Solution: We recall that for a matrix $A$ a complex number $\lambda$ is an eigenvalue of $A$ if there exists a nonzero vector $\mathbf{v}$ such that $A\mathbf{v} = \lambda \mathbf{v}$.
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Let \( \lambda \) be an eigenvalue of \( A \). Then, there exists a nonzero vector \( \mathbf{v} \) such that \( A \mathbf{v} = \lambda \mathbf{v} \). Now, consider the matrix \( A + cI \). We have
\[
(A + cI) \mathbf{v} = A \mathbf{v} + cI \mathbf{v} = \lambda \mathbf{v} + c \mathbf{v} = (\lambda + c) \mathbf{v}.
\]
Thus, \( \lambda + c \) is an eigenvalue of \( A + cI \) with the same eigenvector \( \mathbf{v} \).
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Let $p(t) = \det (A - tI)$ be the characteristic polynomial of $A$ and $q(t)$ be the characteristic polynomial of $A + cI$. Then
\begin{align*}
q(t) & = \det ((A + cI) - tI) \\
& = \det (A - (t-c)I) \\
& = p(t-c).
\end{align*}
Now it remains to show that if $\lambda $ is an eigenvalue of $A$ with multiplicity $m$, then the multiplicity of the eigenvalue $(\lambda +c)$ of $A + cI$ is also $m$.
The characteristic polynomial of \( A \) is given by
\[
p(t) = (t - \lambda)^m g(t),
\]
where \( g(t) \) is a polynomial of degree \( n - m \) and $g(\lambda ) \neq 0$. The characteristic polynomial of \( A + cI \) is given by
\begin{align*}
q(t) & = p(t-c) \\
& = (t - (\lambda + c))^m g(t-c), \quad g(\lambda -c) \neq 0.
\end{align*}
Thus, the algebraic multiplicity of \( \lambda + c \) in \( A + cI \) is \( m \).
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Let $E_\lambda $ denotes the eigenspace of $A$ with eigenvalue $\lambda $ and $E_{\lambda + c}$ denotes the eigenspace of $A + cI$ with eigenvalue $\lambda + c$. Then from part (a), we have
\[
E_{\lambda} \subseteq E_{\lambda + c}.
\]
Note that if $w \in E_{\lambda + c}$, then
\begin{align*}
(A + Ic)w = (\lambda + c)w & \implies Aw + cw = \lambda w + cw \\
& \implies Aw = \lambda w \\
& \implies w \in E_\lambda.
\end{align*}
Hence, $E_\lambda \subseteq E_{\lambda + c}$. Therefore, the geometric multiplicity of $\lambda $ in $A$ is equal to the geometric multiplicity of $\lambda + c$ in $A + cI$.