# Prove or disprove that the matrix is invertible

Let $$P$$ be a $$n \times n$$ matrix with integer entries. Let $$q$$ be a non integer and $$Q = P + qI$$ where $$I$$ is identity matrix.

Prove or disprove that $$Q$$ is invertible.

This is easy if $$P$$ is a $$2 \times 2$$ matrix. In this case it is easy to see that determinant of $$Q$$ is non zero. Let $$P$$ be a matrix

$$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$$

then we see that $$\det Q = (a + q)(d + q) - bc$$. If $$q$$ is an irrational then the determinant is obviously non zero. If $$q$$ is a rational then let $$q = \frac mn$$ where gcd of $$m$$ and $$n$$ is $$1$$. Now it is easy to see that if $$(an+m)(dn+m) = n^2 bc$$ then any prime which divides $$n$$ doesn't divide $$m$$ which implies that it doesn't divide LHS as well. So determinant of $$Q$$ is non zero so we conclude that $$Q$$ is invertible.

Now how to prove for higher dimension matrices?

You basically want to know whether $$-q$$ is an eigenvalue of $$P$$. This is true if and only if $$k_P(-q) = 0$$, where $$k_P(x) = \det(x I-P)$$ is the characteristic polynomial of $$P$$. Since $$P$$ has integer coefficients, we have $$k_P \in \Bbb{Z}[x]$$ and it is a monic polynomial with $$\deg k_P = n$$ so clearly if $$q$$ not an algebraic integer or is an algebraic integer of degree $$> n$$ then $$-q$$ cannot be a zero of $$k_P$$ so $$P+qI$$ is invertible.
On the other hand, if $$q$$ is algebraic of degree $$\le n$$, then let $$p(t) = c_0 + c_1 t + \cdots + c_{k-1}t_{k-1} + t^k \in \Bbb{Z}[x]$$ be the minimal polynomial of $$-q$$ with $$k \le n$$. You can check that the $$n \times n$$ matrix $$P = \begin{bmatrix}0 & 0 & \cdots & 0 & -c_0 & 0 & 0 & \cdots & 0\\ 1 & 0 & \cdots & 0 & -c_1 & 0 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0 & -c_2 & 0 & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 & -c_{k-1} & 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 & 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & \cdots & 0 & 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 0 & 0 & 0 & 0 & \cdots & 1 \end{bmatrix}$$ has $$k_P(x) = p(x)(x-1)^{n-k}$$ so $$k_P(-q)=0$$ and therefore $$P+qI$$ is not invertible.
For your $$2 \times 2$$ example, your analysis is wrong, for example $$\sqrt{2}$$ is an algebraic integer with minimal polynomial $$x^2-2$$ so for $$P = \begin{bmatrix} 0 & 2 \\ 1 & 0 \end{bmatrix}$$ we have that $$P+\sqrt{2} I$$ is not invertible. Also for any $$P$$ and $$q = \frac12$$ (which is not an algebraic integer), we have $$\det \left(P+\frac12 I\right) = \left(a+\frac12\right)\left(d+\frac12\right) - bc \in \Bbb{Z}+\frac14$$ so it is nonzero and hence $$P+\frac12 I$$ is invertible. Similarly, for any $$P$$ and $$q = \sqrt[3]{2}$$ which is an algebraic integer of degree $$3$$, you can check that $$P+\sqrt[3]{2} I$$ is invertible.