Let $V = M_2(\mathbb{R})$ and $W = \lbrace A \in M_2(\mathbb{R}) : A = A^T\rbrace$.
Determine if $W$ is a subspace of $V$.
I am a bit confused about the vector space of $2\times 2$ matrices and how to start over with the transpose function.
Subspace of $M_2(\mathbb{R})$
linear algebra
Related Solutions
(i) You are right, the upper triangular matrices are a vector subspace of the vector space of $2\times 2$ matrices. But you need to prove this. Nothing hard involved, all we need to do is to show that (a) the sum of two upper triangular matrices is upper triangular, and (b) that an upper triangular matrix times a constant is upper triangular.
The full vector space of $2\times 2$ matrices is basically $\mathbb{R}_4$. But instead of writing the entries in the usual $(a,b,c,d)$ format, we write them as a $2\times 2$ array. So the space of $2\times 2$ matrices has a basis consisting of the $4$ matrices that have a $1$ in one place, and $0$'s elsewhere.
Three of these basis elements are upper triangular. So the space of upper triangular matrices has dimension at least $3$. And the dimension is definitely not $4$.
The name $U_2$is harmless, just a name.
(ii) The name is $GL_2$, meaning the general linear group. But don't worry about the name. It is, however, a standard name, and some other year you may learn more about $GL_n$. In this case, we have been told just enough about it to answer the question.
This is not a subspace of the space of $2\times 2$ matrices. To show this, it is enough to show either that there are two invertible matrices whose sum is not invertible, or that there is an invertible $2\times 2$ matrix $M$ and a constant $c$ such that $cM$ is not invertible.
Both are easy, and you only need to do one. Think $c=0$. Or think about adding together two diagonal matrices, one the identity matrix and the other $\dots$.
(iii) You need to show that the sum of two matrices of the shape described is of the shape described, and that a matrix of the shape described, times a constant, is of the shape described.
Basically, this subspace is the same as the set of all $3$-tuples of the shape %(a,b,c,-a)$.
To find the dimension, maybe show that the three matrices which if written "flat" are $(1,0,0,-1)$, $(0,1,0,0)$, and $(0,0,1,0)$ are of the right form, and linearly independent, and in our subspaxe. So the space has dimension at least $3$. But the dimension is not $4$, else it would be the whole space of $2\times 2$ matrices.
Hint. Let $\textsf W$ be a subspace of a finite-dimensional vector space $\textsf V$. If $\{ w_1,\dots,w_k\}$ is a basis for $\textsf W$ and $\{w_1,\dots,w_k,w_{k+1},w_{k+2}\dots,w_n\}$ an extension of this basis to a basis for the whole space $\textsf V$, then $$\{w_{k+1}+\textsf{W}, w_{k+2}+\textsf{W}, \dots, w_n+\textsf{W}\}$$ form a basis for the quotient space $\textsf{V}/\textsf{W}$.
Best Answer
Take $A,B\in W$ and arbitrary scalars $a,b\in\mathbb{R}$, then : \begin{align}(aA+bB)^T&=(aA)^T+(bB)^T\\ &= a^TA^T+b^TB^T\\& = aA+bB\end{align}
so $aA+bB\in W$ and thus it is a vector subspace.