matrix and eigenvalues

Question

Please help me solve this question. This is not a homework assignment. If $\lambda_M$ is the eigenvalue of the matrix M, please prove that $max|\lambda_M|<1$ $$M_{2n\times2n}=\begin{vmatrix} A_{n\times n} & I-A\\I_{n\times n} & 0_{n\times n}\end{vmatrix}$$ NOTE: $max|\lambda_A|<2$

I easily solved this problem in scalar mode but I have no idea about its matrix mode

Answer 1

We write elements of $\mathbb{R}^{2n}$ as ordered pairs $(a, b)$ where $a, b \in \mathbb{R}^n$.

Suppose that $(a, b)$ is an eigenvector of $M$ with eigenvalue $\lambda$. Then we have $M (a, b) = (Aa + (I - A)b, Ia) = (Aa + (I - A)b, a) = \lambda(a, b) = (\lambda a, \lambda b)$.

Then in particular, we have $a = \lambda b$. So we have $M (a, b) = (A \lambda b + (I - A) b, \lambda b)$.

Then we have $A \lambda b + (I - A) b = \lambda a = \lambda^2 b$.

Then we have $(\lambda - 1) A b = (\lambda^2 - 1)b$.

We pause to note that every vector of the form $(b, b)$ is an eigenvector with eigenvalue 1.

Now we suppose that $\lambda \neq 1$. Then in that case, we have $Ab = (\lambda + 1) b$. Then $\lambda + 1$ is an eigenvalue of $A$.

On the other hand, let $\lambda_A$ be an eigenvalue of $A$ with eigenvector $b$. Then we see that, setting $\lambda = \lambda_A - 1$, we have

$\begin{equation} \begin{split} M (\lambda b, b) &= (A \lambda b + (I - A) b, I \lambda b) \\ &= (\lambda (\lambda + 1) b + (1 - (\lambda + 1)) b, \lambda b) \\ &= (\lambda (\lambda + 1) b - \lambda b, \lambda b) \\ &= \lambda ((\lambda + 1) b - b, \lambda b) \\ &= \lambda (\lambda b, b) \end{split} \end{equation}$

So we see that $\{\lambda \in \mathbb{R}$ : $\lambda$ is an eigenvalue of $M\} = \{1\} \cup \{\lambda_A - 1$ : $\lambda_A$ is an eigenvalue of $A\}$.

Now, let's consider the constraint that $\max |\lambda_A| < 2$. This means for all eigenvalues $\lambda_A$, we have $-2 < \lambda_A < 2$. Thus, the strongest constraint we can get on the eigenvalues of $M$ is that for all eigenvalues $\lambda_M$, we have $-3 < \lambda_M \leq 1$.

Answer 2

Let $a$ denote first $n$ coordinates and $b$ the last $n$ coordinates of a potential eigenvector, that is, $$\lambda\pmatrix{a\\b}=\pmatrix{A&I-A\\I&0}\pmatrix{a\\b}=\pmatrix{Aa+b-Ab\\a}\,.$$ It follows that $\lambda b=a$ and $\lambda a=Aa+b-Ab$.
Substituting, we get $$(\lambda^2-1)b=(\lambda-1)Ab$$ so in case $\lambda\ne 1$, $b$ must be an eigenvector of $A$ with eigenvalue $\lambda+1$.
Can you take it from here?

Answer 3

The block matrix in question is similar to $$ \pmatrix{I&-I\\ 0&I}\pmatrix{A&I-A\\ I&0}\pmatrix{I&I\\ 0&I} =\pmatrix{A-I&0\\ I&I}. $$ Its spectrum is therefore the union of the spectrum of $A-I$ and $n$ copies of $1$.