Mixing
Computing the Adjoint using the Definition $langle Tv,wrangle = langle v, T^*wrangle$
$begingroup$
Let $T$ be the linear operator on $mathbb{C}^2$ defined by $$T(a,b)=(2ia+3b,a-b).$$ I am trying to compute the adjoint. The answer is $$T^*(c,d)=(-2ic+d,3c-d)tag{$1$},$$ which can be seen by (i) computing the matrix of $T$ with respect to the standard ordered basis, then (ii) taking the conjugate transpose of this matrix, and then (iii) using the new matrix to write the formula of $T^*$.
However, I'd like to compute the adjoint by using the definition: $langle T(v),wrangle = langle v, T^*(w) rangle$. I have $$langle (a,b),T^*(c,d) rangle= langle T(a,b), (c,d) rangle\=langle (2ia+3b,a-b), (c,d)rangle \=2aic+3bc+ad-bd\=a(2ic+d)+b(3c-d)\=langle (a,b), (2ic+d, 3c-d)rangle.$$
However, this implies $T^*(c,d)=(2ic+d,3c-d),$ which disagrees with equation $(1)$.
How can I obtain the correct answer using the definition?
linear-algebra inner-product-space adjoint-operators
asked Feb 2 at 6:37
Iced PalmerIced Palmer
274211
$endgroup$
add a comment |
$begingroup$
Let $T$ be the linear operator on $mathbb{C}^2$ defined by $$T(a,b)=(2ia+3b,a-b).$$ I am trying to compute the adjoint. The answer is $$T^*(c,d)=(-2ic+d,3c-d)tag{$1$},$$ which can be seen by (i) computing the matrix of $T$ with respect to the standard ordered basis, then (ii) taking the conjugate transpose of this matrix, and then (iii) using the new matrix to write the formula of $T^*$.
However, I'd like to compute the adjoint by using the definition: $langle T(v),wrangle = langle v, T^*(w) rangle$. I have $$langle (a,b),T^*(c,d) rangle= langle T(a,b), (c,d) rangle\=langle (2ia+3b,a-b), (c,d)rangle \=2aic+3bc+ad-bd\=a(2ic+d)+b(3c-d)\=langle (a,b), (2ic+d, 3c-d)rangle.$$
However, this implies $T^*(c,d)=(2ic+d,3c-d),$ which disagrees with equation $(1)$.
How can I obtain the correct answer using the definition?
linear-algebra inner-product-space adjoint-operators
asked Feb 2 at 6:37
Iced PalmerIced Palmer
274211
$endgroup$
1
$begingroup$
Well there must be a definition of the inner product given in the question. Or is it the standard inner product?
$endgroup$
– Yadati Kiran
Feb 2 at 6:46
add a comment |
$begingroup$
Let $T$ be the linear operator on $mathbb{C}^2$ defined by $$T(a,b)=(2ia+3b,a-b).$$ I am trying to compute the adjoint. The answer is $$T^*(c,d)=(-2ic+d,3c-d)tag{$1$},$$ which can be seen by (i) computing the matrix of $T$ with respect to the standard ordered basis, then (ii) taking the conjugate transpose of this matrix, and then (iii) using the new matrix to write the formula of $T^*$.
However, I'd like to compute the adjoint by using the definition: $langle T(v),wrangle = langle v, T^*(w) rangle$. I have $$langle (a,b),T^*(c,d) rangle= langle T(a,b), (c,d) rangle\=langle (2ia+3b,a-b), (c,d)rangle \=2aic+3bc+ad-bd\=a(2ic+d)+b(3c-d)\=langle (a,b), (2ic+d, 3c-d)rangle.$$
However, this implies $T^*(c,d)=(2ic+d,3c-d),$ which disagrees with equation $(1)$.
