eigenvalue of block matrix in terms of original matrix 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.
 A: 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 $|A - \lambda_aI|=0$. 
A: 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.
A: 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!!!
