Problem about Inner product of matrix with two vectors

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1
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If A, B are two lists with length n and M is a n*n Matrices, then I want to compute the inner product of them B.M.A. The result should be a scalar however, what I get is a matrix. Any one knows what happened here? Thanks!
My code is below

A = Subscript[z, 0], Subscript[z, 1], Subscript[z, 2],
 Subscript[z, 3], Subscript[z, 4];
B = Subscript[OverBar[z], 0], Subscript[OverBar[z], 1], Subscript[OverBar[z], 2], 
 Subscript[OverBar[z], 3], Subscript[OverBar[z], 4];

 M = Chop[0.9947525414706575`, 
 0.014318078739690324` + 
 0.012070727204725996` I, -0.022576963511297447` + 
 0.017365770023498133` I, 
 0.028293920126770736` + 0.0041202947110136594` I, 
 0.02281394023398197` - 
 0.0005272617323759155` I, 0.014318078739690316` - 
 0.012070727204725998` I, 1.0089738927082252`, 
 0.02221807191535088` - 
 0.018113027609091715` I, -0.012682665041359427` - 
 0.010424540405231546` I, -0.055382503060447716` + 
 0.035624018712441564` I, -0.02257696351129744` - 
 0.017365770023498137` I, 
 0.022218071915350877` + 0.018113027609091715` I, 
 1.0204773571789212`, -0.0007522563609968502` + 
 0.026534550066808438` I, 
 0.0003575302509902032` - 
 0.0635032495006343` I, 0.028293920126770733` - 
 0.00412029471101366` I, -0.012682665041359427` + 
 0.010424540405231544` I, -0.000752256360996853` - 
 0.026534550066808438` I, 1.0132613424630528`, 
 0.059304690914936876` - 
 0.01648460212183861` I, 0.02281394023398197` + 
 0.0005272617323759158` I, -0.055382503060447716` - 
 0.035624018712441564` I, 
 0.0003575302509902011` + 0.06350324950063431` I, 
 0.059304690914936876` + 0.016484602121838613` I, 
 0.9911763824117488`]

In[31]:= Dimensions[B.M.A]

Out[31]= 5

edited 1 hour ago

Carl Woll

62.4k281158

asked 1 hour ago

cwei

654

1

You must keep in mind that everything is an expression. Your calculation results in a symbolic sum of five terms; see FullForm[A.M.B]. Dimensions is returning the dimensions of the head of the expression (Plus) which has 5 terms.
â€“Â Edmund
34 mins ago

@Edmund Thanks for your explanation! I understand now!
â€“Â cwei
19 mins ago

add a commentÂ |Â

up vote
1
down vote

favorite

A = Subscript[z, 0], Subscript[z, 1], Subscript[z, 2],
 Subscript[z, 3], Subscript[z, 4];
B = Subscript[OverBar[z], 0], Subscript[OverBar[z], 1], Subscript[OverBar[z], 2], 
 Subscript[OverBar[z], 3], Subscript[OverBar[z], 4];

 M = Chop[0.9947525414706575`, 
 0.014318078739690324` + 
 0.012070727204725996` I, -0.022576963511297447` + 
 0.017365770023498133` I, 
 0.028293920126770736` + 0.0041202947110136594` I, 
 0.02281394023398197` - 
 0.0005272617323759155` I, 0.014318078739690316` - 
 0.012070727204725998` I, 1.0089738927082252`, 
 0.02221807191535088` - 
 0.018113027609091715` I, -0.012682665041359427` - 
 0.010424540405231546` I, -0.055382503060447716` + 
 0.035624018712441564` I, -0.02257696351129744` - 
 0.017365770023498137` I, 
 0.022218071915350877` + 0.018113027609091715` I, 
 1.0204773571789212`, -0.0007522563609968502` + 
 0.026534550066808438` I, 
 0.0003575302509902032` - 
 0.0635032495006343` I, 0.028293920126770733` - 
 0.00412029471101366` I, -0.012682665041359427` + 
 0.010424540405231544` I, -0.000752256360996853` - 
 0.026534550066808438` I, 1.0132613424630528`, 
 0.059304690914936876` - 
 0.01648460212183861` I, 0.02281394023398197` + 
 0.0005272617323759158` I, -0.055382503060447716` - 
 0.035624018712441564` I, 
 0.0003575302509902011` + 0.06350324950063431` I, 
 0.059304690914936876` + 0.016484602121838613` I, 
 0.9911763824117488`]

