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Let $A$ be a square matrix of order, say, $4$.

Consider the matrix $$B=\left( \begin{array}{ccc}A &I \\ I & A\end{array} \right)$$ where $I$ is the identity matrix of order $4$.

Let $\lambda$ be an eigenvalue of $A$. Then there exists a nonzero vector $\bf{x}=$ $\left( \begin{array}{c} x_1 &x_2 &x_3 &x_4\end{array} \right)^T$ such that $A\bf{x}=\lambda \bf{x}$

Now, $$\begin{eqnarray*}\left( \begin{array}{ccc}A &I \\ I & A\end{array} \right) \left( \begin{array}{rrrrrrrr}x_1 \\x_2 \\x_3 \\x_4 \\x_1 \\x_2 \\x_3 \\x_4\end{array} \right)&=&\left( \begin{array}{rr}A\bf{x}+I\bf{x} \\ I\bf{x}+A\bf{x}\end{array} \right)\\&=&\left( \begin{array}{rr}\lambda \bf{x}+I\bf{x} \\ I\bf{x}+\lambda \bf{x}\end{array} \right)\\&=& (\lambda+1) \left( \begin{array}{rrrrrrrr}x_1 \\x_2 \\x_3 \\x_4 \\x_1 \\x_2 \\x_3 \\x_4 \end{array} \right)\end{eqnarray*}$$

So $\lambda+1$ is an eigenvalue of $B$. So, in this way we are able to say $4$ eigenvalues of $B$. Can we say anything about the other $4$ eigenvalues of $B$?

Is there anything special about this kind of matrix $B$?

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4 Answers 4

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Whatever $A$ is, if you consider $B- xI_{2n}$ (where $A$ is $n \times n$), you can perform elementary row operations and put it in the form $\left( \begin{array}{clcr} I & (A-xI)\\ 0 & I -(A-xI)^2 \end{array}\right).$ Up to a sign, the determinant factors as $|A-(x-1)I|.|A -(x+1)I|.$ Hence if $\lambda$ occurs as an eigenvalue of $A$ with algebraic multiplicity $r,$ then $\lambda +1$ and $\lambda -1$ are each eigenvalues of $B$ with algebraic multiplicity $r.$

(Additional information, in view of comment: $B - xI$ has the form $\left( \begin{array}{clcr} A-xI & I\\I & A-xI \end{array} \right).$ You exploit the $I$ in the bottom left, and subtract approprate multiples of the last $n$ rows until the $A-xI$ in the top left is cleared. This leaves $I - (A-xI)^2$ in the top right corner. Then you permute the rows. As for the question about "this type of matrix", it depends how general you want to be. One way I might think about it is as a kind of "twisted circulant": you start with a matrix of the form $ \left( \begin{array}{clcr} I & A & \ldots & \ldots & A^{n-1}\\A& A^{2} & \ldots & A^{n-1} & I\\ \ldots & \ldots & \ldots & \ldots \end{array} \right),$ and then multiply it by a permutation matrix obtained as the Kronecker product of an $n$-cycle permutation matrix with $I_n.$ It may be possible to analyse the characteristic polynomials of such matrices in a similar way).

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Can you explain a bit about the way you perform the row operations to bring $B-xI_{2n}$ to the desired form? Can you also answer other question that, Is there anything special about this kind of matrix B? –  Ashok Mar 5 '12 at 14:08
    
Thanks. There is "(" missing in the beginning of the last latex code which I was not able to fix as it is very minor. –  Ashok Mar 6 '12 at 6:00

If $\vec{x}$ is an eigenvector of $A$ with eigenvalue $\lambda$, then:

$$\left(\begin{array}{cc}A & I \\ I & A\end{array}\right) \left(\begin{array}{c} \vec{x} \\ \vec{x}\end{array}\right) = (\lambda + 1) \left(\begin{array}{c} \vec{x} \\ \vec{x}\end{array}\right)$$

but also

$$\left(\begin{array}{cc}A & I \\ I & A\end{array}\right) \left(\begin{array}{c} \vec{x} \\ -\vec{x}\end{array}\right) = (\lambda - 1) \left(\begin{array}{c} \vec{x} \\ -\vec{x}\end{array}\right)$$

In other words, you find two eigenvectors from each eigenvector of $A$.

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I am french (and this is my first post), so it could explain my strange english. λ-1 is an eigen value of A. See x on -x.

I have also a question : How to use latex on this website ?

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Here is a useful observation. For any $k$, let $I_k$ denote the $k \times k$ identity matrix.

Fixing a positive integer $n$, you can check that the $(2n) \times (2n)$ square matrix $$ C = \frac{1}{\sqrt{2}} \begin{pmatrix} I_n & -I_n \\ I_n & I_n \end{pmatrix} $$ satisfies $C^T C = CC^T = I_{2n}$ (here $C^T$ denotes the transpose of $C$). In other words, $C^T = C^{-1}$, or: $C$ is a unitary matrix.

Moreover, you can check that that for any square $n \times n$ matrices $A$ and $B$, you have $$ C^T \begin{pmatrix} A & B \\ B & A \end{pmatrix} C = \begin{pmatrix} A + B & 0 \\ 0 & A - B \end{pmatrix} $$ (where the $0$s on the extreme right hand side denote the $n \times n$ zero matrices).

This shows that the block matrices $\begin{pmatrix} A & B \\ B & A \end{pmatrix}$ and $\begin{pmatrix} A + B & 0 \\ 0 & A - B \end{pmatrix}$ are similar. Similar matrices have the same characteristic polynomials, and in particular, the same eigenvalues, with the same algebraic multiplicities. (The matrix $C$ implementing the similarity even allows you to pass back and forth between eigenvectors of one matrix and eigenvectors of the other, so even the geometric multiplicities of the eigenvalues coincide.)

It should be clear that the eigenvalue list of the block matrix $\begin{pmatrix} A + B & 0 \\ 0 & A - B \end{pmatrix}$ consists of the eigenvalue list of $A + B$, together with the eigenvalue list of $A - B$. This explains why in the special case $B = I$ the eigenvalues of your matrix are the eigenvalues of $A \pm I$, which are, of course, the eigenvalues of $A$, plus or minus $1$.

For more general $A$ and $B$, of course, the eigenvalues of $A \pm B$ are not obtained simply by adding (or subtracting) eigenvalues of $B$ to (or from) the eigenvalues of $A$. (As an example, if $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, then the eigenvalues of $A$ are $0,2$, and $0$ is the only eigenvalue of $B$, but the eigenvalues of $A+B$ are $1 \pm \sqrt{2}$, and $1$ is the only eigenvalue of $A - B$.) But the computation above still shows that the problem of determining the eigenstuff of any $(2n) \times (2n)$ matrix of the form $\begin{pmatrix} A & B \\ B & A \end{pmatrix}$ reduces to determining the eigenstuff of the two $n \times n$ matrices $A + B$ and $A-B$.

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Thanks @leslie townes. Really useful information. –  Ashok Mar 6 '12 at 12:38

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