Is Theorem 4.5 Linear Despite Condition A Failing?

[g.lemaitre]

Homework Statement

this is theorem 4.5

The Attempt at a Solution

The book says that condition A fails. How can you even know? Condition A has a w in it and there is no W in the above question.

 

[Muphrid]

The lowercase $u, v, w$ are arbitrary vectors in some vector space. In the statement of the theorem, they are placeholders. In fact, so are the "vectors" you've been given. Indeed, you could've been given

In Exercises 9-11, let

$$A = \begin{bmatrix} a_1 & a_2 \\ a_3 & a_4 \end{bmatrix}, \quad B = \begin{bmatrix} b_1 & b_2 \\ b_3 & b_4 \end{bmatrix}$$

Check that parts (a)-(d) of theorem 4.5 are satisfied for the following definitions of the inner product of $A$ and $B$:

11. $A \cdot B = |a_1 b_1| + |a_2 b_2| + |a_3 b_3| + |a_4 b_4|$

It would not change the meaning of the problem in the slightest because all these symbols are being used for is to denote what an element of this vector space is and how the inner product is defined between elements.

You should interpret lowercase $u$, $v$, and $w$ as meaning arbitrary vectors in the space, and the theorem should hold for any such vectors in the space. All the problem statement has told you is that, well, vectors in this space are 2x2 matrices.

 

[g.lemaitre]

The lowercase $u, v, w$ are arbitrary vectors in some vector space. In the statement of the theorem, they are placeholders. In fact, so are the "vectors" you've been given. Indeed, you could've been given
It would not change the meaning of the problem in the slightest because all these symbols are being used for is to denote what an element of this vector space is and how the inner product is defined between elements.
You should interpret lowercase $u$, $v$, and $w$ as meaning arbitrary vectors in the space, and the theorem should hold for any such vectors in the space. All the problem statement has told you is that, well, vectors in this space are 2x2 matrices.

But if there are only two vectors then how can it satisfy the condition for linearity

(u+v)*w = u*w + v*w

which is a condition that applies to 3 vectors

 

[Muphrid]

They are two examples of vectors; they are not the only two vectors that exist in that space. The space in question is that of $M_{22}$, the space of all 2x2 matrices. Any 2x2 matrix is a valid vector here. Think of what they gave you as more of a statement about the form a vector in this space can take. It should be clear that you can come up with any third vector, any third 2x2 matrix, just with four arbitrarily-labeled components.

In fact, once they told you that the space was that of 2x2 matrices, they didn't need to give you explicit forms for the vectors at all except to be able to easily write a formula for the inner product.

 

[g.lemaitre]

I think I understand