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I have several subspaces where I have to determine their dimension and whether they are affine or linear? These are my answers- are they correct? Thanks for help!

a) $X = \{ x \in R^n | a^Tx = 0 \}, a \in R^n$ is given

  • linear since any linear combination $\alpha x + \beta y, x, y \in X$ is in X. I believe its affine, too, since actual coefficients $\alpha, \beta$ doesnt really matter in this case because $\alpha(a_1x_1 + \dots + a_nx_n) + \beta(a_1y_1 + a_ny_n) = 0$ every time...
  • dimension? Im guessing its max $n-1$ but Im not sure

b) $X = \{ x \in R^n | a^Tx = c \}, a \in R^n, c \in R$ are given

  • affine because in order for the linear combination to be in $X$, $\alpha_i$ must sum to 1
  • dimension again max $n-1$?

c) $X = \{ x \in R^n | x^Tx = 1 \}$

  • affine (same justification as in b))

d) $X = \{ x \in R^n | a^Tx = I \}, a \in R^n$ is given

  • not sure at all Are my assumptions correct
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How are you defining "affine"? Based on your answers to (b) and (c), you appear to be defining affine as "not linear", but I doubt that's what you want. – user108903 Jan 3 '13 at 12:58
My bad, I just realized taht every affine combination is linear too.. edited – Smajl Jan 3 '13 at 13:04
up vote 2 down vote accepted

a) Are you familiar with concept of null spaces? set of all vectors $x$ such that $Ax=0$ for a given matrix $A$ lies in the null space of $A$.Take $A=a^T$. Every linear set is also affine, but not every affine set is linear. Clue: Add a given constant non-zero vector to any given subspace (vector should not be from that subspace), then the subspace won't contain orgin after that, thus it won't be linear, but it will be affine.

Now, if you are familiar with rank-nullity theorem which says $rank(A)+dim(null(A))=number_of_columns$. Here $A=a^T$ has only one non-zero row, so $rank(A)=1$. $dim(null(A))$ is dimension of null space of A which is what you are looking for.

b) Again, that forms a affine space, but not linear (since, it doesn't contain orgin).

I am not sure how to define dimension in this case, since it is not a subspace. But one can talk about the dimension as if it is a surface. Number of variables is $n$. Number of constraints defining the surface is $1$. So dimension is $n-1$.

c) It is not an affine space. Take $x_1=[1,0,\dots,0]$, $x_2=[0,1,0,\dots,0]$. Both vectors satisfy your conditions. Add them, do you get a point in the same set? Intuitively, the set given is set of points in the n-dimensional sphere (n=2, it is the circle).

d) What is $I$ here?

share|cite|improve this answer
I'd suggest "... but not every affine set is linear" as a slightly clearer wording for your boldfaced mantra :-) – user108903 Jan 3 '13 at 13:07
I agree!! :) and I am changing it – dineshdileep Jan 3 '13 at 13:09
I is the same as E (matrix with 1 on diagonal and 0 elsewhere).. thanks for your asnwer.. and sorry for bad formulation – Smajl Jan 3 '13 at 13:09
$a^T$ is $1 \times n$, $x$ is $n \times 1$, so $a^Tx$ should be $1 \times 1$. So either it is same as question $b$, or some mistake in question formulation. – dineshdileep Jan 3 '13 at 13:12
$a^Tx \in R^{n \times n}$ – Smajl Jan 3 '13 at 13:15

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