Exercise 5.3.9 (Using eigenvalues to compute determinants)

Question

$
ewcommand{\T}{^\mathsf{T}}$
$
ewcommand{\mtbf}[2]{\mathbf{#1}_{#2}}$
\begin{enumerate}[label = \alph*)]
    \item Find the matrix of the orthogonal projection onto the one-dimensional subspace in $\mathbb{R}^n$ spanned by the vector $(1, 1, ..., 1)\T$;
    \item Let $A$ be the $n 	imes n$ matrix with all entries equal 1. Compute its eigenvalues and their multiplicities (use the previous problem);
    \item Compute eigenvalues (and multiplicities) of the matrix $A - I$, i.e., of the matrix with zeros on the main diagonal and ones everywhere else;
    \item Compute $\det (A-I)$.
\end{enumerate}

Jing Huang · Accepted Answer

a) Note that from Remark 3.5, we know $P_E = \sum_{k=1}^n \frac{1}{||\mtbf{v}{k}||^2} \mtbf{v}{k}\mtbf{v}{k}^*$. For this one-dimensional subspace, it is
%%%%%%%%%%%%%%%%%
\begin{equation*}
    P_E = \frac{1}{n} 
    \begin{bmatrix}
    1  &1  &\dots  &1  \
    1  &1  &\dots  &1  \
    \vdots  &\vdots  &\ddots  &\vdots  \
    1  &1  &\dots  &1
    \end{bmatrix}_{n 	imes n}.
\end{equation*}
%%%%%%%%%%%%%%%%
b) Note that $A = nP_E$. Suppose $A \mathbf{x} = \lambda \mathbf{x}$, then $n P_E \mathbf{x} = \lambda \mathbf{x}$. i.e., $n$ times of the eigenvector's projection on the 1D subspace equals $\lambda$ times of itself. It also shows that the eigenvector's orthogonal projection is parallel to itself. Then there are two possibilities:
\begin{itemize}
    \item One is the eigenvector is parallel with basis of the 1D subspace $\mathbf{v}$, i.e., $\mathbf{x} = \alpha \mathbf{v}, \alpha 
eq 0$. In this case, $P_E \mathbf{x = x}$, then $\lambda = n$. The geometric multiplicity is 1 for 1D eigenspace.
    \item The eigenvector is orthogonal to $\mathbf{v}$, i.e., $\mathbf{x} \perp \mathbf{v}$ and $P_E \mathbf{x = 0}, \lambda = 0$. We can totally find $n-1$ linearly independent eigenvectors so the geometric multiplicity of eigenvalue $\lambda = 0$ is $n-1$.
\end{itemize}
c) $\det (A - I - \lambda I) = \det (A - (\lambda + 1) I)$. i.e., an eigenvalue of $A-I$ plus $1$ is an eigenvalue of $A$. Then the eigenvalues of $A-I$ equal to the eigenvalues of $A$ minus $1$. Thus the eigenvalues of $A-I$ are $n-1$ with multiplicity 1 and $-1$ with multiplicity $n-1$.  \
d) $\det(A-I) = (n-1)(-1)^{n-1}$, which equals $n-1$ if $n$ is odd, $1-n$ if $n$ is even.

Exercise 5.3.9 (Using eigenvalues to compute determinants)

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Exercise 5.3.9 (Using eigenvalues to compute determinants)

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