Determine which of the following subsets of R n. with n ge 1. are in fact subspa
ID: 1888139 • Letter: D
Question
Determine which of the following subsets of R n. with n ge 1. are in fact subspaces. Justify your answers. {x R n : xi ge 0 i = 1. ,n}; {x R n : X 1 = 0}; {x R n : x1x2 = 0 n ge 2}: {x R n : x1 + + x n=0}: {x R n : x1 + + x n = 1}; {x R n : Ax = b. A 0, b 0}. Determine which of the following subsets of F n, n, with n ge 1, are in fact subspaces. Justify your answers. {A F n, n : A = A T}; {A F n, n : A invertible}; {A F n, n : A not invertible}; {A F n, n : A upper-triangular}; {A F n, n : A2 = A}: {A F n, n : tr (A) = 0}. Given a matrix X F n, n, define {A F n, n : [A. X] = 0}.Explanation / Answer
I agree that if you want someone to do this rigorously from scratch it is too much for one problem. However, I leave the cases where you have a subspace to be shown on your own (they are hardly difficult, just tedious to type). Those that are NOT subspaces are easy to spot.
(a) Not a subspace since it is not closed. Let ( x_{1} = 1) and (y_{1}=2) of some vectors (x) and (y). Then clearly (x-y) is not in this subspace.
(b) Is a subspace. You've only fixed what (x_{1}) is. There is no linear combination of vectors that would make (x_{1} eq 0), thus closure passes. The (0) matrix is obviously here, as are additive inverses (0 is its own additive inverse).
(c) Not a subspace, fails closure. Suppose (u,v) vectors in our subspace. Noting first that (0) must be equal to (u_{1}), (u_{2}) or both (equivalently for (v), then the the only issue is if (u_{1}=0,u_{2}=a) and (v_{1}=b,v_{2}=0), with (a,b eq 0) (ie the 0's in are not in the same spot in both vectors). Then (u-v) will yield (-b eq 0) and (-a eq 0) in the first and second spots of our new vector. Thus, with (a,b eq 0), we have a vector not in this subspace.
(d) Is a subspace. The zero vector is obviously in here. Closure works because if (u=(u_{1},. . .,u_{n})) and (v=(v_{1},. . .,v_{n})), such that (u_{1}=. . .u_{n}=0) and (v_{1}+. . .v_{n}=0), then (u+v=(u_{1}+v_{1},. . .u_{n}=v_{n}). It is clear that the sum of your entries will be 0. Scalar multiplication works as well, as you are just scaling 0, which is always 0 (and consequently gives you additive inverses). Speaking of which. . .
(e) Is not a subspace. Fails closure by scalar multiplication. If ( u = (u_{1},. . .,u_{n})) is in your subspace, and we scale it by (r in mathbb{R}) such that (r eq 1), then the sum of the entries will be (r eq 1). Thus (ru) is not in your subspace.
(f) Is not a subspace. Fails closure by scalar multiplication. Let (r in mathbb{R}) such that eq 1). Then (A(rx) = rAx=rb eq b). Thus we fail closure. An easier check would be that (x = 0 ) can not be in this subspce, but closure is "funner".
Next set:
(a) Is a subspce. Closure is quite easy to check. The zero matrix is obviously equal to it's transpose. The only "non-obvious" one might be additive inverses, but it should be obvious that scaling a matrix by negative one, will still have it equal to it's transpose - it's just all negatives.
(b) Not a subspace. Does not contain the zero matrix (since it is NOT invertible).
(c) Not a subspace. Fails closure. Take (2 imes 2) matrices. Let (A=(1,0,0,0), B=(0,1,0,0), C=(0,0,1,0)) and (D=(0,0,0,1)). Each of those have determinant (0), and are thus not-invertible. However their sum is just the matrix with all (1) in its entries, which IS invertible. Can extend to (n imes n) matrices.
(d) Is a subspace. Will pass closure obviously (the lower side will still be 0 under scalar multiplication and addition). Additive inverses are also clearly there as is the 0 matrix (which is upper - and lower - triangular).
(e) May not be a subspace. We would want it to pass closure, meaning (A+B=(A+B)^{2}). However ( (A+B)^{2}=AA+BA+AB+BB). This will only equal (A+B) if (AB+BA=0); ie are additive inverses. Obviously, with (A,B) arbitrary matrices, this wont always work. Thus not a subspace.
(f) Is a subspace. Is much like (d) in the first problem. This will pass closure, as well as additive inverses. The zero matrix is obviously in here.
(g) I am not familiar with the operation ([A,X]). You will need to define this.
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