# Help: All subspaces of 2x2 diagonal matrices

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1. Sep 24, 2015

### kostoglotov

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 $R^2$

2. Sep 24, 2015

### WWGD

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 $(a, 0,0,b) :a, b \in \mathbb R$.

3. Sep 24, 2015

### FireGarden

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.

4. Sep 24, 2015

### WWGD

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 $\mathbb R^4$.

5. Sep 25, 2015

### FireGarden

Oh, I didn't read the requirement for the matrices to be diagonal. We still get some of the 1 dimensional subspaces and the zero subspace anyway - the second and third entries must be zero to be diagonal, but we could just as well fix the first and/or fourth to be zero, and we will still have a diagonal matrix. I'm not sure why the answer claims there are 3 dimensional subspaces in this case though..