I have to find two variables $a,b \in {\bf Z}$ that the given $n \times n$ matrix A becomes orthogonal. \begin{equation*} A = \begin{pmatrix} 1&2 \\ a&b \end{pmatrix} \end{equation*} I know that a $n \times n$ matrix is called orthogonal if $A^TA$ $=$ $id$ which means: \begin{equation*} \begin{pmatrix} 1&a \\ 2&b \end{pmatrix} \cdot \begin{pmatrix} 1&2 \\ a&b \end{pmatrix} = \begin{pmatrix} 1+a^2&2+ab \\ 2+ab&4+b^2 \end{pmatrix} \stackrel{?}{=} \begin{pmatrix} 1&0 \\ 0& 1 \end{pmatrix} \end{equation*}
How to find two variables $a,b \in {\bf Z}$ that the matrix $A$ is orthogonal
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linear-algebra
matrices
orthogonality
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0You are already done, because this means that $A^TA\neq I_2$ for all $a,b$. – 2017-01-03
3 Answers
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The system of equations has no solution at all, since $1+a^2=1$ implies $a=0$ and hence $2+ab=2\neq 0$. hence $A^TA\neq I_2$ for all $a,b$. This is, given the computation in the question, much more immediate than to use that the norm of a column vector must be $1$ for an orthogonal matrix.
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0I see, i had the same thought with $a = 0$ but i was confused because the question was to find a,b, but it is impossible for $a,b\in Z$, as shown. – 2017-01-03
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There is no solution since the norm of the column vector of an orthogonal matrix must be $1$. (Note that the second column is $(2,b)^T$.)
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You are right. Now solve the system of equation in terms of $a$ and $b$. However, if $b$ is restricted to be an integer, the equation $$4 + b^2 = 1$$ does not have a solution.
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0It does not seem to have a solution at all, since the top-left element forces $a^2 = 0$ and then the top-right element $2 + ab = 0$ becomes $2 + 0 = 0$. Or am I missing something? – 2017-01-03
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0@DietrichBurde sorry was confused by the "if $b$ is restricted to be an integer", it seemed to matter. – 2017-01-03
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0@CompuChip What I mean is that $4+b^2=1$ does not give a contradiction, say, over the complex numbers, but your argument (and mine) does. – 2017-01-03