Prove that a bilinear form is nondegenerate
0
$begingroup$
Let $n$ be a positive integer, $V=mathbb{C}^{ntimes n}$ and $$f(A,B)=nmathrm{Tr}(AB) - mathrm{Tr}(A)mathrm{Tr}(B).$$ Let $$W={ Ain V, |, mathrm{Tr}(A)=0 } $$ and let $f_1=f|_{W}$.
Prove that $f_1$ is nondegenerate.
What I have done:
We have that $mathrm{Tr}(A)=mathrm{Tr}(B)=0 $, then $f_1(A,B)=nmathrm{Tr}(AB)$. We have to prove that: $nmathrm{Tr}(A,B)=0 ; forall beta in W $ implies $A=0$. But I don't know where to go from there.
linear-algebra abstract-algebra vector-spaces bilinear-form
edited Feb 8 at 23:50
user26857
39.5k124283
asked Dec 18 '18 at 19:13
evaristegdevaristegd
1419
$endgroup$
add a comment |
0
$begingroup$
Let $n$ be a positive integer, $V=mathbb{C}^{ntimes n}$ and $$f(A,B)=nmathrm{Tr}(AB) - mathrm{Tr}(A)mathrm{Tr}(B).$$ Let $$W={ Ain V, |, mathrm{Tr}(A)=0 } $$ and let $f_1=f|_{W}$.
Prove that $f_1$ is nondegenerate.
What I have done:
We have that $mathrm{Tr}(A)=mathrm{Tr}(B)=0 $, then $f_1(A,B)=nmathrm{Tr}(AB)$. We have to prove that: $nmathrm{Tr}(A,B)=0 ; forall beta in W $ implies $A=0$. But I don't know where to go from there.
linear-algebra abstract-algebra vector-spaces bilinear-form
edited Feb 8 at 23:50
user26857
39.5k124283
asked Dec 18 '18 at 19:13
evaristegdevaristegd
1419
$endgroup$
add a comment |
0
0
0
$begingroup$
Let $n$ be a positive integer, $V=mathbb{C}^{ntimes n}$ and $$f(A,B)=nmathrm{Tr}(AB) - mathrm{Tr}(A)mathrm{Tr}(B).$$ Let $$W={ Ain V, |, mathrm{Tr}(A)=0 } $$ and let $f_1=f|_{W}$.
Prove that $f_1$ is nondegenerate.
What I have done:
We have that $mathrm{Tr}(A)=mathrm{Tr}(B)=0 $, then $f_1(A,B)=nmathrm{Tr}(AB)$. We have to prove that: $nmathrm{Tr}(A,B)=0 ; forall beta in W $ implies $A=0$. But I don't know where to go from there.
linear-algebra abstract-algebra vector-spaces bilinear-form
edited Feb 8 at 23:50
user26857
39.5k124283
asked Dec 18 '18 at 19:13
evaristegdevaristegd
1419
$endgroup$
Let $n$ be a positive integer, $V=mathbb{C}^{ntimes n}$ and $$f(A,B)=nmathrm{Tr}(AB) - mathrm{Tr}(A)mathrm{Tr}(B).$$ Let $$W={ Ain V, |, mathrm{Tr}(A)=0 } $$ and let $f_1=f|_{W}$.
Prove that $f_1$ is nondegenerate.
What I have done:
We have that $mathrm{Tr}(A)=mathrm{Tr}(B)=0 $, then $f_1(A,B)=nmathrm{Tr}(AB)$. We have to prove that: $nmathrm{Tr}(A,B)=0 ; forall beta in W $ implies $A=0$. But I don't know where to go from there.
linear-algebra abstract-algebra vector-spaces bilinear-form
linear-algebra abstract-algebra vector-spaces bilinear-form
edited Feb 8 at 23:50
user26857
39.5k124283
asked Dec 18 '18 at 19:13
evaristegdevaristegd
1419
edited Feb 8 at 23:50
user26857
39.5k124283
asked Dec 18 '18 at 19:13
evaristegdevaristegd
1419
edited Feb 8 at 23:50
user26857
39.5k124283
edited Feb 8 at 23:50
user26857
39.5k124283
edited Feb 8 at 23:50
user26857
39.5k124283
39.5k124283
asked Dec 18 '18 at 19:13
evaristegdevaristegd
1419
asked Dec 18 '18 at 19:13
evaristegdevaristegd
1419
asked Dec 18 '18 at 19:13
evaristegdevaristegd
1419
1419
add a comment |
add a comment |
1 Answer
1
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oldest
votes
0
$begingroup$
Hint: What is $f_1(A, ^toverline{A})$?
answered Dec 18 '18 at 19:27
MindlackMindlack
4,910211
$endgroup$
$begingroup$
I assume you meant to write the $t$ superscript after the comma. The transpose of $A$ ($A^{t}$) would still belong to W, and we can say the same for $overline{A}$. But I'm not sure what can we say about the product. I don't think the trace of the product would be zero.
