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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!

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  • $\begingroup$ «uninvertible» does not exist: we say «singular». $\endgroup$ – Mariano Suárez-Álvarez May 30 '15 at 19:36
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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.

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$sp\{\left[ \begin{matrix} 1 & 0 \\ 1 & 0 \\ \end{matrix} \right], \left[ \begin{matrix} 0 & 1 \\ 1 & -1 \\ \end{matrix} \right] \} = sp\{\left[ \begin{matrix} \frac{1}{2} & 0 \\ \frac{1}{2} & 0 \\ \end{matrix} \right], \left[ \begin{matrix} -\frac{1}{2} & 1 \\ \frac{1}{2} & -1 \\ \end{matrix} \right] \}$

that should do it.

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