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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 $U$?

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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.

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I'm a little confuse with the part that if the identity matrix $I$ is a subspace, then $cI$ has to be in the subspace for all numbers $c$. If an identity matrix is multiplied by a constant $c$, it is still diagonal. Adding a identity matrix to another diagonal matrix is still a diagonal matrix. Then this satisfies the linearity rules but isn't it still a subspace? eg: $$3\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}=\begin{bmatrix}3 & 0\\ 0 & 3\end{bmatrix}$$ $$\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}+\begin{bmatrix}4 & 0\\ 0 & 4\end{bmatrix}=\begin{bmatrix}5 & 0\\ 0 & 5\end{bmatrix}$$ –  xenon Sep 24 '11 at 14:55
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The collection of $K$ of all matrices of the shape $0$ off diagonal, constant on diagonal is a subspace. This collection contains the matrices in your comment above, and many more. The identity matrix by itself is not a subspace. This space $K$ is a subspace of the space $D$ of (all) diagonal matrices, which is a subspace of the space of (all) upper triangular matrices, which is a subspace of the space of all $2\times 2$ matrices. Remember, with one trivial exception, any of our vector spaces has infinitely many elements. Don't confuse a single matrix with a space of matrices. –  André Nicolas Sep 24 '11 at 15:10
    
oh yea.. you are right. I think I got the single matrix and space of matrices confused. Thanks! –  xenon Sep 24 '11 at 15:15
    
Yes, my post was almost totally concerned with trying to explain that single object is never (well, almost never!) a subspace. It was clear that this was the source of your difficulty. A space is a space, lots of stuff. –  André Nicolas Sep 24 '11 at 15:19

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.

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