Given $\mathbf v=\begin{bmatrix}v_1\\v_2\\\vdots\\v_n\end{bmatrix}$ and $\mathbf A=\mathbf{vv}^\top$, find the matrix $(\mathbf A+\mathbf I)^{-1}$ and its eigenvectors and eigenvalues.

This is a follow up of questions of the previous one.

Previous example told us $\mathbf A+\mathbf I$ has eigenvalue $\lambda_1=(\|\mathbf v\|^2+1)$ with corresponding eigenvector $\mathbf v_1=\mathbf v$. And eigenvalue $1$ with multiplicity $n-1$.

We also know that if a matrix $\mathbf B$ has eigenvector $\mathbf v$ and eigenvalue $\lambda$, then the inverse matrix $\mathbf B^{-1}$ will have same eigenvector $\mathbf v$ and eigenvalue $\frac1\lambda$.

Based on the above knowledge, we can easily to get eigenvalue of $(\mathbf A+\mathbf I)^{-1}$ will be $1$ with multiplicty $n-1$ and $\frac1{(\|\mathbf v\|^2+1)}$. They share the same eigenvectors.

Now, I want to know how does the matrix $(\mathbf A+\mathbf I)^{-1}$ look like by using minimal polynomial. I am guessing I can use the same techniques here.

First, I find out the characteristic polynomial for $(\mathbf A+\mathbf I)$.

$f(\lambda)=(\lambda-1)^{n-1}(\lambda-1-\|\mathbf v\|^2)$

Then we know the possible candidate for minimal polynomial will be $(\lambda-1)^{k}(\lambda-1-\|\mathbf v\|^2)$ where $k \ge 1$.

How do I know the $k$ for minimal polynomial from here?

Is minimal polynomial a good starting point to find the inverse of the matrix?

  • 1
    $\begingroup$ Look up the Sherman Morrison formula. $\endgroup$
    – copper.hat
    Jan 5 '18 at 17:57
  • $\begingroup$ Note that $\lambda_1 = 1+ \|v\|^2$. You are missing the square. $\endgroup$
    – copper.hat
    Jan 5 '18 at 18:01

This is a standard derivation:

Suppose $(I + u v^T)x = y$ then $x + u v^T x = y$. We can get an explicit expression for $v^T x$ in terms of $y$ by premultiplying by $v^T$ to get $v^T x + v^T u v^T x = v^T y$ and so $v^T x = {v^T y \over 1+v^T u}$. Hence $x = (I - {u v^T \over 1 + v^T u})y$ and so $(I + u v^T)^{-1} = I - {u v^T \over 1 + v^T u}$.

In the above case, this reduces to $I - {v v^T \over 1 + \|v\|^2}$.

If one takes $x \bot v$ we see that $(I + v v^T)x = x$, so the matrix has $n-1$ eigenvalues at $1$ and since $(I + v v^T)v = (1+\|v\|^2) v$ we see that there is one eigenvalue at $1+ \|v\|^2$.


First for finding inverse matrix we go through below equalities: $$vv^t=vv^t({{1+v^tv}\over{1+v^tv}})=vv^t({{1}\over{1+v^tv}}+{{v^tv}\over{1+v^tv}})={{vv^t}\over{1+v^tv}}+{{v(v^tv)v^t}\over{1+v^tv}}$$ so $$I+vv^t-{{vv^t}\over{1+v^tv}}-{{v(v^tv)v^t}\over{1+v^tv}}=I$$ where $I$ indicates on identity matrix. By factorizing we have: $$(I+vv^t)(I-{{vv^t}\over{1+v^tv}})=I$$ or equivalently: $$(I+vv^t)^{-1}=I-{1\over{1+v^tv}}{vv^t}$$ Now note that according to rank-nullity theorem we can divide $\Bbb R^n$ to two subsets: $$A={\lbrace{w:v^tw=0}\rbrace},B={\lbrace{w:w=av}\rbrace}\to\Bbb R^n=A\cup B$$ For $w\in A$ we have $$v^tw=0\to{(I-{1\over{1+v^tv}}{vv^t})w=w-{1\over{1+v^tv}}{vv^t}w=w}\to\lambda=1$$ and for those of B $$w=av\to{(I-{1\over{1+v^tv}}{vv^t})w=av-{1\over{1+v^tv}}{vv^t}av}=av({1-{{v^tv}\over{1+v^tv}}})={1\over{1+v^tv}}av\to\lambda={1\over{1+v^tv}}$$ Then the eigenvalues are $\lambda=1$ and $\lambda={1\over{1+v^tv}}$ and the eigenvectors are all $v\in\Bbb R^n$

  • $\begingroup$ Why in the first equation $v v^t(v^t)v=v v^t v v^t$? Isn't $v^t v $ just a scalar? $\endgroup$
    – Ying
    Jan 6 '18 at 3:38
  • $\begingroup$ Brilliant! Is this a common techniques (first three equation) to solve this kind of questions? $\endgroup$
    – Ying
    Jan 6 '18 at 3:46
  • $\begingroup$ That's kind of you!.....At first, build an intuition for solving such kind of problems and try to prove it. $\endgroup$ Jan 6 '18 at 8:54

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