Finding basis made of uninvertible matrices Let there be transformation $T: \mathbb R_3[X] \rightarrow M_{2 \times 2}(\mathbb R)$, $T(ax^3+bx^2+cx+d)=\left[ \begin{matrix}
        a+d & b-2c \\
        a+b-2c+d & 2c-b  \\
               \end{matrix} \right] $
Find a basis of $Im(T)$ made of non-invertible matrices. 
So we get that $Im(T)= sp\{\left[ \begin{matrix}
        1 & 0 \\
        1 & 0  \\
               \end{matrix} \right], \left[ \begin{matrix}
        0 & 1 \\
        1 & -1  \\
               \end{matrix} \right], \left[ \begin{matrix}
        0 & -2 \\
        -2 & 2  \\
               \end{matrix} \right],\left[ \begin{matrix}
        1 & 0 \\
        1 & 0  \\
               \end{matrix} \right]\}=sp\{\left[ \begin{matrix}
        1 & 0 \\
        1 & 0  \\
               \end{matrix} \right], \left[ \begin{matrix}
        0 & 1 \\
        1 & -1  \\
               \end{matrix} \right] \}$
And that's also the basis. The first matrix is not invertible, but the second one  is. How do I find a basis made of non-invertible matrices?
Thank you for your time! 
 A: $sp\{\left[ \begin{matrix}
        1 & 0 \\
        1 & 0  \\
               \end{matrix} \right], \left[ \begin{matrix}
        0 & 1 \\
        1 & -1  \\
               \end{matrix} \right] \} = 
sp\{\left[ \begin{matrix}
        \frac{1}{2} & 0 \\
        \frac{1}{2} & 0  \\
               \end{matrix} \right], \left[ \begin{matrix}
        -\frac{1}{2} & 1 \\
        \frac{1}{2} & -1  \\
               \end{matrix} \right] \}$ 
that should do it.
A: Solve for $a$ where
$$\det\left(a\cdot\begin{bmatrix}
 1 & 0 \\
 1 & 0
\end{bmatrix}+
\begin{bmatrix}
 0 & 1 \\
 1 & -1
\end{bmatrix}\right)=0$$
The solution to that is $a=-\frac 12$. The resulting matrix with zero determinant is
$$\begin{bmatrix}
 -\frac 12 & 1 \\
  \frac 12 & -1
\end{bmatrix}$$
which can replace the second matrix in your basis and answer your question.
This works because replacing a vector in a basis with a sum of that vector with a linear combination of other vectors in the basis does not change the span of the basis. And of course a non-invertible (singular) matrix has determinant zero.