$endgroup$
– evaristegd
Dec 18 '18 at 20:20
$begingroup$
I meant: "I assume you meant to write the superscript before the comma"
$endgroup$
– evaristegd
Dec 18 '18 at 20:27
$begingroup$
I used my native country notation. So it is the trace of the matrix product of $A$ and the transpose of $overline{A}$. When you write it explicitly, you get $sum_{i,j}{|A_{i,j}|^2}$.
$endgroup$
– Mindlack
Dec 18 '18 at 23:44
$begingroup$
I see, so that would be greater than zero for every non-zero A. But I'm not sure of how to proceed after that.
$endgroup$
– evaristegd
Dec 19 '18 at 1:57
$begingroup$
Do you remember the definition of “nondegenerate”?
$endgroup$
– Mindlack
Dec 19 '18 at 17:02
|
show 1 more comment
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
Hint: What is $f_1(A, ^toverline{A})$?
answered Dec 18 '18 at 19:27
MindlackMindlack
4,910211
$endgroup$
$begingroup$
I assume you meant to write the $t$ superscript after the comma. The transpose of $A$ ($A^{t}$) would still belong to W, and we can say the same for $overline{A}$. But I'm not sure what can we say about the product. I don't think the trace of the product would be zero.
$endgroup$
– evaristegd
Dec 18 '18 at 20:20
$begingroup$
I meant: "I assume you meant to write the superscript before the comma"
$endgroup$
– evaristegd
Dec 18 '18 at 20:27
$begingroup$
I used my native country notation. So it is the trace of the matrix product of $A$ and the transpose of $overline{A}$. When you write it explicitly, you get $sum_{i,j}{|A_{i,j}|^2}$.
$endgroup$
– Mindlack
Dec 18 '18 at 23:44
$begingroup$
I see, so that would be greater than zero for every non-zero A. But I'm not sure of how to proceed after that.
$endgroup$
– evaristegd
Dec 19 '18 at 1:57
$begingroup$
Do you remember the definition of “nondegenerate”?
$endgroup$
– Mindlack
Dec 19 '18 at 17:02
|
show 1 more comment
0
$begingroup$
Hint: What is $f_1(A, ^toverline{A})$?
answered Dec 18 '18 at 19:27
MindlackMindlack
4,910211
$endgroup$
$begingroup$
I assume you meant to write the $t$ superscript after the comma. The transpose of $A$ ($A^{t}$) would still belong to W, and we can say the same for $overline{A}$. But I'm not sure what can we say about the product. I don't think the trace of the product would be zero.
$endgroup$
– evaristegd
Dec 18 '18 at 20:20
$begingroup$
I meant: "I assume you meant to write the superscript before the comma"
$endgroup$
– evaristegd
Dec 18 '18 at 20:27
$begingroup$
I used my native country notation. So it is the trace of the matrix product of $A$ and the transpose of $overline{A}$. When you write it explicitly, you get $sum_{i,j}{|A_{i,j}|^2}$.
$endgroup$
– Mindlack
Dec 18 '18 at 23:44
$begingroup$
I see, so that would be greater than zero for every non-zero A. But I'm not sure of how to proceed after that.
$endgroup$
– evaristegd
Dec 19 '18 at 1:57
$begingroup$
Do you remember the definition of “nondegenerate”?
$endgroup$
– Mindlack
Dec 19 '18 at 17:02
|
show 1 more comment
0
0
0
$begingroup$
Hint: What is $f_1(A, ^toverline{A})$?
answered Dec 18 '18 at 19:27
MindlackMindlack
4,910211
$endgroup$
Hint: What is $f_1(A, ^toverline{A})$?
answered Dec 18 '18 at 19:27
MindlackMindlack
4,910211
answered Dec 18 '18 at 19:27
MindlackMindlack
4,910211
answered Dec 18 '18 at 19:27
MindlackMindlack
4,910211
answered Dec 18 '18 at 19:27
MindlackMindlack
4,910211
4,910211
$begingroup$
I assume you meant to write the $t$ superscript after the comma. The transpose of $A$ ($A^{t}$) would still belong to W, and we can say the same for $overline{A}$. But I'm not sure what can we say about the product. I don't think the trace of the product would be zero.
