Linear Algebra: Linearity of inner product

[teddyayalew]

Homework Statement

An inner product is linear in both components.

Homework Equations

<x,y> = <conjugate(y),conjugate(x)>
<x+y,z> = <x,z> +<y,z>

The Attempt at a Solution

I thought it was true . It is obvious that it is linear for the first component by definition
Attempt to show it is for second component:

<x,y+z> = <conjugate(y+z),conjugate(x)>
=<conjugate(y),conjugate(x)> + <conjugate(z),conjugate(x)>
= <x,y> + <x,y>


But the answer is false. I am having trouble understanding why it is not linear in both components. The answer key says that the second component is conjugate- linear.

 

[Ray Vickson]

Homework Statement

An inner product is linear in both components.

Homework Equations

<x,y> = <conjugate(y),conjugate(x)>
<x+y,z> = <x,z> +<y,z>

The Attempt at a Solution

I thought it was true . It is obvious that it is linear for the first component by definition
Attempt to show it is for second component:

<x,y+z> = <conjugate(y+z),conjugate(x)>
=<conjugate(y),conjugate(x)> + <conjugate(z),conjugate(x)>
= <x,y> + <x,y>


But the answer is false. I am having trouble understanding why it is not linear in both components. The answer key says that the second component is conjugate- linear.

What is the vector space you are working in? What is the relevant definition of inner product?

RGV

 

[teddyayalew]

It doesn't say which space. It was just a true or false statement, but I know that the entries of the vectors can be from the complex space. And it doesn't tell me which definition of the inner product but I believe it is the standard inner product.

 

[Ray Vickson]

It doesn't say which space. It was just a true or false statement, but I know that the entries of the vectors can be from the complex space. And it doesn't tell me which definition of the inner product but I believe it is the standard inner product.

In your OP YOU said "It is obvious that it is linear for the first component by definition", implying that you had some definition in mind--otherwise, why use the word 'definition'? So, are you now saying that is not true?

If we assume you are speaking of C^n with &lt;x,y&gt; = \sum_{i=1}^n \bar{x_i} y_i, it is definitely linear in both x and y. Your "equation"
&lt;\text{conjugate}(y),\text{conjugate}(x)&gt; = &lt;x,y&gt; is false, if by "conjugate" you mean "complex conjugate". It is not even true in one dimension, with &lt;x,y&gt; = \bar{y}x.

RGV