What about it? You are not saying anything about that ratio.
Okay, so x is inand .
No, that was not what you wrote. What you should have written is for some fixed value of c.Here, in set notation, I have defined that the subset in question, W, has a zero vector "together" with all vectors, x.
You are aware that a "linear combination of the zero vector" with any vector is just that vector times a scalar aren't you? And, in particular, that " " is just ?By "together", I presume they mean the linear combination of the zero vector with all vectors.
That makes no sense. and are numbers, there is no question of "constant" or "variable" for numbers. What this means is that if and are two vectors in this set thenNext, I observe that has a constant value. What does it mean to be a constant value? I asked myself. To be a constant value, and must also be constant values; they do not contain variable terms.
All you are saying is that if one of the components is 0, the other must also be 0.In addition, if we suppose that be equal to 0, then the quotient is undefined—which also is not a constant value. The latter is my argument for why this subset is not a subspace of the indiciated vector space.
The vector x = 0 + (0,2) would not belong in W because is undefined. Which also qualifies that it is not a constant value.
The problem is that you are completely mistaken as to what " has a constant value" means. You seem to be under the impression that it simply means "is a finite number" and that is not true. "Constant" means "always has the same value". (1, 1) and (-1, 1) can NOT be in such a subspace because the ratio of 1 to 1 is 1 and the ratio of -1 to 1 is -1. Those are not the same number so the is NOT constant. I strongly recommend you review basic definitions since you are using a number of words, including "constant" in incorrect ways, as well as using meaningless notation such as "[tex]0+ (x_1, x_2)[tex]".If we imagined two vectors which do belong in W, namely x = 0 + (1,1) and y = 0 + (-1,1). These two vectors indeed belong in W because (1,1) and (-1,1) are vectors in that have constant values together with the zero vector.
However, the linear combination of the two produces a vector that does not belong in W, thereby violating the subspace critera,
Subspace criterion. If every vector of the form
belongs to W whenever and belong to W, and and are arbitrary sclars, then W is a subspace of V.
I take this rule strictly because the subspace criterion states that is is a subspace if every vector is of the form.