How is diagonal matrices a subspace of upper triangular matrices? I'm kind of confused that the diagonal matrices is a subspace of the upper triangular matrices. Suppose I have the following matrices:
$$U=\begin{bmatrix}
a & b\\ 
0 & d
\end{bmatrix}
; \;
D=\begin{bmatrix}
a & 0\\ 
0 & d
\end{bmatrix}$$
I read in the book that the matrix $D$ is a subspace of $U$. But how's it so? At first, I thought the reason why the diagonal matrices are subspaces of the upper triangular matrices was because the the matrix $D$ is just a case of matrix $U$ with $b=0$ and that's why it is a subspace of it.
But on a second thought, if my assumption was true, then everything, even the identity matrix is a subspace of the upper triangular matrices and the upper triangular matrices would be subspace of any 2 by 2 matrices. This doesn't make sense at all.
So how is this being looked at that $D$ is a subspace of $U$?
 A: Generally to check that a given subset is a Subspace of a vector space you should check to things:
a)
The operation $*$ of the vector space $U$ (in your case i assume to be addition between $2\times2$ matrices) is closed in $D$ i.e. it is well define as a bilinear map $$+:D\times D\to D,$$ and this is easy to check.
b)
You have to check that multiplication by scalar elements of the field (usually your matrices are defined over a field, i.e real/complex matrices, or more generally $K$ - matrices where $K$ is a field), is well define as a bilinear map $$\cdot:K\times D \to D.$$
Now you should check that both these two operations are well defined on $D$ as I have described previously.
But then you can onclude that $D$ is a subspace of $U$.
Note that in this case, since you are working with square matrices, you can as well have a product defined in $U$, namely $$*:U\times U\to U,$$ which is the usual product between matrices. Equipped with this operation, $U$ becomes more than a vector space, it becomes an Algebra. The interesting thing is that, if you consider the restriction to the subspace $D$, you obtain another bilinear map $$*:D\times D\to D,$$ and so in this case you can view $D$ as a subalgebra of $U$, not only as a subspace.
Now let's come to your doubts:
The identity matrix is not a subspace: for example what happens if you multiply by a scalar element of the field? you obtain something different from the identity but this contradicts b).
Finally the space of the upper triangular matrices is a subspace of $M_2(K)$, the vector space of the $2\times 2$ matrices over a generic field $K$. But, as i pointed out before, this is more than just a subspace, it is a subalgebra of $M_2(K)$.
Recalling everything.. You have the follwing $$D\leq U\leq M_2(K).$$ Where $\leq$ means "to be a subspace of." In your case this can be read even as "to be a subalgebra of." Hope everything is clearer now.
A: I don't know whether you are dealing with $2\times 2$ matrices or general $n \times $n$ matrices.  The result is true in either case.
It may not be clear to you what these spaces are.  Define addition of matrices by adding corresponding entries. So for example 
$$\begin{bmatrix}
1 & 2\\ 
0 & 3
\end{bmatrix}
+
\begin{bmatrix}
5 & 3\\ 
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
6 & 5\\ 
0 & 4
\end{bmatrix}
$$
If $c$ is a constant (a scalar, a number) then you multiply a matrix by $c$ by multiplying each entry by $c$. So for example
$$3\begin{bmatrix}
1 & 2\\ 
0 & 3
\end{bmatrix}
=
\begin{bmatrix}
3 & 6\\ 
0 & 9
\end{bmatrix}
$$
A vector space of matrices is a collection $V$ of matrices (of the same size) such that if $A$ and $B$ are matrices in the collection, then so is the sum $A+B$, and also if $c$ is any scalar, then $cA$ is in the collection.  
So typically a vector space of matrices will have many matrices in it. The only vector space of matrices that consists of a single matrix is the space whose only element is the all $0$'s matrix.  
In particular, the identity matrix by itself ($1$'s down the main diagonal, $0$'s elsewhere) is not a subspace of the collection of $2\times 2$ matrices, for if the identity matrix $I$ is in the subspace, then $cI$ has to be in the subspace for all numbers $c$.  The collection of all matrices which are $0$ off diagonal, and have all diagonal terms equal is a subspace of the space of all matrices.  Maybe that will take care of part of your objection.
Let $V$ be any vector space, and take a collection $U$ of some of the elements of $V$.  Then $U$ is called a subspace of $V$ if $U$ by itself is a vector space, meaning that the sum of any two elements of $U$ is in $U$, and any constant times an element of $U$ is in $U$.
You quoted something to the effect that a certain $D$ is a subspace of the space of upper triangular matrices.  That's not true.  The collection of all matrices of the shape you described, with everything off diagonal equal to $0$, is a subspace.  So $D$ is supposed to be not a single matrix, it is a largish collection of matrices. 
Now let's look at your particular problem. Let $V$ be the collection of all upper triangular matrices. Is this a vector space?  Take any two upper triangular matrices $A$ and $B$. Is $A+B$ upper triangular? Yes. If $c$ is a constant, and $A$ is upper triangular, is $cA$ upper triangular? Yes. So $V$ is a vector space.
Let $D$ be the collection of all diagonal matrices? Is this a vector space? Yes, the sum of two diagonal matrices is diagonal, a constant times a diagonal matrix is a diagonal matrix. $D$ is a subspace of the upper triangular matrices, because any diagonal matrix is in particular upper triangular, it is a special upper triangular matrix.
