Singular matrices and complex entries

[member 428835]

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]

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.

 

[member 428835]

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]

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]

Right, so the only solution is when $x=i$?

$x = -i$ is a solution.

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

 

[fresh_42]

$x = -i$ is a solution.

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

Oops. Right.

 

[member 428835]

$x = -i$ is a solution.

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

Typo on my part, yea sorry.

 

[WWGD]

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]

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]

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.