Question

Determine which of the following subsets of R^n are subspaces of R^n (n>2).
a) The symmetric matrices.
b) The nonsingular matrices
c) The diagonal matrices.
d) {x/(Summation(from j=1 to n)) of xj = 0}

Answer 1

First, what is a symmetric matrix?, its definition already implies it has to be a nxn matrix. Second,nonsingular, same argument, it is nonsingular if and only if it has nonzero determinant, and only square matrices have nonzero determinant. Same for diagonal,only square matrices have diagonals,and same for d).

Answer 2

Although this definitions are correct, you're still to show that you have a closure under addition and scalar multiplication. If you manage to do that for the above defined subsets, then you have shown that they are subspaces.

Answer 3

Let \(C= \lbrace A \in \text{Mat}_{n \times n} ( \mathbb{R}) \mid A^T=A\rbrace\) be the set of symmetric Matrices.
So for a) Your argumentation can be like that Of course \(0^{n \times n}\) is a symmetric square matrix. So we don't need to worry about our zero element.

Now suppose that \(A \in C\) and \(B \in C\). You're supposed to show that \(A+B\) is also an element of \(C\). Is that the case? Well clearly you have \[\large (A+B)^T=A^T+B^T=A+B \implies A+B \in C \]
You can do the same for scalars, which is trivial in the case of transposing and you have shown that you have a subset.

Answer 4

i think c and d are the only subsets that are subspaces of R^n. what do you think?

Answer 5

I think above I have just shown you that the symmetric matrices form a subspace of \(\mathbb{R}^n\) (Although in my understanding it would be way better to denote \(\mathbb{R}^{n \times n }\) here)

Answer 6

Is there something that confuses you about it @BGrg007 ?

Answer 7

Lets look at b). It is a very famous group, usually denoted as \(\text{GL}_n( \mathbb{F}) \) where \(\mathbb{F}\) denotes a field, it is called the general linear group. Now is it a subspace? Can you show that the sum of two invertible matrices does not necessarily lead to an invertible matrix?

Answer 8

someone told me that if i added an identity matrix I with its opposite matrix(1's for 0's, 0's for 1's), i would get a singular matrix would make the nonsingular matrix not a subspace. true?

Answer 9

exactly, if you add \(I_n + -I_n = 0_N\) and the determinant of the null matrix is equal to 0, therefore not invertible.

Answer 10

so, so far you have a) is a subspace, b is not, for c you only need wording, you can of course implement some formulas and use congruent matrices, but frankly I believe that will make things only look unnecessarily complicated.

Answer 11

ok. as for d, the summation of each column is 0? idk how to figure this out

Answer 12

Can you maybe rewrite d? Or is the way as written in your question? \[\large \frac{x}{\sum_{j=1}^{n}x_j} \] or is it just \(\sum x_j=0\)

Answer 13

oh im sorry. its x "such that". the vertical bar, not slash bar. silly me

Answer 14

oh it's a set \(D= \lbrace x \in \mathbb{R}^n \mid \sum x_j=0 \rbrace \)

Answer 15

What have you tried? For the addition? It sure looks like a subset to me :)

Answer 16

b