In this question, \(\mathbf{A,\mathbf{B\)
}}and \(\mathbf{X\) are non-zero \(2\times2\) real matrices.}
Are the following assertions true or false? You must provide a proof or a counterexample in each case.
The equation \(\mathbf{AX=0}\) has a non-zero solution \(\mathbf{X}\)
if and only if \(\det\mathbf{A}=0.\)
For any \(\mathbf{A}\) and \(\mathbf{B}\) there are at most two matrices
\(\mathbf{X}\) such that \(\mathbf{X}^{2}+\mathbf{AX}+\mathbf{B}=\mathbf{0}.\)
This is also false, using the same matrices from part (i), we find:
\begin{align*}
(\mathbf{A - B})(\mathbf{A + B}) &= \mathbf{A}^2-\mathbf{BA}+\mathbf{AB}-\mathbf{B}^2 \\
&= \mathbf{A}^2-\mathbf{B}^2+\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} \\
&\neq \mathbf{A}^2-\mathbf{B}^2
\end{align*}
This is true. Claim: The equation \(\mathbf{AX=0}\) has a non-zero solution \(\mathbf{X}\) if and only if \(\det\mathbf{A}=0.\)
Proof: \((\Rightarrow)\) Suppose \(\det\mathbf{A} \neq 0\) then \(\mathbf{A}\) has an inverse, and so we must have \(\mathbf{A}^{-1}\mathbf{AX} = \mathbf{0} \Rightarrow \mathbf{X} = \mathbf{0}\).
\((\Leftarrow)\) Suppose \(\det \mathbf{A} = 0\) then \(ad-bc=0\), so consider the matrix \(\mathbf{X} = \begin{pmatrix} d & d\\ -c & -c\end{pmatrix}\) (or if this is zero, \(\mathbf{X} = \begin{pmatrix} a & a\\ -b & -b\end{pmatrix}\))
This is false. Consider \(\mathbf{A} = \mathbf{B} = \mathbf{0}\), then \(\mathbf{X} = \begin{pmatrix} 0 & x \\ 0 & 0\end{pmatrix}\) has the property that \(\mathbf{X}^2 = \mathbf{0}\) for all \(x\), so at least more than 2 values