In[31]:= Dimensions[B.M.A]

Out[31]= 5

edited 1 hour ago

Carl Woll

62.4k281158

asked 1 hour ago

cwei

654

1

You must keep in mind that everything is an expression. Your calculation results in a symbolic sum of five terms; see FullForm[A.M.B]. Dimensions is returning the dimensions of the head of the expression (Plus) which has 5 terms.
â€“Â Edmund
34 mins ago

@Edmund Thanks for your explanation! I understand now!
â€“Â cwei
19 mins ago

add a commentÂ |Â

up vote
1
down vote

favorite

A = Subscript[z, 0], Subscript[z, 1], Subscript[z, 2],
 Subscript[z, 3], Subscript[z, 4];
B = Subscript[OverBar[z], 0], Subscript[OverBar[z], 1], Subscript[OverBar[z], 2], 
 Subscript[OverBar[z], 3], Subscript[OverBar[z], 4];

 M = Chop[0.9947525414706575`, 
 0.014318078739690324` + 
 0.012070727204725996` I, -0.022576963511297447` + 
 0.017365770023498133` I, 
 0.028293920126770736` + 0.0041202947110136594` I, 
 0.02281394023398197` - 
 0.0005272617323759155` I, 0.014318078739690316` - 
 0.012070727204725998` I, 1.0089738927082252`, 
 0.02221807191535088` - 
 0.018113027609091715` I, -0.012682665041359427` - 
 0.010424540405231546` I, -0.055382503060447716` + 
 0.035624018712441564` I, -0.02257696351129744` - 
 0.017365770023498137` I, 
 0.022218071915350877` + 0.018113027609091715` I, 
 1.0204773571789212`, -0.0007522563609968502` + 
 0.026534550066808438` I, 
 0.0003575302509902032` - 
 0.0635032495006343` I, 0.028293920126770733` - 
 0.00412029471101366` I, -0.012682665041359427` + 
 0.010424540405231544` I, -0.000752256360996853` - 
 0.026534550066808438` I, 1.0132613424630528`, 
 0.059304690914936876` - 
 0.01648460212183861` I, 0.02281394023398197` + 
 0.0005272617323759158` I, -0.055382503060447716` - 
 0.035624018712441564` I, 
 0.0003575302509902011` + 0.06350324950063431` I, 
 0.059304690914936876` + 0.016484602121838613` I, 
 0.9911763824117488`]

In[31]:= Dimensions[B.M.A]

Out[31]= 5

edited 1 hour ago

Carl Woll

62.4k281158

asked 1 hour ago

cwei

654

A = Subscript[z, 0], Subscript[z, 1], Subscript[z, 2],
 Subscript[z, 3], Subscript[z, 4];
B = Subscript[OverBar[z], 0], Subscript[OverBar[z], 1], Subscript[OverBar[z], 2], 
 Subscript[OverBar[z], 3], Subscript[OverBar[z], 4];

