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.

 
LoadingAdd to solve later

Sponsored Links

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.


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Linear Algebra
Introduction to Linear Algebra at the Ohio State University quiz problems and solutions
Quiz 10. Find Orthogonal Basis / Find Value of Linear Transformation

(a) Let $S=\{\mathbf{v}_1, \mathbf{v}_2\}$ be the set of the following vectors in $\R^4$. \[\mathbf{v}_1=\begin{bmatrix} 1 \\ 0 \\ 1 \\...

Close