I have been given the matrix $$B = \begin{pmatrix} 1 & 3 & 3 \\ 2 & 3 & 2\\ 2 & 1 & 4\end{pmatrix}$$

and I've been given that its eigenvalues are: $7, 2, -1$

Firsly I was asked to find the column and row eigenvectors corresponding to the Perron eigenvalue. I know the Perron eigenvalue is $7$ so was able to find the eigenvectors to be $$ v = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \ w^{T} =\begin{pmatrix} \frac{5}{9} & \frac{2}{3} & 1 \end{pmatrix}$$

The next part asks:

Find a rank $1$ non negative matrix C such that the matrix $B+C$ will have eigenvalues $13,2,-1$.

I know from a theorem I've done in class that since $C$ is a rank $1$ matrix it can be expressed as $vy^{T}$ such that $B+C$ has the same eigenvalues as $B$ except that $\lambda_{1}$ of $B$ is replaced by $\lambda_{1} +y^{T}v$.

In this example then I have let $y = \begin{pmatrix} a & b & c \end{pmatrix}^{T}$ and from this I have $13 = 7 + \begin{pmatrix} a & b & c \end{pmatrix}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$, therefore $a + b + c = 6$.

I'm uncertain how to solve to find $C$ from this, do I need to use the row eigenvector?

  • $\begingroup$ You may just pick $C=\frac{6vw^T}{w^Tv}$. $\endgroup$ – user1551 May 16 at 19:13

Let $v_2$ be an eigenvector for eigenvalue $2$ and $v_3$ be an eigenvector for eigenvalue $-1$. Vector $y$ can be found as a solution of the following system of three equations with three unknowns $$(B+vy^T)v=13v\,,\quad (B+vy^T)v_2=2v_2\,\quad (B+vy^T)v_3=-v_3\,.$$


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