some subspace questions

fasterthanjoao

I'm preparing for an end of year linear algebra test which briefly covers things about subsets of Matnxn and their relations to subspaces of Matnxn; I found the following question:

Which of the following subsets of Matn×n are subspaces of Matn×n?

(i) Symmetric matrices (i.e. matrices A such that AT = A).
(ii) Skew-symmetric matrices (i.e. matrices A such that AT = −A).
(iii) Orthogonal matrices (i.e. matrices A such that AT = A−1).
(iv) Singular matrices.
(v) Invertible matrices.
(vi) Echelon matrices.
(vii) Reduced echelon matrices.

----
I'd appreciate some guidance on how to tell which of the above apply.. is it the same as showing that vectors in a subset are subspaces when they're closed under addition and linear multiplication?

thanks.

Timbuqtu

Yes, the matrices are the vectors of the vectorspace Mat(nxn).

For example, (i) is a subspace, because:
If A and B are symmetric and r a scalar, then A^t=A, B^t=B. So (A+B)^t=A^t + B^t = A+B is symmetric and (r A)^t = r A^t = r A is symmetric.

matt grime

Exactly where do you struggle? It would be helpful for you to write out your attempt at showing one of them is, or isn't, a subspace, so we can see where you have problems.

1. Is it the definition of a subspace?
2. Is it in seeing whether these will or will not be subspaces?
3. Is it in writing out a proof?

fasterthanjoao

2. Is it in seeing whether these will or will not be subspaces?

--
which also, clearly, means I can't write a proof. It seems as though all thats needed is to check whether they're closed under addition and multiplication. I don't see how the fact that a matrix is singular or non is relevant. The fact that a matrix can be reduced to echelon form shows the vectors it represents are linearly independent but is there anything else that needs said from there? Can the same be said of a reduced echelon form? the only difference is that the echelon form represents a less trivial basis?

shmoe

The usual subspace tests apply, your matrices are 'vectors'. You're thinking of Matn×n as a vector space after all.

If you add two invertible matrices do you always end up with an invertible matrix?

matt grime

Firstly, it does not refer anywhere to the set of matrices that *can* be put in row echelon form, it refers to the matrices that *are* in row echelon form. Every matrix *can* be put in (reduced) row echelon form by row operations.

If I remember correctly reduced echelon form is where the first non-zero entry in any row is a 1. Take some matrix of that form. Do you think that every scalar mutliple of such a matrix will have a 1 as its first nonzero entry?

One obvious thing to check is if 0 is in the subspace.


So, step 1. Try and prove that the definitions hold. If you can't prove it from the definitions of the *subset* try and find a counter example. In simple cases like this if you can't do the former easily, then it must fail to be a subspace. Find the/a reason.

It is impossible to prove that A invertible and B invertible implies A+B is invertible, isn't it? So try and find a counter example. (Of course it fails for an easier reason than that to be a subspace, as I've already alluded to.) Whereas it is easy to show that if A is symmetric and B is symmetric then so is A+B (and tA for any scalar t).

Answers in maths (beyond highschool) shouldn't just don't jump out at you I'm afraid (unless you happen to be exceptional), you need to sit and stare at the questions for a while and play around with things until you see what's going on. If you don't see what singular or nonsingular has to do with the vector space structure then it's a good bet that the singular or nonsingular matrices are not a subspace. Singularity is a property of multiplication and therefore not necessarily going to be respected by addition.

mathwonk

basically inverse images of linear functions and linear operations, are flat like subspaces, and inverse images under more complicated operations are not.

forming a transpose is a linear operation so the inverse image of zero under A-At is a subspace. multiplication is not linear, so the inverse image of the identity under the product AAt is not a subspace, also we are taking the inverse image of 1 not zero.

the singular matrices are the inverse image of zero, but under the determinant function. does that function seem linear to you? what is the degree of the determinant as a function of the entries?

etc etc...