Let \(S\) be the set of matrices of the form
\[
\begin{pmatrix}a & a\\
a & a
\end{pmatrix},
\]
where \(a\) is any real non-zero number. Show that \(S\) is closed under
matrix multiplication and, further, that \(S\) is a group under matrix
multiplication.
Let \(G\) be a set of \(n\times n\) matrices which is a group
under matrix multiplication, with identity element \(\mathbf{E}.\)
By considering equations of the form \(\mathbf{BC=D}\) for suitable
elements \(\mathbf{B},\) \(\mathbf{C}\) and \(\mathbf{D}\) of \(G\), show
that if a given element \(\mathbf{A}\) of \(G\) is a singular matrix
(i.e. \(\det\mathbf{A}=0\)), then all elements of \(G\) are singular.
Give, with justification, an example of such a group of singular matrices
in the case \(n=3.\)
Solution:
Let $\mathbf{A} = \begin{pmatrix}1 & 1\\
1 & 1
\end{pmatrix}\(, then we need to show that \)(a\mathbf{A})(b\mathbf{A})\( is of the form \)cA\( where \)a, b, c \neq 0$.
Since $\mathbf{A}^2 = \begin{pmatrix}2 & 2\\
2 & 2
\end{pmatrix} = 2\mathbf{A}\( this is certainly the case, since \)(a\mathbf{A})(b\mathbf{A}) = 2ab\mathbf{A}$.
To check that we have a group be need to check:
Closure (done)
Associativity (inherited from matrix multiplication)
Identity (\(\frac12 \mathbf{A}\))
Inverses the inverse of \(a\mathbf{A}\) is \(\frac{1}{4a}\mathbf{A}\)
Suppose \(\mathbf{A}\) is singular (ie \(\det\mathbf{A}=0\)), then \(\mathbf{AA^{-1}B=B}\) (where inverse is the group inverse rather than the matrix inverse) for any matrix \(\mathbf{B}\). Taking determinants we have:
\(\det(\mathbf{AA^{-1}B}) = \det(B) \Rightarrow \det(A) \det(A^{-1}B) = \det(B) \Rightarrow 0 = \det(B)\), ie all matrices are singular.
Consider the set of non-zero multiples of \(\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}\), then the same logic as part (i) will suffice
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