 M = Chop[0.9947525414706575`, 
 0.014318078739690324` + 
 0.012070727204725996` I, -0.022576963511297447` + 
 0.017365770023498133` I, 
 0.028293920126770736` + 0.0041202947110136594` I, 
 0.02281394023398197` - 
 0.0005272617323759155` I, 0.014318078739690316` - 
 0.012070727204725998` I, 1.0089738927082252`, 
 0.02221807191535088` - 
 0.018113027609091715` I, -0.012682665041359427` - 
 0.010424540405231546` I, -0.055382503060447716` + 
 0.035624018712441564` I, -0.02257696351129744` - 
 0.017365770023498137` I, 
 0.022218071915350877` + 0.018113027609091715` I, 
 1.0204773571789212`, -0.0007522563609968502` + 
 0.026534550066808438` I, 
 0.0003575302509902032` - 
 0.0635032495006343` I, 0.028293920126770733` - 
 0.00412029471101366` I, -0.012682665041359427` + 
 0.010424540405231544` I, -0.000752256360996853` - 
 0.026534550066808438` I, 1.0132613424630528`, 
 0.059304690914936876` - 
 0.01648460212183861` I, 0.02281394023398197` + 
 0.0005272617323759158` I, -0.055382503060447716` - 
 0.035624018712441564` I, 
 0.0003575302509902011` + 0.06350324950063431` I, 
 0.059304690914936876` + 0.016484602121838613` I, 
 0.9911763824117488`]

In[31]:= Dimensions[B.M.A]

Out[31]= 5

matrix

edited 1 hour ago

Carl Woll

62.4k281158

asked 1 hour ago

cwei

654

edited 1 hour ago

Carl Woll

62.4k281158

asked 1 hour ago

cwei

654

edited 1 hour ago

Carl Woll

62.4k281158

edited 1 hour ago

Carl Woll

62.4k281158

edited 1 hour ago

Carl Woll

62.4k281158

asked 1 hour ago

cwei

654

asked 1 hour ago

cwei

654

asked 1 hour ago

cwei

654

1

You must keep in mind that everything is an expression. Your calculation results in a symbolic sum of five terms; see FullForm[A.M.B]. Dimensions is returning the dimensions of the head of the expression (Plus) which has 5 terms.
â€“Â Edmund
34 mins ago

@Edmund Thanks for your explanation! I understand now!
â€“Â cwei
19 mins ago

add a commentÂ |Â

1

You must keep in mind that everything is an expression. Your calculation results in a symbolic sum of five terms; see FullForm[A.M.B]. Dimensions is returning the dimensions of the head of the expression (Plus) which has 5 terms.
â€“Â Edmund
34 mins ago

@Edmund Thanks for your explanation! I understand now!
â€“Â cwei
19 mins ago

You must keep in mind that everything is an expression. Your calculation results in a symbolic sum of five terms; see FullForm[A.M.B]. Dimensions is returning the dimensions of the head of the expression (Plus) which has 5 terms.
â€“Â Edmund
34 mins ago

@Edmund Thanks for your explanation! I understand now!
â€“Â cwei
19 mins ago

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
4
down vote

accepted

Dimensions works for arbitrary heads. If you restrict to only allowing a List head:

Dimensions[B.M.A, AllowedHeads->List]

answered 1 hour ago

Carl Woll

62.4k281158

Thank you for your reply! However, what I really want is to make sure this result is a scalar. This is only an Intermediate result. I need to use it do more complicated linear algebra computation.
â€“Â cwei
58 mins ago

1

@cwei I fail to see why my code doesn't suffice. Perhaps you could use ListQ[B.M.A] instead.
â€“Â Carl Woll
56 mins ago

Or ArrayQ. Maybe also TensorQ?
â€“Â Henrik Schumacher
49 mins ago

@CarlWoll Problem solved! Your answer is correct! Thanks!
â€“Â cwei
40 mins ago

@HenrikSchumacher Yes! They work as well! Thanks!
â€“Â cwei
40 mins ago

add a commentÂ |Â

up vote
1
down vote

Here's a simple example:

a = RandomReal[-1, 1, 10];
b = RandomReal[-1, 1, 10];
m = RandomReal[-1, 1, 10, 10];
a.m.b

This does give a scalar answer, as expected.

answered 41 mins ago

bill s

51.8k375146

Thanks! Yes! It works for numbers. But for an expression involved with variables. Dimension[expr] does not work properly if you don't specify head explicitly.
â€“Â cwei
32 mins ago

