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A is a $4*4$ matrix with eigenvalues $\lambda_A$. Consider a block matrix $B = \left( \begin{array}{ccc} A & I \\ I & A \end{array} \right) $. Then how can we find eigenvalue $\lambda_B$ of matrix B in terms of $\lambda_A$ ?
I know that to find $\lambda_B$, we have to do $|B-\lambda_BI| = 0$, and by properties of block matrices, we have equivalently $|(A-\lambda_BI)^2 - I^2| = 0$.
But I am stuck at this point, I don't know how to proceed further.

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It looks like you are trying to find a determinant of block matrix using the rule for 2x2 matrices, but it is incorrect.

According to http://en.wikipedia.org/wiki/Determinant#Block_matrices

$$ |B-\lambda_bI| = \left| \left(\begin{matrix}A-\lambda_bI & I \\ I & A - \lambda_bI\end{matrix}\right) \right|= |A-\lambda_bI - I||A-\lambda_bI + I|= \\ =|A-(\lambda_b+1)I||A-(\lambda_b - 1)I| = 0 $$

Then you can determine $\lambda_b$ using the fact that $\lambda_a$ is a solution of $|A - \lambda_aI|=0$.

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  • $\begingroup$ ...learing to $\sigma(B)=\{\lambda \in \mathbb{C}: \lambda + 1 \in \sigma(A) \mbox{ or } \lambda - 1 \in \sigma(A)$. $\endgroup$ – Roland Feb 6 '14 at 8:12
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Or to look at it from the point of view of normal forms, if $A=TJT^{-1}$, then using

$$\begin{pmatrix}T&0\\0&T\end{pmatrix}^{-1}\begin{pmatrix}A&I\\I&A\end{pmatrix}\begin{pmatrix}T&0\\0&T\end{pmatrix}=\begin{pmatrix}J&I\\I&J\end{pmatrix},$$

the block matrix can be decomposed into smaller 2x2 blocks if $J$ is diagonal, and correspondingly larger structured blocks if not. In any event, the eigenvalues are determined as the eigenvalues of the 2x2 blocks $\begin{pmatrix}λ_a&1\\1&λ_a\end{pmatrix}$, leading of course to the same eigenvalues as in Steelravens answer.

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I think I can see two ways to proceed with this. But before going further, I wish to point out that what I am about to say applies to any $N \times N$ matrix $A$, so I will remove the restriction that $\text{size} (A) = 4$ until further notice. Having clarified that point, the first option is to work directly from the definitions of eigenvalue and eigenvector, as follows: suppose $\lambda_A$ is an eigenvalue of $A$, with eigenvector $v_A \ne 0$, so that we have

$Av_A = \lambda_A v_A, \tag{1}$

and note that the $2N$-vectors

$V_{A+} = \begin{pmatrix} v_A \\ v_A \end{pmatrix} \tag{2}$

and

$V_{A-} = \begin{pmatrix} v_A \\ -v_A \end{pmatrix} \tag {3}$

are both eigenvectors of $B$, thus:

$BV_{A+} = \begin{bmatrix} A & I \\ I & A \end{bmatrix} \begin{pmatrix} v_A \\ v_A \end{pmatrix} = \begin{pmatrix} (A + I) v_A \\ (I + A) v_A \end{pmatrix} = \begin{pmatrix} (\lambda_A + 1) v_A \\ (\lambda_A + 1) v_A \end{pmatrix} = (\lambda_A + 1) V_{A+}, \tag{4}$

and thus:

$BV_{A-} = \begin{bmatrix} A & I \\ I & A \end{bmatrix} \begin{pmatrix} v_A \\ -v_A \end{pmatrix} = \begin{pmatrix} (A - I) v_A \\ (I - A) v_A \end{pmatrix} = \begin{pmatrix} (\lambda_A - 1) v_A \\ -(\lambda_A - 1) v_A \end{pmatrix} = (\lambda_A - 1) V_{A-}. \tag{5}$

(4) and (5) show that the eigenvalues of $B$ are at least those of the form $\lambda_A \pm 1$, but could there be more? We may investigate this possibility via a sort of unravelling of (4)-(5), in the sense that if

$V =\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} \ne 0, \tag{6}$

where $v_1$, $v_2$ are $N$-vectors, is an eigenvector of $B$ with eigenvalue $\mu$, that is,

$BV = \mu V, \tag{7}$

then if we write out (7) in terms of $v_1$, $v_2$ using the definition of $B$ in terms of $A$ we see that

$BV = \begin{bmatrix} A & I \\ I & A \end{bmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} Av_1 + v_2 \\ v_1 + A v_2 \end{pmatrix} = \mu \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}, \tag{8}$

so that (7)-(8) breaks out into the two equations

$Av_1 + v_2 = \mu v_1 \tag{9}$

and

$Av_2 + v_1 = \mu v_2. \tag{10}$

We analyze (9)-(10) by first addressing the cases $v_2 = \pm v_1$; in the event that $v_2 = v_1$, (9)-(10) become the same equation

$Av_i + v_i = \mu v_i \tag{11}$

or

$Av_i = (\mu - 1)v_i \tag{12}$

for $i = 1, 2$. If, on the other hand, $v_2 = -v_1$, (9)-(10) coalesce into

$Av_i - v_1 = \mu v_i \tag{13}$

or

$Av_i = (\mu + 1)v_i. \tag{14}$

Having dispensed with the possibility that $v_2 = \pm v_1$, we next proceed to the case $v_2 \ne \pm v_1$; then adding (9) and (10) we obtain

$A(v_1 + v_2) + (v_1 + v_2) = \mu (v_1 + v_2) \tag{15}$

whereas subtracting yields

$A(v_1 - v_2) - (v_1 - v_2) = \mu (v_1 - v_2). \tag{16}$

(15) and (16) are equivalent to

$A(v_1 + v_2) = (\mu - 1) (v_1 + v_2) \tag{17}$

and

$A(v_1 - v_2) = (\mu + 1)(v_1 - v_2), \tag{18}$

respectively. It should be observed that, by (6), at least one of $v_1, v_2 \ne 0$, so that in the case $v_2 = \pm v_1$, (12) and (14) are bona fide eigen-equations for the $v_i$ and $\mu$. Likewise, when $v_2 \ne \pm v_1$, neither $v_1 + v_2 = 0$ nor $v_1 - v_2 = 0$ may hold, so that (17)-(18) are legitimate eigen-equations as well.

A careful scrutiny of (1)-(5) reveals that $\lambda_A \pm 1$ are eigenvalues of $B$ for every eigenvalue $\lambda_A$ of $A$. Furthermore, if $\mu$ is an eigenvalue of $B$, so that (8) holds for some $v_1, v_2$, we see from (9)-(18) that at least one of $\lambda_A = \mu \pm 1$ is in fact an eigenvalue of $A$; then by setting up an equation of the from (4) or (5) we see that $\mu = \lambda_A \mp 1$ in fact arises from an eigenvalue of $A$, so every eigenvalue $\mu$ of $B$ is of this form. The eigenvalues of $B$ are precisely the $\lambda_A \pm 1$, where $\lambda_A$ ranges over the eigenvalues of $A$.

Finally, a few comments concerning the eigenvectors of $A$ and $B$.

Almost Done!!! Your patience is appreciated! Meanwhile,

Hope this helps; Cheerio,

and as always,

***Fiat Lux!!!

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