I Singular matrices and complex entries

joshmccraney

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?

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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.

joshmccraney

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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2018 Award
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.

fresh_42

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2018 Award
$x = -i$ is a solution.

I'm assuming $x$ is a complex number.
Oops. Right.

joshmccraney

$x = -i$ is a solution.

I'm assuming $x$ is a complex number.
Typo on my part, yea sorry.

WWGD

Gold Member
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.

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PeroK

Homework Helper
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2018 Award
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

WWGD

Gold Member
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.

"Singular matrices and complex entries"

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