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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.

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  • $\begingroup$ 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 $y^T x$. $\endgroup$ – Joppy Mar 13 '18 at 11:25
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    $\begingroup$ 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. $\endgroup$ – user1551 Mar 13 '18 at 11:32
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    $\begingroup$ 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? $\endgroup$ – BAYMAX Mar 13 '18 at 11:35
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    $\begingroup$ Yes, you know that $dimKerA+dimImA=n$, and also $KerA$ is the eignspace of 0. So $rkA=dimImA=1$ implies $dimKerA=n-1$. $\endgroup$ – Espace' etale Mar 13 '18 at 12:16
  • $\begingroup$ Related: math.stackexchange.com/q/3736628/339790 $\endgroup$ – Rodrigo de Azevedo Jul 6 '20 at 12:19
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Let $\rm A = u v^*$ be a rank-$1$ matrix. Using Sylvester's determinant identity,

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

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

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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.

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

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