# Determine whether the set $H$ of all matrices form a subspace of $M_{2 \times 2}$

Determine if the set $Z$ of all matricies form $\left[ \begin{array}{cc} a & b \\ 0 & d \end{array} \right]$ is a subspace of $M_{2 \times 2}$ (the set of all $2 \times 2$ matrices).

% This is something I came up with. Can someone look at it and let me know any useful corrections/suggestions to the question please.

Without specification as to the nature of $a,b$ and $d$, it is assumed that $a,b,d \in \mathbb{R}$

Hence, $H$ is determined to be a subspace of $M_{2 \times 2}$ because it is closed under scalar addition and scalar multiplication and contains the zero vector when $a=b=d=0$.

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@Thomas: This seems correct -- the set appears to be closed under addition as well as the scalar multiplication, and contains the zero element. – InterestedGuest Mar 11 '11 at 8:11

What you have done looks reasonable except for one tiny point. When you say "because it is closed under scalar addition and scalar multiplication", you should delete the first (but not the second) scalar, as you are trying to show that adding two elements give a third member of the subspace.

So you should end with

Hence, $H$ is determined to be a subspace of $M_{2 \times 2}$ because it is closed under addition and scalar multiplication and contains the zero vector when $a=b=d=0$.

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As Tobias and Calle pointed out in comments elsewhere, $H$ is also closed under matrix multiplication. This does not affect whether H is a subspace, but does mean it is also a subring of $M_{2 \times 2}$. – Henry Mar 12 '11 at 17:19

Let $A= \left[ \begin{array}{cc} a & b \\ 0 & d \end{array} \right]$ and $B= \left[ \begin{array}{cc} x & y \\ 0 & z \end{array} \right]$. Then $(AB)_{2,1}=dz$ and not $0$. It doesn't seem to be closed under multiplication on its own set.

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You need to put $...$ or $$...$$ around your $\TeX$ expressions. – Henry Mar 11 '11 at 8:30
You are correct that $H$ is not closed under matrix multiplication, but to be a subspace it only needs to be closed under scalar multiplication. – Henry Mar 11 '11 at 8:34
Now i'm learning. Oh, i see. I'm not all that aware of all the definitions, you're right. – Andr Mar 11 '11 at 8:51
Actually, the set is closed under matrix multiplication as well, so it is a subring (and a subalgebra). It is just not an ideal. – Tobias Kildetoft Mar 11 '11 at 9:32
$(AB)_{2,1} = 0$ actually. $(AB)_{2,2} = dz$. – Calle Mar 11 '11 at 12:07