Problem 1516

To prove that $\text{tr}(A^*) \ne -1$ if and only if the matrix $I_n + A^*$ is invertible, we will consider different cases based on the rank of matrix $A$.

1. **Case 1: $n = 2$**

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. The adjugate matrix $A^*$ is given by:
$$A^* = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$
The determinant of $I + A^*$ is:
$$\det(I + A^*) = \det \begin{bmatrix} 1 + d & -b \\ -c & 1 + a \end{bmatrix}$$
By direct computation, we have:
$$\det(I + A^*) = (1 + d)(1 + a) - (-b)(-c) = 1 + a + d + ad - bc$$
Since $A$ is not invertible, $\det(A) = ad - bc = 0$. Therefore:
$$\det(I + A^*) = 1 + a + d$$
This determinant is nonzero if and only if $a + d \ne -1$. Since $\text{tr}(A^*) = a + d$, we conclude that $\text{tr}(A^*) \ne -1$ if and only if $I + A^*$ is invertible.

2. **Case 2: $n \ge 3$ and $\text{rank}(A) \le n-2$**

If $\text{rank}(A) \le n-2$, then any $n-1$ rows of $A$ are linearly dependent. This implies that any $(n-1) \times (n-1)$ submatrix of $A$ has dependent rows, and hence $A^* = 0$. In this case:
$$\text{tr}(A^*) = 0 \ne -1$$
and:
$$I + A^* = I$$
which is clearly invertible. Thus, $\text{tr}(A^*) \ne -1$ if and only if $I + A^*$ is invertible.

3. **Case 3: $n \ge 3$ and $\text{rank}(A) = n-1$**

Since $\text{rank}(A) = n-1$, the null space of $A$ is one-dimensional. Every column of $A^*$ lies within the null space of $A$, implying $\text{rank}(A^*) = 1$. Let $B = A^*$. For rank 1 matrices, we have:
$$B^2 = (\text{tr}(B))B$$
This means the algebra generated by $B$ consists of $\text{span}\{I, B\}$. If $I + B$ is invertible, it must lie within this algebra. Suppose the inverse of $I + B$ is of the form $I + \alpha B$. Then:
$$(I + B)(I + \alpha B) = I + (1 + \alpha)B + \alpha (\text{tr}(B))B = I + (1 + \alpha + \alpha \text{tr}(B))B$$
For this to equal $I$, we must have:
$$1 + \alpha + \alpha \text{tr}(B) = 0 \implies \alpha (1 + \text{tr}(B)) = -1$$
If $\text{tr}(B) = -1$, there is no solution for $\alpha$, and $I + B$ is not invertible. If $\text{tr}(B) \ne -1$, we can solve for:
$$\alpha = \frac{-1}{1 + \text{tr}(B)}$$
and verify that $I + \alpha B$ is indeed the inverse of $I + B$. Thus, $\text{tr}(A^*) \ne -1$ if and only if $I + A^*$ is invertible.

$\blacksquare$