# Help: All subspaces of 2x2 diagonal matrices

## Main Question or Discussion Point

The exercise is: (b) describe all the subspaces of D, the space of all 2x2 diagonal matrices.

I just would have said I and Z initially, since you can't do much more to simplify a diagonal matrix. I cannot understand how D is $R^4$, let alone the rest of the answer. I kind of get why there'd be orthogonal subspaces in that case, since it's diagonal...but that's just grasping at straws.

I can see how we might take the columns of D and form linear combinations from them, but those column vectors are in $R^2$

Related Linear and Abstract Algebra News on Phys.org
WWGD
Gold Member
Maybe they are using the identification of a matrix $(a_{ij})$ with the $i \times j$-ple (i.e., a point in $\mathbb R^{i \times j}$) given by : $(a_{11}, a_{12},...., a_{ij})$ , i.e., you use a double-alphabet ordering to do the identification. $2 \times 2$ diagonal matrices are then identified with the set $(a, 0,0,b) :a, b \in \mathbb R$.

I cannot understand how D is $R^4$,
In short, the set of 2x2's with real entries is just a silly way of writing $\mathbb{R}^4$.

A 2x2 matrix is of course going to be a set of 4 independent real numbers. Independent in the sense that the elements do not constrain one another. We add component-wise, and we perform scalar multiplication component-wise. Really, this is exactly how we work with row/column vectors. We've just written them down differently. Thinking of them as actual matrices is misleading, I think. The question's solution then follows by describing (very generally) that the subspaces are just (any!) spaces of dimension 0,1,2 and 3. 1D subspaces always have to pass through the zero vector, that's nothing special about this case.

WWGD
In short, the set of 2x2's with real entries is just a silly way of writing $\mathbb{R}^4$.
But the diagonal matrices are already a subspace of $\mathbb R^4$ whose 2nd, 3rd entries are both $0$. That makes it into a 2-dimensional subspace of $\mathbb R^4$.
But the diagonal matrices are already a subspace of $\mathbb R^4$ whose 2nd, 3rd entries are both $0$. That makes it into a 2-dimensional subspace of $\mathbb R^4$.