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 $U$?
Best Answer
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.