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

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.

• If $A$ is a rank one matrix, it is $xy^T$ for some vectors $x$ and $y$. In this case, the trace of $A$ is $y^T x$. – Joppy Mar 13 '18 at 11:25
• If $A$ has rank one, all except one eigenvalues are zero (think rank-nullity theorem). That's why $\prod_k(1+\lambda_k)=1+\sum_k\lambda_k$, because all higher-order terms in the product are zero. – user1551 Mar 13 '18 at 11:32
• hmm, Nice but is that true always? is it correct to say that if $A$ has rank 1 then all eigenvalues of $A$ are zero except one eigenvalue which is non-zero? – BAYMAX Mar 13 '18 at 11:35
• Yes, you know that $dimKerA+dimImA=n$, and also $KerA$ is the eignspace of 0. So $rkA=dimImA=1$ implies $dimKerA=n-1$. – Espace' etale Mar 13 '18 at 12:16
• – Rodrigo de Azevedo Jul 6 '20 at 12:19

## 3 Answers

Let $$\rm A = u v^*$$ be a rank-$$1$$ matrix. Using Sylvester's determinant identity,

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

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 $$\text{tr}(A)=\lambda$$.

The Jordan normal forma of 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.

• What are the Jordan blocks in the 2nd matrix? – Rodrigo de Azevedo Mar 13 '18 at 12:51