$Av=lambda v Rightarrow A^*v=bar lambda v$ (general case)
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Suppose $V$ is a finite-dimensional complex inner product space and $v_1,v_2,...,v_3$ is an orthonormal basis of $V$.
Define $A:V to V$ by $Av_i=lambda_i v_i$ for some $lambda_i in mathbb{C}.$
Show that $A^*v_i=bar lambda v_i.$
My attempt:
$(Av_i,v_i)=(lambda_iv_i,v_i)=(v_i,barlambda_i v_i)=(v_i,A^*v_i)$
$Rightarrow(v_i,(A^*-bar lambda_i)v_i)=0 Rightarrow (A^*-bar lambda_i)v_i perp v_i $
How to show that $(A^*-barlambda_i)v_i=0$ ?
linear-algebra matrices adjoint-operators
asked Nov 16 at 14:21
XYZ
1858
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up vote
3
down vote
favorite
Suppose $V$ is a finite-dimensional complex inner product space and $v_1,v_2,...,v_3$ is an orthonormal basis of $V$.
Define $A:V to V$ by $Av_i=lambda_i v_i$ for some $lambda_i in mathbb{C}.$
Show that $A^*v_i=bar lambda v_i.$
My attempt:
$(Av_i,v_i)=(lambda_iv_i,v_i)=(v_i,barlambda_i v_i)=(v_i,A^*v_i)$
$Rightarrow(v_i,(A^*-bar lambda_i)v_i)=0 Rightarrow (A^*-bar lambda_i)v_i perp v_i $
How to show that $(A^*-barlambda_i)v_i=0$ ?
linear-algebra matrices adjoint-operators
asked Nov 16 at 14:21
XYZ
1858
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose $V$ is a finite-dimensional complex inner product space and $v_1,v_2,...,v_3$ is an orthonormal basis of $V$.
Define $A:V to V$ by $Av_i=lambda_i v_i$ for some $lambda_i in mathbb{C}.$
Show that $A^*v_i=bar lambda v_i.$
My attempt:
$(Av_i,v_i)=(lambda_iv_i,v_i)=(v_i,barlambda_i v_i)=(v_i,A^*v_i)$
$Rightarrow(v_i,(A^*-bar lambda_i)v_i)=0 Rightarrow (A^*-bar lambda_i)v_i perp v_i $
How to show that $(A^*-barlambda_i)v_i=0$ ?
linear-algebra matrices adjoint-operators
asked Nov 16 at 14:21
XYZ
1858
Suppose $V$ is a finite-dimensional complex inner product space and $v_1,v_2,...,v_3$ is an orthonormal basis of $V$.
Define $A:V to V$ by $Av_i=lambda_i v_i$ for some $lambda_i in mathbb{C}.$
Show that $A^*v_i=bar lambda v_i.$
My attempt:
$(Av_i,v_i)=(lambda_iv_i,v_i)=(v_i,barlambda_i v_i)=(v_i,A^*v_i)$
$Rightarrow(v_i,(A^*-bar lambda_i)v_i)=0 Rightarrow (A^*-bar lambda_i)v_i perp v_i $
How to show that $(A^*-barlambda_i)v_i=0$ ?
linear-algebra matrices adjoint-operators
linear-algebra matrices adjoint-operators
asked Nov 16 at 14:21
XYZ
1858
asked Nov 16 at 14:21
XYZ
1858
edited Nov 16 at 14:38
asked Nov 16 at 14:21
XYZ
1858
asked Nov 16 at 14:21
XYZ
1858
asked Nov 16 at 14:21
XYZ
1858
1858
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
Just extend what you have a bit:
begin{align}
langle sum_j a_j v_j,A^*sum_k b_k v_k rangle = &langle A sum_{j}a_j v_j, sum_{k}b_k v_krangle \
= & langle sum_{j}lambda_j a_j v_j,sum_k b_k v_krangle \
= &sum_jlambda_j a_joverline{b_j} \
= &langlesum_j a_j v_j,sum_k overline{lambda_k}b_kv_krangle \
end{align}
answered Nov 16 at 15:06
DisintegratingByParts
57.9k42477
@user1561 : Like the way I have redone it?
– DisintegratingByParts
Nov 16 at 17:01
1
Yes. $phantom{}$+1.
– user1551
Nov 16 at 17:23
add a comment |
up vote
0
down vote
Here is another method:
Let $M=[A]_beta$, where $beta={v_i}_{1le ile n}$. Then $det,(M-lambda I)=0impliesdet,(M-lambda I)^*=0$.
answered Nov 16 at 15:07
614177
416
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Just extend what you have a bit:
begin{align}
langle sum_j a_j v_j,A^*sum_k b_k v_k rangle = &langle A sum_{j}a_j v_j, sum_{k}b_k v_krangle \
= & langle sum_{j}lambda_j a_j v_j,sum_k b_k v_krangle \
= &sum_jlambda_j a_joverline{b_j} \
= &langlesum_j a_j v_j,sum_k overline{lambda_k}b_kv_krangle \
end{align}
answered Nov 16 at 15:06
DisintegratingByParts
57.9k42477
@user1561 : Like the way I have redone it?
