I Singular matrices and complex entries

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Hi PF!

Let's say we have a matrix that looks like $$
A = \begin{bmatrix}
1-x & 1+x \\
i & 1
\end{bmatrix} \implies\\ \det(A) = (1-x) -i(1+x).
$$
I want ##A## to be singular, so ##\det(A) = 0##. Is this impossible?
 

fresh_42

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No. It would be possible over ##\mathbb{Z}_2## but then ## i ## doesn't make sense, as ##x^2+1\in \mathbb{Z}_2[x]## isn't irreducible.
 
1,766
69
No. It would be possible over ##\mathbb{Z}_2## but then ## i ## doesn't make sense, as ##x^2+1\in \mathbb{Z}_2[x]## isn't irreducible.
Right, so the only solution is when ##x=i##?
 

fresh_42

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Right, so the only solution is when ##x=i##?
No, this would yield ##\det(A)=2-2i## if we assume ##A\in \mathbb{M}_2(\mathbb{C})##. But before you carry on: What is ##i ##? Where is ##x## supposed to be from? And does ##1## represent the multiplicative neutral, i.e. ##1\neq 0\,?##
 

PeroK

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Right, so the only solution is when ##x=i##?
##x = -i## is a solution.

I'm assuming ##x## is a complex number.
 

WWGD

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Hi PF!

Let's say we have a matrix that looks like $$
A = \begin{bmatrix}
1-x & 1+x \\
i & 1
\end{bmatrix} \implies\\ \det(A) = (1-x) -i(1+x).
$$
I want ##A## to be singular, so ##\det(A) = 0##. Is this impossible?
I thunk by the Fundamental theorem of Algebra it must have a root. Not that I am disagreeing with the proposed solution x=-i, just a comment.
 

PeroK

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I thunk by the Fundamental theorem of Algebra it must have a root. Not that I am disagreeing with the proposed solution x=-i, just a comment.
Whether the determinant can be zero or not depends on whether you get a quadratic, linear equation (in ##x##) or a constant. If, for example, you change the matrix to:
$$
A = \begin{bmatrix}
1-x & 1+x \\
-1 & 1
\end{bmatrix}
$$
Then there is no solution.
 

WWGD

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Whether the determinant can be zero or not depends on whether you get a quadratic, linear equation (in ##x##) or a constant. If, for example, you change the matrix to:
$$
A = \begin{bmatrix}
1-x & 1+x \\
-1 & 1
\end{bmatrix}
$$
Then there is no solution.
Ok, good point, I did not consider hat possibility. Edit: But the complex geometry behind is not as obvious as if x were purely Real.
 

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