2$ matrices. Define $langle A, Brangle = mathrm{tr}(A^TDB)$, where $D = begin{pmatrix}7&1\ 1&1end{pmatrix}$. Prove that $<A, B>$ defines an inner product on $V$.
So I do see that the inner product is symmetric and bilinear as $D$ is equal to its transpose. But I struggling to prove the idea of positive definiteness as that would require that $D$ have positive eigenvalues.
But after calculating the eigenvalues, I get that they are $4+sqrt{10}$ and $4-sqrt{10}$. Since the second eigenvalue is negative, I am not sure how this inner product is positive definite.
Any help would be highly appreciated!
linear-algebra matrices inner-product-space trace
edited Dec 17 '18 at 2:35
Arturo Magidin
265k34590918
asked Dec 17 '18 at 1:03
sktsasussktsasus
1,015418
$endgroup$
Let $V$ be the vector space of all $2$ by $2$ matrices. Define $langle A, Brangle = mathrm{tr}(A^TDB)$, where $D = begin{pmatrix}7&1\ 1&1end{pmatrix}$. Prove that $<A, B>$ defines an inner product on $V$.
So I do see that the inner product is symmetric and bilinear as $D$ is equal to its transpose. But I struggling to prove the idea of positive definiteness as that would require that $D$ have positive eigenvalues.
But after calculating the eigenvalues, I get that they are $4+sqrt{10}$ and $4-sqrt{10}$. Since the second eigenvalue is negative, I am not sure how this inner product is positive definite.
Any help would be highly appreciated!
linear-algebra matrices inner-product-space trace
linear-algebra matrices inner-product-space trace
edited Dec 17 '18 at 2:35
Arturo Magidin
265k34590918
asked Dec 17 '18 at 1:03
sktsasussktsasus
1,015418
edited Dec 17 '18 at 2:35
Arturo Magidin
265k34590918
asked Dec 17 '18 at 1:03
sktsasussktsasus
1,015418
edited Dec 17 '18 at 2:35
Arturo Magidin
265k34590918
edited Dec 17 '18 at 2:35
Arturo Magidin
265k34590918
edited Dec 17 '18 at 2:35
Arturo Magidin
265k34590918
265k34590918
asked Dec 17 '18 at 1:03
sktsasussktsasus
1,015418
asked Dec 17 '18 at 1:03
sktsasussktsasus
1,015418
asked Dec 17 '18 at 1:03
sktsasussktsasus
1,015418
1,015418
3
$begingroup$
The second isn't negative. $sqrt{10}<sqrt{16}=4$.
$endgroup$
– jgon
Dec 17 '18 at 1:08
$begingroup$
Oh dear how embarrassing...Thanks for letting me know.
$endgroup$
– sktsasus
Dec 17 '18 at 1:12
1
$begingroup$
No worries, I've certainly made similar mistakes before.
$endgroup$
– jgon
Dec 17 '18 at 1:14
add a comment |
3
$begingroup$
The second isn't negative. $sqrt{10}<sqrt{16}=4$.
$endgroup$
– jgon
Dec 17 '18 at 1:08
$begingroup$
Oh dear how embarrassing...Thanks for letting me know.
$endgroup$
– sktsasus
Dec 17 '18 at 1:12
1
$begingroup$
No worries, I've certainly made similar mistakes before.
$endgroup$
– jgon
Dec 17 '18 at 1:14
3
3
$begingroup$
The second isn't negative. $sqrt{10}<sqrt{16}=4$.
$endgroup$
– jgon
Dec 17 '18 at 1:08
$begingroup$
The second isn't negative. $sqrt{10}<sqrt{16}=4$.
$endgroup$
– jgon
Dec 17 '18 at 1:08
$begingroup$
Oh dear how embarrassing...Thanks for letting me know.
$endgroup$
– sktsasus
Dec 17 '18 at 1:12
$begingroup$
Oh dear how embarrassing...Thanks for letting me know.
$endgroup$
– sktsasus
Dec 17 '18 at 1:12
1
1
$begingroup$
No worries, I've certainly made similar mistakes before.
$endgroup$
– jgon
Dec 17 '18 at 1:14
$begingroup$
No worries, I've certainly made similar mistakes before.
$endgroup$
– jgon
Dec 17 '18 at 1:14
add a comment |
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3
$begingroup$
The second isn't negative. $sqrt{10}<sqrt{16}=4$.
$endgroup$
– jgon
Dec 17 '18 at 1:08
$begingroup$
Oh dear how embarrassing...Thanks for letting me know.
$endgroup$
– sktsasus
Dec 17 '18 at 1:12
1
$begingroup$
No worries, I've certainly made similar mistakes before.
$endgroup$
– jgon
Dec 17 '18 at 1:14