I was showing you that your result is a scalar, irrespective of what Dimensions is telling you.
â€“Â bill s
30 mins ago

Yes! I understand your example.
â€“Â cwei
21 mins ago

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
4
down vote

accepted

Dimensions works for arbitrary heads. If you restrict to only allowing a List head:

Dimensions[B.M.A, AllowedHeads->List]

answered 1 hour ago

Carl Woll

62.4k281158

Thank you for your reply! However, what I really want is to make sure this result is a scalar. This is only an Intermediate result. I need to use it do more complicated linear algebra computation.
â€“Â cwei
58 mins ago

1

@cwei I fail to see why my code doesn't suffice. Perhaps you could use ListQ[B.M.A] instead.
â€“Â Carl Woll
56 mins ago

Or ArrayQ. Maybe also TensorQ?
â€“Â Henrik Schumacher
49 mins ago

@CarlWoll Problem solved! Your answer is correct! Thanks!
â€“Â cwei
40 mins ago

@HenrikSchumacher Yes! They work as well! Thanks!
â€“Â cwei
40 mins ago

add a commentÂ |Â

up vote
4
down vote

accepted

Dimensions works for arbitrary heads. If you restrict to only allowing a List head:

Dimensions[B.M.A, AllowedHeads->List]

answered 1 hour ago

Carl Woll

62.4k281158

Thank you for your reply! However, what I really want is to make sure this result is a scalar. This is only an Intermediate result. I need to use it do more complicated linear algebra computation.
â€“Â cwei
58 mins ago

1

@cwei I fail to see why my code doesn't suffice. Perhaps you could use ListQ[B.M.A] instead.
â€“Â Carl Woll
56 mins ago

Or ArrayQ. Maybe also TensorQ?
â€“Â Henrik Schumacher
49 mins ago

@CarlWoll Problem solved! Your answer is correct! Thanks!
â€“Â cwei
40 mins ago

@HenrikSchumacher Yes! They work as well! Thanks!
â€“Â cwei
40 mins ago

add a commentÂ |Â

up vote
4
down vote

accepted

Dimensions works for arbitrary heads. If you restrict to only allowing a List head:

Dimensions[B.M.A, AllowedHeads->List]

answered 1 hour ago

Carl Woll

62.4k281158

Dimensions works for arbitrary heads. If you restrict to only allowing a List head:

Dimensions[B.M.A, AllowedHeads->List]

answered 1 hour ago

Carl Woll

62.4k281158

answered 1 hour ago

Carl Woll

62.4k281158

answered 1 hour ago

Carl Woll

62.4k281158

answered 1 hour ago

Carl Woll

62.4k281158

Thank you for your reply! However, what I really want is to make sure this result is a scalar. This is only an Intermediate result. I need to use it do more complicated linear algebra computation.
â€“Â cwei
58 mins ago

1

@cwei I fail to see why my code doesn't suffice. Perhaps you could use ListQ[B.M.A] instead.
â€“Â Carl Woll
56 mins ago

Or ArrayQ. Maybe also TensorQ?
â€“Â Henrik Schumacher
49 mins ago

@CarlWoll Problem solved! Your answer is correct! Thanks!
â€“Â cwei
40 mins ago

@HenrikSchumacher Yes! They work as well! Thanks!
â€“Â cwei
40 mins ago

add a commentÂ |Â

Thank you for your reply! However, what I really want is to make sure this result is a scalar. This is only an Intermediate result. I need to use it do more complicated linear algebra computation.
â€“Â cwei
58 mins ago

1

@cwei I fail to see why my code doesn't suffice. Perhaps you could use ListQ[B.M.A] instead.
â€“Â Carl Woll
56 mins ago

Or ArrayQ. Maybe also TensorQ?
â€“Â Henrik Schumacher
49 mins ago

@CarlWoll Problem solved! Your answer is correct! Thanks!
â€“Â cwei
40 mins ago

@HenrikSchumacher Yes! They work as well! Thanks!
â€“Â cwei
40 mins ago

Thank you for your reply! However, what I really want is to make sure this result is a scalar. This is only an Intermediate result. I need to use it do more complicated linear algebra computation.
â€“Â cwei
58 mins ago

