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

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 $$v = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \ w^{T} =\begin{pmatrix} \frac{5}{9} & \frac{2}{3} & 1 \end{pmatrix}$$

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?

• You may just pick $C=\frac{6vw^T}{w^Tv}$. – 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 $$(B+vy^T)v=13v\,,\quad (B+vy^T)v_2=2v_2\,\quad (B+vy^T)v_3=-v_3\,.$$