Can we show that all $2 \times 2$ matrices are sums of matrices with determinant 1?

I was considering several cases when I realized that $$ \begin{bmatrix} a&b\\c&d \end{bmatrix}= \begin{bmatrix} a&1\\-1&0 \end{bmatrix} + \begin{bmatrix}1&b\\0&1 \end{bmatrix} + \begin{bmatrix} -1&0\\c&-1 \end{bmatrix} +\begin{bmatrix}0&-1\\1&d\end{bmatrix}. $$ This avoids the problem of the first technique that requires dealing with the cases where some entries are zero.

As for the second technique, I cannot imagine what is the idea.


It is sufficient to show that the matrices $\left(\begin{matrix} a & 0 \\ 0 & 0 \end{matrix}\right)$, $\left(\begin{matrix} 0 & b \\ 0 & 0 \end{matrix}\right)$, $\left(\begin{matrix} 0 & 0 \\ c & 0 \end{matrix}\right)$, and $\left(\begin{matrix} 0 & 0 \\ 0 & d \end{matrix}\right)$ can be written as the sum of matrices with determinant $1$.

Now, notice that \begin{align}\left(\begin{matrix} a & 0 \\ 0 & 0 \end{matrix}\right) &= \left(\begin{matrix} \frac{a}{2} & 1 \\ -1 & 0 \end{matrix}\right) + \left(\begin{matrix} \frac{a}{2} & -1 \\ 1 & 0 \end{matrix}\right) \\ \left(\begin{matrix} 0 & b \\ 0 & 0 \end{matrix}\right) &= \left(\begin{matrix} 1 & \frac{b}{2} \\ 0 & 1 \end{matrix}\right) + \left(\begin{matrix} -1 & \frac{b}{2} \\ 0 & -1 \end{matrix}\right) \\ \left(\begin{matrix} 0 & 0 \\ c & 0 \end{matrix}\right) &= \left(\begin{matrix} 1 & 0 \\ \frac{c}{2} & 1 \end{matrix}\right) + \left(\begin{matrix} -1 & 0 \\ \frac{c}{2} & -1 \end{matrix}\right) \\ \left(\begin{matrix} 0 & 0 \\ 0 & d \end{matrix}\right) &=\left(\begin{matrix} 0 & 1 \\ -1 & \frac{d}{2} \end{matrix}\right) + \left(\begin{matrix} 0 & -1 \\ 1 & \frac{d}{2} \end{matrix}\right). \end{align}

Tags:

Linear Algebra

Matrices

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