@cwei I fail to see why my code doesn't suffice. Perhaps you could use ListQ[B.M.A] instead.
â€“Â Carl Woll
56 mins ago

Or ArrayQ. Maybe also TensorQ?
â€“Â Henrik Schumacher
49 mins ago

@CarlWoll Problem solved! Your answer is correct! Thanks!
â€“Â cwei
40 mins ago

@HenrikSchumacher Yes! They work as well! Thanks!
â€“Â cwei
40 mins ago

add a commentÂ |Â

up vote
1
down vote

Here's a simple example:

a = RandomReal[-1, 1, 10];
b = RandomReal[-1, 1, 10];
m = RandomReal[-1, 1, 10, 10];
a.m.b

This does give a scalar answer, as expected.

answered 41 mins ago

bill s

51.8k375146

Thanks! Yes! It works for numbers. But for an expression involved with variables. Dimension[expr] does not work properly if you don't specify head explicitly.
â€“Â cwei
32 mins ago

I was showing you that your result is a scalar, irrespective of what Dimensions is telling you.
â€“Â bill s
30 mins ago

Yes! I understand your example.
â€“Â cwei
21 mins ago

add a commentÂ |Â

up vote
1
down vote

Here's a simple example:

a = RandomReal[-1, 1, 10];
b = RandomReal[-1, 1, 10];
m = RandomReal[-1, 1, 10, 10];
a.m.b

This does give a scalar answer, as expected.

answered 41 mins ago

bill s

51.8k375146

Thanks! Yes! It works for numbers. But for an expression involved with variables. Dimension[expr] does not work properly if you don't specify head explicitly.
â€“Â cwei
32 mins ago

I was showing you that your result is a scalar, irrespective of what Dimensions is telling you.
â€“Â bill s
30 mins ago

Yes! I understand your example.
â€“Â cwei
21 mins ago

add a commentÂ |Â

up vote
1
down vote

Here's a simple example:

a = RandomReal[-1, 1, 10];
b = RandomReal[-1, 1, 10];
m = RandomReal[-1, 1, 10, 10];
a.m.b

This does give a scalar answer, as expected.

answered 41 mins ago

bill s

51.8k375146

Here's a simple example:

a = RandomReal[-1, 1, 10];
b = RandomReal[-1, 1, 10];
m = RandomReal[-1, 1, 10, 10];
a.m.b

This does give a scalar answer, as expected.

answered 41 mins ago

bill s

51.8k375146

answered 41 mins ago

bill s

51.8k375146

answered 41 mins ago

bill s

51.8k375146

answered 41 mins ago

bill s

51.8k375146

Thanks! Yes! It works for numbers. But for an expression involved with variables. Dimension[expr] does not work properly if you don't specify head explicitly.
â€“Â cwei
32 mins ago

I was showing you that your result is a scalar, irrespective of what Dimensions is telling you.
â€“Â bill s
30 mins ago

Yes! I understand your example.
â€“Â cwei
21 mins ago

add a commentÂ |Â

Thanks! Yes! It works for numbers. But for an expression involved with variables. Dimension[expr] does not work properly if you don't specify head explicitly.
â€“Â cwei
32 mins ago

I was showing you that your result is a scalar, irrespective of what Dimensions is telling you.
â€“Â bill s
30 mins ago

Yes! I understand your example.
â€“Â cwei
21 mins ago

Thanks! Yes! It works for numbers. But for an expression involved with variables. Dimension[expr] does not work properly if you don't specify head explicitly.
â€“Â cwei
32 mins ago

I was showing you that your result is a scalar, irrespective of what Dimensions is telling you.
â€“Â bill s
30 mins ago

Yes! I understand your example.
â€“Â cwei
21 mins ago

add a commentÂ |Â

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