How can I obtain the correct answer using the definition?
linear-algebra inner-product-space adjoint-operators
asked Feb 2 at 6:37
Iced PalmerIced Palmer
274211
$endgroup$
Let $T$ be the linear operator on $mathbb{C}^2$ defined by $$T(a,b)=(2ia+3b,a-b).$$ I am trying to compute the adjoint. The answer is $$T^*(c,d)=(-2ic+d,3c-d)tag{$1$},$$ which can be seen by (i) computing the matrix of $T$ with respect to the standard ordered basis, then (ii) taking the conjugate transpose of this matrix, and then (iii) using the new matrix to write the formula of $T^*$.
However, I'd like to compute the adjoint by using the definition: $langle T(v),wrangle = langle v, T^*(w) rangle$. I have $$langle (a,b),T^*(c,d) rangle= langle T(a,b), (c,d) rangle\=langle (2ia+3b,a-b), (c,d)rangle \=2aic+3bc+ad-bd\=a(2ic+d)+b(3c-d)\=langle (a,b), (2ic+d, 3c-d)rangle.$$
However, this implies $T^*(c,d)=(2ic+d,3c-d),$ which disagrees with equation $(1)$.
How can I obtain the correct answer using the definition?
linear-algebra inner-product-space adjoint-operators
linear-algebra inner-product-space adjoint-operators
asked Feb 2 at 6:37
Iced PalmerIced Palmer
274211
asked Feb 2 at 6:37
Iced PalmerIced Palmer
274211
asked Feb 2 at 6:37
Iced PalmerIced Palmer
274211
asked Feb 2 at 6:37
Iced PalmerIced Palmer
274211
asked Feb 2 at 6:37
Iced PalmerIced Palmer
274211
274211
1
$begingroup$
Well there must be a definition of the inner product given in the question. Or is it the standard inner product?
$endgroup$
– Yadati Kiran
Feb 2 at 6:46
add a comment |
1
$begingroup$
Well there must be a definition of the inner product given in the question. Or is it the standard inner product?
$endgroup$
– Yadati Kiran
Feb 2 at 6:46
1
1
$begingroup$
Well there must be a definition of the inner product given in the question. Or is it the standard inner product?
$endgroup$
– Yadati Kiran
Feb 2 at 6:46
$begingroup$
Well there must be a definition of the inner product given in the question. Or is it the standard inner product?
$endgroup$
– Yadati Kiran
Feb 2 at 6:46
add a comment |
1 Answer
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oldest
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$begingroup$
You forgot to take the complex conjugate.
$$
langle((2ia + 3b, a-b), (c,d)rangle=-2aic+3bc+ad-bd
$$
answered Feb 2 at 6:48
PriyathamPriyatham
2,2241028
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add a comment |
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$begingroup$
You forgot to take the complex conjugate.
$$
langle((2ia + 3b, a-b), (c,d)rangle=-2aic+3bc+ad-bd
$$
answered Feb 2 at 6:48
PriyathamPriyatham
2,2241028
$endgroup$
add a comment |
$begingroup$
You forgot to take the complex conjugate.
$$
langle((2ia + 3b, a-b), (c,d)rangle=-2aic+3bc+ad-bd
$$
answered Feb 2 at 6:48
PriyathamPriyatham
2,2241028
$endgroup$
add a comment |
$begingroup$
You forgot to take the complex conjugate.
$$
langle((2ia + 3b, a-b), (c,d)rangle=-2aic+3bc+ad-bd
$$
answered Feb 2 at 6:48
PriyathamPriyatham
2,2241028
$endgroup$
You forgot to take the complex conjugate.
$$
langle((2ia + 3b, a-b), (c,d)rangle=-2aic+3bc+ad-bd
$$
answered Feb 2 at 6:48
PriyathamPriyatham
2,2241028
answered Feb 2 at 6:48
PriyathamPriyatham
2,2241028
answered Feb 2 at 6:48
PriyathamPriyatham
2,2241028
answered Feb 2 at 6:48
PriyathamPriyatham
2,2241028
2,2241028
add a comment |
add a comment |
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$begingroup$
Well there must be a definition of the inner product given in the question. Or is it the standard inner product?
$endgroup$
– Yadati Kiran
Feb 2 at 6:46