Linear Algebra-Eigenvalue/vectors

Question

Let $J_n$ be the $n\times n$ matrix with all entries being ones, where $n\geq2$.

(a) Show that the vector of all ones in $\mathbb{R}^n$ is an eigenvector of $J_n$. What is its corresponding eigenvalue?

(b) Show that $0$ is an eigenvalue of $J_n$. What is the geometric multiplicity of this eigenvalue?

(c) For which values of $n\geq2$ is $J_n$ diagonalizable? Justify your answer.

Answer 1

Recall the definition, $$ J\mathrm{x} =\lambda \mathrm{x}, $$ where $\mathrm{x}$ is a non-zero vector from $\mathrm{R}^n$. Note that you have here $n$ equations of the form $$ \sum_{i=1}^n x_i = \lambda x_i. $$ Clearly, $\lambda = 0$ satisfies all the $n$ equations, and you'll have $\sum_{i=1}^n x_i =0$, i.e., $n-1$ degrees of freedom (=geometric multiplicity of $0$). Another option is $\lambda = n$. Try to figure out why it corresponds to eigenvector of the form $(1,...,1)^T$. Another useful property that might help you to get those values is $\det(J) = 0 = \prod_{i=1}^n \lambda_i $ and $tr(J)=n = \sum_{i=1}^n \lambda_i%$.

For the third question - recall that every symmetric real matrix is diagonalizable.

Answer 2

Finding eigenvalues: You can easily observed that the rank of the matrix $J_n$ is $1$. Therefore $0$ must be an eigenvalue of $J_n$,$\forall n \geq2$. Since the sum of each rows is $n$, so $n$ is also an eigenvalue of $J_n$. Now you think is their any eigenvalue other than $0$ and $n$?

The answer is: No. Note that $dim(E_\lambda)=n-rank(J_n-\lambda I)$. So $dim(E_0)=n-rank(J_n)=n-1$. Thus geometric multiplicity of $0$ is $n-1$. Also since $A.M.\geq G.M.$, so the algebraic multiplicity of $0$ is $\geq n-1$. Also $n$ is an eigenvalue of $J_n$, so algebraic multiplicity of $0$ is equal to $n-1$.

Now you notice that the eigenvector corresponding to eigen value $n$ is solution of the system $(J_n-nI)X=0$, which gives $X=(1,1,1,\cdots 1)^T$.

Since $G.M.$ of $0$ is $n-1$, so there are $n-1$ Jordan block corresponding to $0$ in Jordan canonical form of $J_n$. So the minimal polynomial of $J_n$ is $m(t)=t(t-n)$ for $n\geq 2$, which is product of distinct factors. Therefore $J_n$ is diagonalizable. Or simply, since $J_n$ is real symmetric matrix so it is diagonalizable.