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I want to consider the following matrix:

\begin{bmatrix}\boldsymbol{I}_n & \boldsymbol{I}_n \\\boldsymbol{I}_n & \boldsymbol{I}_n\end{bmatrix}

By doing several numerical examples, I recognized that this matrix has $n$ eigenvalues equal to zero and $n$ eigenvalues equal to $2$. Is there any way to prove this for an arbitrary number $n$?

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    $\begingroup$ Hint: compute $A^2$. $\endgroup$ – Gribouillis Sep 8 '17 at 12:05
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    $\begingroup$ $( {\bf 1}_2 {{\bf 1}_2}^T )\otimes {\bf I}_n$. Then use the results how Kronecker products inherits their factors' eigenvalues. $\endgroup$ – mathreadler Sep 8 '17 at 12:35
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Hint: Write it as $$( {\bf 1}_2 {{\bf 1}_2}^T )\otimes {\bf I}_n$$Then use the results how Kronecker products inherits their factor's eigenvalues.

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Hint. Note that $A^2=2A$ and $\mbox{dim}(\mbox{Ker}(A))=n$.

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The characteristic polynomial of the given $2 \times 2$ block matrix is

$$\begin{array}{rl} \det \left( s \begin{bmatrix} \mathrm I_n & \mathrm O_n\\ \mathrm O_n & \mathrm I_n\end{bmatrix} - \begin{bmatrix} \mathrm I_n & \mathrm I_n\\ \mathrm I_n & \mathrm I_n\end{bmatrix} \right) &= \det \begin{bmatrix} (s-1) \mathrm I_n & -\mathrm I_n\\ -\mathrm I_n & (s-1) \mathrm I_n\end{bmatrix}\\ &= \det \left( (s-1)^2 \mathrm I_n - \mathrm I_n\right)\\ &= \det \left( (s^2 - 2s) \mathrm I_n \right)\\ &= \det \left( s (s - 2) \mathrm I_n \right)\\ &= s^n (s-2)^n\end{array}$$

where we used the fact that multiples of the identity matrix always commute. Thus, the eigenvalues of the given $2 \times 2$ block matrix are $0$ and $2$, both with multiplicity $n$.

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Suppose $\operatorname{char}{F}\neq 2$. Let $(e_i)_{i=1}^n$ be a basis of $F^n$. They are also all eigenvectors of $I_n$ with eigenvalue $1$ (since all elements of $F^n$ are). Then $((e_i, \pm e_i))_{i=1}^n$ are a basis of $F^n \oplus F^n$, and all eigenvectors of $A$: $$ \begin{pmatrix} I_n & I_n \\ I_n & I_n \end{pmatrix} \begin{pmatrix} e_i \\ \pm e_i \end{pmatrix} = \begin{pmatrix} I_n e_i \pm I_n e_i \\ I_n e_i \pm I_n e_i \end{pmatrix} = \begin{pmatrix} (1 \pm 1)e_i \\ (1 \pm 1) e_i \end{pmatrix} = \begin{cases} 2(e_i,e_i) & + \\ (0,0) & - \end{cases}, $$ so $(e_i,e_i)$ is an eigenvector of $A$ with eigenvalue $2$, and $(e_i,-e_i)$ one with eigenvalue $0$.

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Given the structure of the matrix $$ A = \begin{bmatrix} I_n & I_n\\ I_n & I_n \end{bmatrix} $$ you can simply calculate the determinant of $$ A - \lambda I_{2n} = \begin{bmatrix} (1-\lambda)I_n & I_n\\ I_n & (1-\lambda)I_n \end{bmatrix}. $$ If $\lambda = 1$, the columns of $A - I_{2n}$ are the canonical basis vectors in $\mathbb{R}^{2n}$ in which case its determinant must be nonzero. Now consider the case $\lambda \neq 1$. Row reducing $A - \lambda I_{2n}$ we get the diagonal matrix $$ \begin{bmatrix} (1-\lambda)I_n & I_n\\ 0 & \lambda(\lambda - 2)/(1-\lambda)I_n \end{bmatrix} $$ whose determinant is $\lambda(2-\lambda)$. Hence, the eigenvalues of $A$ are $0$ and $2$. To see that each eigenspace has dimension $n$, we observe that $A$ has $n$ linearly independent columns and hence $\dim(\ker(A)) = n$.

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Make row operations: $$\begin{vmatrix}I_n&I_n\\I_n&I_n\end{vmatrix}=\begin{vmatrix}I_n& I_n\\0&0\end{vmatrix}=0.$$

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