Is the subspace of $2 \times 2$ matrices over $\mathbb{R}$ consisting of diagonalizable matrices with determinant $0$ a Topological Manifold? The question is as stated in the title.
Consider the space $M(2,\mathbb{R})$ of $2\times 2$ real matrices, and the subspace $X$ of $M(2,\mathbb{R})$ consisting of all the diagonalizable matrices with determinant $0$. Is $X$ a topological manifold without boundary?
Intuitively, it is not a topological manifold, but it is difficult for me to construct an argument explaining why.
Edit: Perhaps it should be clearer that I'm specifically concerned with the question of whether it is a topological manifold. As Jason Devito's answer suggests, it is not a smooth submanifold with the subspace topology.
 A: Building on Jason's answer that $X$ is not a smooth manifold with the subspace topology, we show $X$ is not a topological manifold.
Indeed, noted in Jason's answer, letting $0$ denote the zero matrix, $X-\{0\}$ is a smooth submanifold of $M(2,\mathbb{R})$ of dimension 3.
Another way to see this is to recall that the space $Y$ of matrices with rank exactly $1$ is a smooth submanifold of codimension $(2-1)(2-1) = 1$, and so dimension equal to $3$. Obeserve that $X-\{0\} = Y\cap \text{trace} ^{-1}(\mathbb{R}-\{0\})$. Indeed, every non-zero matrix of $X$ has non-zero trace and rank equal to $1$. Conversely, any matrix with rank equal to 1 and non-zero trace must have 2 distinct eigenvalues so is diagonalizable.
Now let $U$ be a neighbourhood of $0$. I claim that $U-\{0\}$ is disconnected. Indeed, take any matrix $A$ in $U$ with positive trace, and any matrix $B$ in $U$ with negative trace. This can always be done since the matrix $\begin{bmatrix}
\lambda &0 \\
0 & 0
\end{bmatrix}$ lies in $U$ for sufficiently small $\lambda$.
Consider a continuous path $\gamma$ in $U$ connecting $A$ and $B$. Then by the intermediate value theorem, $\text{trace}(\gamma(c)) = 0$ for some $c$ so that $\gamma$ contains the $0$ matrix in its image.
Thus $U-\{0\}$ is disconnected for any neighbourhood $U$ containing $0$.
But if $X$ were a topological manifold, then by invariance of dimension it must be locally homeomorphic to $\mathbb{R}^3$. Therefore it has a neighborhood $U$ around $0$ that is homeomorphic to $\mathbb{R}^3$ and so cannot be disconnected by removing a single point $0$.
This is a contradiction so $X$ is not a topological manifold.
A: No, it is not a manifold.
First note that $X$ is path connected:  If $A\in X$, then $tA\in X$ for all real $t$, so we can join any element $A\in X$ to $0 = \begin{bmatrix} 0 & 0\\ 0 & 0\end{bmatrix}\in X$.  In addition, this implies that $X$ is a union of lines through $0\in M(2,\mathbb{R})$.
Now, assume for a contradiction that $X$ is a manifold.   And consider the $0$ matrix $0\in X$ as well as the four curves $\gamma_1(t) = \begin{bmatrix} t & 0 \\ 0 & 0\end{bmatrix}$, $\gamma_2(t) =\begin{bmatrix} 0 & 0\\ 0 & t\end{bmatrix}$, $\gamma_3(t) = \begin{bmatrix} t & t\\ 0 & 0\end{bmatrix}$, and $\gamma_4(t) = \begin{bmatrix} t & 0\\ t & 0\end{bmatrix}.$
I will leave it to you to verify that $\gamma_i(0) = 0$ for any $i$, and that $\gamma_i(t)\in X$ for all $t$ and for any $i$.
Thus, $\gamma_i'(0)\in T_0 X$, which easily implies that $T_0 X$ is $4$-dimensional.   Thus, there is a neighborhood $U$ (in $M(2,\mathbb{R})$Q) about $0$ which lies entirely in $X$.  But $X$ is a union of lines through $0$, and every such line hits $U$.  Thus, we conclude $X = M(2,\mathbb{R})$, giving an obvious contradiction.  Thus, $X$ cannot be a manifold at the point $0$.
One may object that $0$ is quite special here.  I agree.  In fact, we have the following result.

If $Y = X\setminus \{0\}$, then $Y$ is a manifold.

To see this, first consider $Z = \{A\in M(2,\mathbb{R}): A\neq 0, \det A = 0\}$.  As shown in Robert Israel's comment to Georges Elencwagj's answer to this MSE question, $Z$ is a smoothly embedded submanifold in $M(2,\mathbb{R})$.
We then claim that $Y$ is the intersection of $Z$ with an open subset $V$ of $M(2,\mathbb{R})$, which will establish the fact that $Y$ is a smoothly embedded submanifold.
Let $V = \left\{\begin{bmatrix} a&b\\c&d\end{bmatrix}\in M(2,\mathbb{R}): a+d\neq 0\right\}$.  Then we claim that $Y = Z\cap V$.
To see this, suppose $A = \begin{bmatrix} a&b\\c&d\end{bmatrix}\in Z$ but $A\notin V$.  Then $A$ has an eigenvalue of $0$ with algebraic multiplicity $2$.  However, since $A\in Z$, $A$ has rank at least $1$, so its Jordan form must also have rank at least $1$.  This implies the Jordan form is $\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix}$, so $A$ is not diagonalizable.  That is, $A\notin Y$.
On the other hand, assume that $A=\begin{bmatrix} a&b\\c&d\end{bmatrix}\in Z\cap V$.  Then $a+d\neq 0$, so $A$ has distinct eigenvalues of $0$ and $a+d$.  Thus, $A$ is diagonalizable, so $A \in Y$.