– DisintegratingByParts
Nov 16 at 17:01
1
Yes. $phantom{}$+1.
– user1551
Nov 16 at 17:23
add a comment |
up vote
2
down vote
Just extend what you have a bit:
begin{align}
langle sum_j a_j v_j,A^*sum_k b_k v_k rangle = &langle A sum_{j}a_j v_j, sum_{k}b_k v_krangle \
= & langle sum_{j}lambda_j a_j v_j,sum_k b_k v_krangle \
= &sum_jlambda_j a_joverline{b_j} \
= &langlesum_j a_j v_j,sum_k overline{lambda_k}b_kv_krangle \
end{align}
answered Nov 16 at 15:06
DisintegratingByParts
57.9k42477
@user1561 : Like the way I have redone it?
– DisintegratingByParts
Nov 16 at 17:01
1
Yes. $phantom{}$+1.
– user1551
Nov 16 at 17:23
add a comment |
up vote
2
down vote
up vote
2
down vote
Just extend what you have a bit:
begin{align}
langle sum_j a_j v_j,A^*sum_k b_k v_k rangle = &langle A sum_{j}a_j v_j, sum_{k}b_k v_krangle \
= & langle sum_{j}lambda_j a_j v_j,sum_k b_k v_krangle \
= &sum_jlambda_j a_joverline{b_j} \
= &langlesum_j a_j v_j,sum_k overline{lambda_k}b_kv_krangle \
end{align}
answered Nov 16 at 15:06
DisintegratingByParts
57.9k42477
Just extend what you have a bit:
begin{align}
langle sum_j a_j v_j,A^*sum_k b_k v_k rangle = &langle A sum_{j}a_j v_j, sum_{k}b_k v_krangle \
= & langle sum_{j}lambda_j a_j v_j,sum_k b_k v_krangle \
= &sum_jlambda_j a_joverline{b_j} \
= &langlesum_j a_j v_j,sum_k overline{lambda_k}b_kv_krangle \
end{align}
answered Nov 16 at 15:06
DisintegratingByParts
57.9k42477
edited Nov 16 at 17:00
answered Nov 16 at 15:06
DisintegratingByParts
57.9k42477
answered Nov 16 at 15:06
DisintegratingByParts
57.9k42477
answered Nov 16 at 15:06
DisintegratingByParts
57.9k42477
57.9k42477
@user1561 : Like the way I have redone it?
– DisintegratingByParts
Nov 16 at 17:01
1
Yes. $phantom{}$+1.
– user1551
Nov 16 at 17:23
add a comment |
@user1561 : Like the way I have redone it?
– DisintegratingByParts
Nov 16 at 17:01
1
Yes. $phantom{}$+1.
– user1551
Nov 16 at 17:23
@user1561 : Like the way I have redone it?
– DisintegratingByParts
Nov 16 at 17:01
@user1561 : Like the way I have redone it?
– DisintegratingByParts
Nov 16 at 17:01
1
1
Yes. $phantom{}$+1.
– user1551
Nov 16 at 17:23
Yes. $phantom{}$+1.
– user1551
Nov 16 at 17:23
add a comment |
up vote
0
down vote
Here is another method:
Let $M=[A]_beta$, where $beta={v_i}_{1le ile n}$. Then $det,(M-lambda I)=0impliesdet,(M-lambda I)^*=0$.
answered Nov 16 at 15:07
614177
416
add a comment |
up vote
0
down vote
Here is another method:
Let $M=[A]_beta$, where $beta={v_i}_{1le ile n}$. Then $det,(M-lambda I)=0impliesdet,(M-lambda I)^*=0$.
answered Nov 16 at 15:07
614177
416
add a comment |
up vote
0
down vote
up vote
0
down vote
Here is another method:
Let $M=[A]_beta$, where $beta={v_i}_{1le ile n}$. Then $det,(M-lambda I)=0impliesdet,(M-lambda I)^*=0$.
answered Nov 16 at 15:07
614177
416
Here is another method:
Let $M=[A]_beta$, where $beta={v_i}_{1le ile n}$. Then $det,(M-lambda I)=0impliesdet,(M-lambda I)^*=0$.
answered Nov 16 at 15:07
614177
416
edited Nov 16 at 16:00
answered Nov 16 at 15:07
614177
416
answered Nov 16 at 15:07
614177
416
answered Nov 16 at 15:07
614177
416
416
add a comment |
add a comment |
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