# Vector Space Axioms which fail certain matrices

gaborfk
Second week in Linear Algebra...

My homework involves of identifying all failing Vector Space Axioms for various sets of vector spaces. I did fine with a "regular" set like (x,y,z) which has an operation like k(x,y,z)=(kx,y,z). I have worked through all 10 of the axioms, comparing left sides with right sides to figure out which ones fail.

However, now I have to determine if three matrices fail any Vector Space Axioms. I have no clue on how to do matrices.

Here they are:

$$\left(\begin{array}{cc}a&1\\1&b\end{array}\right)$$

$$\left(\begin{array}{cc}a&0\\0&b\end{array}\right)$$

$$\left(\begin{array}{cc}a&a+b\\a+b&b\end{array}\right)$$

I would appreciate any step by step example of another matrix for like axiom 5 which has to do with finding the zero vector, 0+u=u+0=u where 0 is not zero, but a zero vector or for this instance, a zero matrix. I am not looking for a solution, but a method of solving these vector spaces.

Thank you very much

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## Answers and Replies

amcavoy
If I remember correctly, if the following three hold true, then the rest will as well:

1. For each $\vec{u},\vec{v}\in W$, $\vec{u}+\vec{v}\in W$.

2. For all $c\in\mathbb{R}$ and $\vec{u}\in W$, $c\vec{u}\in W$.

3. The vector space contains the zero vector.

Muzza
No, consider for example W = R^2 with the usual addition operator, but we define c * (x, y) = (x^2, y). It fails distributivity.

Science Advisor
Homework Helper
First, I have a problem with the wording. It's not the axioms that "fail", its the set of matrices! Yes, you could say "which axioms fail to be true for this set of matrices" but I would prefer "which axioms does this set of matices fail to satisfy".

In any case- do the obvious. Write down all of the axioms for a vectors space, try each of you generic matrices in each and see which are true and which aren't.

For example, one axiom is "closure" of addition. If you add two matrices of the first kind given:
$$\left(\begin{array}{cc}a&1\\1&b\end{array}\right) + \left(\begin{array}{cc}c&1\\1&d\end{array}\right)$$
Is that still of the same form?

gaborfk
No it is not. It is in the form of

$$\left(\begin{array}{cc}x&2\\2&y\end{array}\right)$$

So that means it is open for addition and scalar multiplication...

However, how would I go about checking if there is a zero vector?

Thank you

amcavoy
Well, what happens when you plug in x=0, y=0? The constants remain, so what does that tell you?

Science Advisor
Homework Helper
If U is a subspace of vector space V, then U "inherits" its operations from V- adding an multiplying are the same in both so the additive and multiplicative identities ('0' and '1') must be the same. All you need to do to check if U has a additive identity is to check if the 0 matrix is in it.
By they way, in order to be a vector space, a set of vectors must satisfy ALL the axioms. Once you have show that the space is not closed under addition or multiplication (I would not say "open" under the operations!) it is not necessary to check the other axioms. Even if they were true, the set could not be a subspace.