Prove that Tr(B) is rank(B) if B is any 2x2 matrix such that $B^2 = B$

Question

How do I prove this? I proved in the last part that Tr(B) was an integer, but I'm lost on how to prove this fact.

I'm guessing the first step would be to put B in rref form, but I'm not sure how I'd go from there

Where Tr() is trace and rank() is the rank.

Answer 1

It's not just $2 \times2$ matrices. If $B^2=B$ then $B$ is diagonalisable with eigenvalues either $0$ or $1$. The multiplicity of $1$ as an eigenvalue is the rank of $B$.

Answer 2

This is a guide, not a full answer.

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Break it into the cases of rank 0, 1, 2. Rank 0 is pretty simple. Rank 1 is easy to do if you think about what rank means for the matrix elements (think about the two rows independently). Rank 2 is easy because it has full rank so it is invertible. Apply an inverse to your equation for $B$ and see what happens.

Answer 3

An outline with just a few gaps you can fill in. This approach tries to only use the very basic definition of rank as the count of the number of pivot positions. No treating matrices as linear operators on vectors. It's not very elegant in this form, but it is elementary.

There are only three possibilities for the rank: $0$, $1$, or $2$.

• If the rank of $B$ is $0$, then there is a trivial reason why you can conclude its trace is $0$.
• If the rank of $B$ is $2$, then $B$ must be the identity matrix. Why? Because through row operations, $B\sim I$. And row operations are equivalent to left-multiplying by elementary matrices. So $B=E_1\cdots E_kI=E_1\cdots E_k$. But $B^2=B$ implies $(E_1\cdots E_k)(E_1\cdots E_k)=E_1\cdots E_k$. Since the elementary matrices are invertible, this means $E_1\cdots E_k=I$. So we've established $B=E_1\cdots E_k=I$. So now it's trivial why the trace must be $2$.
• If the rank of $B$ is $1$, then either $B\sim\begin{bmatrix}1&k\\0&0\end{bmatrix}=:U,$ or $B\sim\begin{bmatrix}0&1\\0&0\end{bmatrix}=:V$. Similar to the rank 2 case, consider $$B=E_1\cdots E_kU\qquad\text{or}\qquad B=E_1\cdots E_kV$$ and then do this trick with $U$ and $V$, factoring each as a $2\times1$ by a $1\times2$: $$B=E_1\cdots E_k\begin{bmatrix}1\\0\end{bmatrix}\begin{bmatrix}1&k\end{bmatrix}\qquad\text{or}\qquad B=E_1\cdots E_k\begin{bmatrix}1\\0\end{bmatrix}\begin{bmatrix}0&1\end{bmatrix}$$ $$B=\left(E_1\cdots E_k\begin{bmatrix}1\\0\end{bmatrix}\right)\begin{bmatrix}1&k\end{bmatrix}\qquad\text{or}\qquad B=\left(E_1\cdots E_k\begin{bmatrix}1\\0\end{bmatrix}\right)\begin{bmatrix}0&1\end{bmatrix}$$ $$B=J\begin{bmatrix}1&k\end{bmatrix}\qquad\text{or}\qquad B=J\begin{bmatrix}0&1\end{bmatrix}$$ where $J$ is the left column of $E_1\cdots E_k$ and therefore cannot be a column of all zeros. Since $B^2=B$, you have $$J\begin{bmatrix}1&k\end{bmatrix}J\begin{bmatrix}1&k\end{bmatrix}=J\begin{bmatrix}1&k\end{bmatrix}\qquad\text{or}\qquad J\begin{bmatrix}0&1\end{bmatrix}J\begin{bmatrix}0&1\end{bmatrix}=J\begin{bmatrix}0&1\end{bmatrix}$$ $$J^TJ\begin{bmatrix}1&k\end{bmatrix}J\begin{bmatrix}1&k\end{bmatrix}=J^TJ\begin{bmatrix}1&k\end{bmatrix}\qquad\text{or}\qquad J^TJ\begin{bmatrix}0&1\end{bmatrix}J\begin{bmatrix}0&1\end{bmatrix}=J^TJ\begin{bmatrix}0&1\end{bmatrix}$$ $$c\begin{bmatrix}1&k\end{bmatrix}J\begin{bmatrix}1&k\end{bmatrix}=c\begin{bmatrix}1&k\end{bmatrix}\qquad\text{or}\qquad c\begin{bmatrix}0&1\end{bmatrix}J\begin{bmatrix}0&1\end{bmatrix}=c\begin{bmatrix}0&1\end{bmatrix}$$ with $c>0$ (because $J$ was not all zeroes) so you may divide by $c$ to get: $$\begin{bmatrix}1&k\end{bmatrix}J\begin{bmatrix}1&k\end{bmatrix}=\begin{bmatrix}1&k\end{bmatrix}\qquad\text{or}\qquad \begin{bmatrix}0&1\end{bmatrix}J\begin{bmatrix}0&1\end{bmatrix}=\begin{bmatrix}0&1\end{bmatrix}$$ There is a very similar way to act on the right side to get to: $$\begin{bmatrix}1&k\end{bmatrix}J=1\qquad\text{or}\qquad \begin{bmatrix}0&1\end{bmatrix}J=1$$ which implies $$\operatorname{tr}\left(\begin{bmatrix}1&k\end{bmatrix}J\right)=1\qquad\text{or}\qquad \operatorname{tr}\left(\begin{bmatrix}0&1\end{bmatrix}J\right)=1$$ But within the argument of trace, you may commute to get: $$\operatorname{tr}\left(J\begin{bmatrix}1&k\end{bmatrix}\right)=1\qquad\text{or}\qquad \operatorname{tr}\left(J\begin{bmatrix}0&1\end{bmatrix}\right)=1$$ $$\operatorname{tr}(B)=1\qquad\text{or}\qquad \operatorname{tr}(B)=1$$