# Book: Exercise2

Square root of identity matrix.

Consider the {$2x2$} identity matrix. What is its square root?

{$$\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}=\begin{pmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix}\begin{pmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix}$$}

This yields equations of the form:

{$${a_{11}}^2 + a_{12}a_{21} = 1$$}

{$$a_{12}a_{22} + a_{11}a_{12} = 0$$}

The latter equation means {$a_{12}=0$} or {$a_{22} = -a_{11}$}.

Similarly, by symmetry, {$a_{21}=0$} or {$a_{22} = -a_{11}$}.

Combining, we have ({$a_{12}=0$} and {$a_{21}=0$}) or {$a_{22} = -a_{11}$}.

Solving further, this yields the following two possibilities:

{$\begin{pmatrix}\pm1 & 0\\ 0 & \pm1\end{pmatrix}$} or {$\begin{pmatrix}\sqrt{1-ab} & a\\ b & -\sqrt{1-ab}\end{pmatrix}$} or {$\begin{pmatrix}-\sqrt{1-ab} & a\\ b & \sqrt{1-ab}\end{pmatrix}$}

But the first case and the second case match when a or b = 0. Thus the answer is:

{$\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}$} or {$\begin{pmatrix}-1 & 0\\ 0 & -1\end{pmatrix}$} or {$\begin{pmatrix}\sqrt{1-ab} & a\\ b & -\sqrt{1-ab}\end{pmatrix}$} or {$\begin{pmatrix}-\sqrt{1-ab} & a\\ b & \sqrt{1-ab}\end{pmatrix}$}

Note that {$a$} and {$b$} can be any complex number. However, if we want a real matrix, then we must have {$a$} and {$b$} real such that {$ab \leq 1$}.

Parsiųstas iš http://www.ms.lt/sodas/Book/Exercise2
Puslapis paskutinį kartą pakeistas 2019 sausio 19 d., 19:49