If Every Vector is Eigenvector, then Matrix is a Multiple of Identity Matrix

Problems and solutions in Linear Algebra

Problem 357

Let $A$ be an $n\times n$ matrix. Assume that every vector $\mathbf{x}$ in $\R^n$ is an eigenvector for some eigenvalue of $A$.
Prove that there exists $\lambda\in \R$ such that $A=\lambda I$, where $I$ is the $n\times n$ identity matrix.

 
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Proof.

Let us write $A=(a_{ij})$.
Let $\mathbf{e}_i$ be the unit vector in $\R^n$ whose $i$-th entry is $1$, and $0$ elsewhere.
Then the vector $\mathbf{e}_i$ is an eigenvector corresponding to some eigenvalue $\lambda_{i}$. That is, we have
\[A\mathbf{e}_i =\lambda_{i} \mathbf{e}_i.\]

More explicitly, this shows that
\[\begin{bmatrix}
a_{1i} \\
a_{2i} \\
\vdots \\
a_{n-1 i}\\
a_{ni}
\end{bmatrix}=
\begin{bmatrix}
0 \\
\vdots \\
\lambda_{i} \\
\vdots \\
0
\end{bmatrix}.\] Therefore, it follows that $a_{ki}=0$ if $k\neq i$ and $a_{i i}=\lambda_{i}$.
So we have
\[A=\begin{bmatrix}
\lambda_1 & 0 & \dots & 0 \\
0 &\lambda_2 & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & \lambda_n
\end{bmatrix}.\]

Now we consider the vector $\mathbf{v}=\begin{bmatrix}
1 \\
1 \\
\vdots \\
1
\end{bmatrix} \in \R^n$ with all entries $1$. Then $\mathbf{v}$ is an eigenvector of an eigenvalue $\lambda$, that is, $A\mathbf{v}=\lambda \mathbf{v}$.
It follows that we have
\begin{align*}
\begin{bmatrix}
\lambda \\
\lambda \\
\vdots \\
\lambda
\end{bmatrix}
=
\lambda \mathbf{v}=A\mathbf{v}=
\begin{bmatrix}
\lambda_1 & 0 & \dots & 0 \\
0 &\lambda_2 & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & \lambda_n
\end{bmatrix}
\begin{bmatrix}
1 \\
1 \\
\vdots \\
1
\end{bmatrix}
=\begin{bmatrix}
\lambda_1 \\
\lambda_2 \\
\vdots \\
\lambda_n
\end{bmatrix}
\end{align*}

As a result, we must have $\lambda=\lambda_1=\cdots =\lambda_n$.
Thus, we obtain
\begin{align*}
A=\begin{bmatrix}
\lambda_1 & 0 & \dots & 0 \\
0 &\lambda_2 & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & \lambda_n
\end{bmatrix}
=\begin{bmatrix}
\lambda & 0 & \dots & 0 \\
0 &\lambda & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & \lambda
\end{bmatrix}
=\lambda I
\end{align*}
as required.


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