Let $A$ be an $n$ by $n$ matrix. Let $D$ be an $n$ by $n$ diagonal matrix with distinct diagonal entries, and let $u$ be an $n$ by $1$ column vector with all non-zero entries. Let $Aq=\lambda q$ with eigenvalue $\lambda$ and non-zero eigenvector $q$. Let $A=D+uu^T$.

  1. Show that $D-\lambda I$ is non-singular and $u^Tq\neq 0$.

  2. Show that the characteristic polynomial for $A$ is $f(\lambda)=1+u^T(D-\lambda I)^{-1}u$.

  3. Show that the eigenvalue for $A$ is greater than the largest eigenvalue for $D$. Consider $f(\lambda)$ on $(\lambda_{\text{max}},\infty)$.

I used the fact that $\lambda $ is an eigenvalue for $A$, and expanded the determinant along the columns to show that $D-\lambda I$ is non-singular i.e., $\det(D-\lambda I)\not=0.$ It is given that all the entries in $u$ are non-zero and $q$ is a non-zero eigenvector.

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    $\begingroup$ (iv) Show your work! What have you tried? Where are you stuck? $\endgroup$ – megas Dec 20 '14 at 3:18
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    $\begingroup$ I'm curious, since $\lim_{|\lambda| \to \infty} f(\lambda) = 1$, how can it be a characteristic polynomial? $\endgroup$ – copper.hat Dec 20 '14 at 5:05
  • $\begingroup$ Can you fix the question please? There must be a mistake for the formula for $f$. $\endgroup$ – copper.hat Dec 20 '14 at 5:41
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    $\begingroup$ You may have edited the expression, but you haven't changed it. My comment above still applies. $\endgroup$ – copper.hat Dec 20 '14 at 21:31
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    $\begingroup$ As you see, the characteristic polynomial is a polynomial of degree $n$ in $\lambda$. As $|\lambda| \rightarrow \infty$ the term $\lambda^{n}$ dominates all other terms, no matter the coefficients. So as cooper.hat pointed out, you would expect that $\lim_{\lambda \rightarrow \infty}f(\lambda)$ should be $\infty$, and not $1$. $\endgroup$ – megas Dec 20 '14 at 21:59

I am going to give a hint for part 1 to get you started. Based on the approach you followed, I am not sure whether you have solved that part. Note, for example that the fact that all entries of $u$ are nonzero, combined with the fact that $q$ is a nonzero vector does not automatically imply that $u^{T}q \neq 0$; the two vectors could have positive and negative entries yielding inner product equal to zero.

You know that $Aq = \lambda q$. Moreover, $A$ has a very specific structure: $A=D-uu^{T}$. Hence, you know that $$ \left(D-uu^{T}\right) q = \lambda q \quad \Leftrightarrow \quad Dq-u (u^{T}q) = \lambda q. $$ We want to show that $u^{T}q \neq 0$. What if it was equal to zero? If you assume $u^{T}q=0$, you must be able to reach a contradiction. There is only a few things you are given in this exercise; use them!

Once you know that $u^{T}q$ is equal to some constant $c \neq 0$, then the above implies that $$ Dq-u\cdot c = \lambda q \quad \Leftrightarrow \quad \left(D - \lambda I\right)q = u\cdot c. $$ Note that $D - \lambda I$ is a diagonal matrix. What would it mean for a diagonal matrix to be singular? Why can it not be true in this case?


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