Does $\det(I+A) = 1 + \mbox{tr}(A)$ hold if $A$ is a rank-$1$ complex matrix?

Question

If $A$ is a complex $n \times n$ matrix of rank $1$, then $$\det(I+A) = 1 + \mbox{tr}(A)$$

How to approach this problem?

Rank-$1$ matrices have special properties. Also, thinking about the determinant of a matrix as the product of its eigenvalues and the trace of the matrix as the sum of its eigenvalues,

$$\prod_{k=1}^{n}(1+\lambda_{k}) = 1 +\sum_{k=1}^{n} \lambda_{k}$$

I could not proceed.

Answer 1

The Jordan normal form of a rank $1$ matrix either has the form$$\begin{pmatrix}\lambda&0&0&\ldots&0\\0&0&0&\ldots&0\\&\vdots&&\ddots&\vdots\\0&0&0&\ldots&0\end{pmatrix}$$for some $\lambda\neq0$, or the form$$\begin{pmatrix}0&1&0&\ldots&0\\0&0&0&\ldots&0\\&\vdots&&\ddots&\vdots\\0&0&0&\ldots&0\end{pmatrix}.$$and the formula that you want to prove holds in both cases.

Answer 2

Let $\rm A = u v^*$ be a rank-$1$ matrix. Using the Weinstein-Aronszajn determinant identity,

$$\det \left( \mathrm I_n + \mathrm A \right) = \det \left( \mathrm I_n + \mathrm u \mathrm v^* \right) = 1 + \mathrm v^* \mathrm u = 1 + \mbox{tr} \left( \mathrm v^* \mathrm u \right) = 1 + \mbox{tr} \left( \,\mathrm u \mathrm v^* \right) = 1 + \mbox{tr} \left( \mathrm A \right)$$

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linear-algebra matrices rank-1-matrices determinant trace

Answer 3

There is a much simpler solution:

1. $A$ has only one nonzero eigenvalue, say $\lambda$.

2. The eigenvalues of $A+I$ are the eigenvalues of $A$ plus $1$ (so $n-1$ eigenvalues of $A+I$ are 1, and 1 eigenvalue is $\lambda+1$).

3. $\det(A+I)$ is the product of the eigenvalues: $\lambda+1$.

4. The trace is the sum of the eigenvalues, so $\text{tr}(A)=\lambda$.