$endgroup$
– evaristegd
Dec 18 '18 at 20:20
$begingroup$
I meant: "I assume you meant to write the superscript before the comma"
$endgroup$
– evaristegd
Dec 18 '18 at 20:27
$begingroup$
I used my native country notation. So it is the trace of the matrix product of $A$ and the transpose of $overline{A}$. When you write it explicitly, you get $sum_{i,j}{|A_{i,j}|^2}$.
$endgroup$
– Mindlack
Dec 18 '18 at 23:44
$begingroup$
I see, so that would be greater than zero for every non-zero A. But I'm not sure of how to proceed after that.
$endgroup$
– evaristegd
Dec 19 '18 at 1:57
$begingroup$
Do you remember the definition of “nondegenerate”?
$endgroup$
– Mindlack
Dec 19 '18 at 17:02
|
show 1 more comment
$begingroup$
I assume you meant to write the $t$ superscript after the comma. The transpose of $A$ ($A^{t}$) would still belong to W, and we can say the same for $overline{A}$. But I'm not sure what can we say about the product. I don't think the trace of the product would be zero.
$endgroup$
– evaristegd
Dec 18 '18 at 20:20
$begingroup$
I meant: "I assume you meant to write the superscript before the comma"
$endgroup$
– evaristegd
Dec 18 '18 at 20:27
$begingroup$
I used my native country notation. So it is the trace of the matrix product of $A$ and the transpose of $overline{A}$. When you write it explicitly, you get $sum_{i,j}{|A_{i,j}|^2}$.
$endgroup$
– Mindlack
Dec 18 '18 at 23:44
$begingroup$
I see, so that would be greater than zero for every non-zero A. But I'm not sure of how to proceed after that.
$endgroup$
– evaristegd
Dec 19 '18 at 1:57
$begingroup$
Do you remember the definition of “nondegenerate”?
$endgroup$
– Mindlack
Dec 19 '18 at 17:02
$begingroup$
I assume you meant to write the $t$ superscript after the comma. The transpose of $A$ ($A^{t}$) would still belong to W, and we can say the same for $overline{A}$. But I'm not sure what can we say about the product. I don't think the trace of the product would be zero.
$endgroup$
– evaristegd
Dec 18 '18 at 20:20
$begingroup$
I assume you meant to write the $t$ superscript after the comma. The transpose of $A$ ($A^{t}$) would still belong to W, and we can say the same for $overline{A}$. But I'm not sure what can we say about the product. I don't think the trace of the product would be zero.
$endgroup$
– evaristegd
Dec 18 '18 at 20:20
$begingroup$
I meant: "I assume you meant to write the superscript before the comma"
$endgroup$
– evaristegd
Dec 18 '18 at 20:27
$begingroup$
I meant: "I assume you meant to write the superscript before the comma"
$endgroup$
– evaristegd
Dec 18 '18 at 20:27
$begingroup$
I used my native country notation. So it is the trace of the matrix product of $A$ and the transpose of $overline{A}$. When you write it explicitly, you get $sum_{i,j}{|A_{i,j}|^2}$.
$endgroup$
– Mindlack
Dec 18 '18 at 23:44
$begingroup$
I used my native country notation. So it is the trace of the matrix product of $A$ and the transpose of $overline{A}$. When you write it explicitly, you get $sum_{i,j}{|A_{i,j}|^2}$.
$endgroup$
– Mindlack
Dec 18 '18 at 23:44
$begingroup$
I see, so that would be greater than zero for every non-zero A. But I'm not sure of how to proceed after that.
$endgroup$
– evaristegd
Dec 19 '18 at 1:57
$begingroup$
I see, so that would be greater than zero for every non-zero A. But I'm not sure of how to proceed after that.
$endgroup$
– evaristegd
Dec 19 '18 at 1:57
$begingroup$
Do you remember the definition of “nondegenerate”?
$endgroup$
– Mindlack
Dec 19 '18 at 17:02
$begingroup$
Do you remember the definition of “nondegenerate”?
$endgroup$
– Mindlack
Dec 19 '18 at 17:02
|
show 1 more comment
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