Question_4

Question_4

(a)

We have:

\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix}a&b\\ c&d\end{pmatrix} = \begin{pmatrix}a&b\\0&0\end{pmatrix}

and:

\begin{pmatrix}a&b\\ c&d\end{pmatrix}\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix}a&0 \\ c & 0 \end{pmatrix}

So A commutes with \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} iff:

\begin{pmatrix}a&b \\ 0& 0\end{pmatrix} = \begin{pmatrix}a&0 \\ c &0 \end{pmatrix}

That is, iff b = 0, c = 0. (and a=a, but that’s a tautology)

(b)

We have:

\begin{pmatrix}a & b \\ c&d\end{pmatrix} \begin{pmatrix}1&1\\0&0\end{pmatrix}= \begin{pmatrix}a&a \\ c&c \end{pmatrix}

and:

\begin{pmatrix}1&1\\0&0\end{pmatrix} \begin{pmatrix}a & b \\ c&d\end{pmatrix} = \begin{pmatrix}a + c & b +d \\ 0 & 0\end{pmatrix}

So A commutes with \begin{pmatrix}1&1\\0&0\end{pmatrix} iff:

\begin{pmatrix}a + c & b +d \\ 0 & 0\end{pmatrix} = \begin{pmatrix}a&a \\ c&c \end{pmatrix}

That is, iff c = 0, a = b + d. So the set of matrices commuting with \begin{pmatrix}1&1\\0&0\end{pmatrix} is:

\left\{\begin{pmatrix}b + d & b \\ 0 & d\end{pmatrix} \mid b, d \in \mathbb R\right\}

© Clearly for a 2 \times 2 matrix to commute with all 2 \times 2 matrices, it must commute with both \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} and \begin{pmatrix}1&1\\0&0\end{pmatrix}. For a matrix \begin{pmatrix}a&b \\ c & d\end{pmatrix} to commute with the former it must have b=c=0, and for it to also commute with the latter it must also have a = b + d = d. So the only candidates for matrices commuting with all matrices are of the form:

\begin{pmatrix}a&0\\0&a\end{pmatrix} = aI

for a \in \mathbb R.

Clearly, for any 2 \times 2 matrix A we have:

A(aI) = (Aa)I = a(AI) = aA and (aI)A = a(IA) = aA

So the set of 2 \times 2 matrices commuting with every 2 \times 2 matrix is \{aI \mid a \in \mathbb R\}, as could be expected.

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Exciting times!!!