# Let $A$ be the $n\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$? Is $A$ diagonalizable?

Let $A$ be the $n\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$ and a basis for each eigenspace? Is $A$ diagonalizable?

Having some trouble with this one. I tried using the fact that $\lambda$ is an eigenvalue of $A$ iff there exist non-zero solutions to $Ax=\lambda x$. Well, clearly the $1\times n$ non-zero vector $x$ with all entries equal is an eigenvector of $A$, with corresponding eigenvalue $\lambda=n$, since $Ax=nx$. But I can't seem to get much further than that in terms of finding eigenvalues for $A$, any hints/suggestions?

• What is the rank of $A$? What does the rank-nullity theorem tell you about the nullity of $A$? What does that imply about the eigenvalue zero? What does the trace say about the eigenvalues? – JMoravitz Mar 14 '16 at 0:03
• – JMoravitz Mar 14 '16 at 0:06
• So the idea is that in general for a matrix with a $1$ in every entry, $p(\lambda)=(-1)^n\lambda^{n}+(-1)^{n-1}n\lambda^{n-1}$, so the only eigenvalues are $\lambda =0,n$? – Borat Sagdiyev Mar 14 '16 at 0:17

Observe:

• $Rank(A)=1$ since all columns are identical and nonzero (rank=number of linearly independent columns)
• $nullity(A)=n-1$ by the rank-nullity theorem (rank+nullity=n)
• $trace(A)=n$ by direct calculation (trace=sum of diagonal entries)

Since $nullity(A)=n-1=nullity(A-0I)$ is the dimension of the eigenspace associated with the eigenvalue zero, we know that we have zero as an eigenvalue of algebraic multiplicity at least $n-1$. This implies that zero is either of algebraic multiplicity $n-1$ or it is of algebraic multiplicity $n$.

Since $n=trace(A)=\sum \lambda\cdot algmu(\lambda)$, we know that zero is not of multiplicity $n$ and that the remaining eigenvalue must be $n$ itself.

As for the question of if it is diagonalizable, as mentioned earlier, we know that $geomu(0)=n-1$ and that $n$ is an eigenvalue (and thus must have geometric multiplicity at least one like every eigenvalue) and has geometric multiplicity exactly one (as it can't exceed its algebraic multiplicity). Since the sum of the geometric multiplicities is $(n-1)+1=n$, it is seen to be diagonalizable.

• $A$ is diagonalizable since $dim(E_0)+dim(E_n)=n$, $dim(E_0)=n-1=mult_0$, and $dim(E_n)=1=mult_n$, right? But how would you solve $(A-nI)x=0$ to find $E_n$? – Borat Sagdiyev Mar 14 '16 at 3:05
• @Borat by inspection is a valid method which you already did in your original post. If you were to do it manually, consider adding all rows together in the row reduction process and see where that gets you. – JMoravitz Mar 14 '16 at 3:13
• Oh, I see. So $E_n=span\{(1,...,1)\}$? – Borat Sagdiyev Mar 14 '16 at 3:18

Note that $A$ is real and symmetric, so it is diagonalizable. We have $A^2=nA$, so all eigenvalues of $A$ satisfy $\lambda^2=n\lambda$: this tells us that the only eigenvalues are $0$ and $n$. Since the trace of $A$ (i.e., the sum of the eigenvalues) is $n$, we deduce that the multiplicity of $n$ is one and that the multiplicity of $0$ is $n-1$.

As for eigenspaces, if $e$ is the vector with all entries equal to one, we have $Ae=ne$, so $e$ spans the eigenspace of $n$. The eigenspace for $0$ consists of the kernel of $A$: $0=Ax=(x_1+\cdots+x_n,\ldots,x_1+\cdots+x_n)^t$. We can make a basis of this subspace by taking the vectors $$\begin{bmatrix}1\\-1\\0\\0\\0\\ \vdots\\ 0\end{bmatrix},\ \ \begin{bmatrix}0\\1\\-1\\0\\0\\ \vdots\\ 0\end{bmatrix},\ \ \begin{bmatrix}0\\0\\1\\-1\\0\\ \vdots\\ 0\end{bmatrix},\ \ \ldots\ \ ,\ \ \begin{bmatrix}0\\0\\0\\ \vdots \\0\\1\\-1\end{bmatrix}.$$

Here is a matrix $P$ that I made up some time ago. Note that $P$ is not orthogonal, although the columns are pairwise orthogonal and therefore independent. The columns are eigenvectors for your matrix, in the 10 by 10 case. Multiply your matrix of all ones (10 by 10) on the left, this matrix on the right. What happens? $$\left( \begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 \end{array} \right).$$

If $\mathbf{1}$ denotes the column vector of all ones, what is $\mathbf{1} \mathbf{1}^T$?

Now, match this to the outer product form of the eigendecomposition to get the eigenvectors and eigenvalues.

Consider the matrix $B = A - \lambda I$, where $\lambda \in \text{Spec}(A)$. Then $\lambda_{1} = n - \lambda$ is an eigenvalue of $B$ with an associated eigenvector $v_{1} = \begin{pmatrix} 1 \\ 1 \\ 1\\ \cdot \\ \cdot \\ \cdot \\ 1 \\1 \\1\\ \end{pmatrix}$.

The remaining eigenvalues of $B$ are $\lambda_{2} = \lambda_{3} =...=\lambda_{n} = -\lambda$ and the corresponding eigenvectors are:

$v_{2} = \begin{pmatrix} 1 \\ -1 \\ 0\\ \cdot \\ \cdot \\ \cdot \\ 0 \\0 \\0\\ \end{pmatrix}$, $v_{3} = \begin{pmatrix} 0 \\ 1 \\ -1\\ \cdot \\ \cdot \\ \cdot \\ 0 \\0 \\0\\ \end{pmatrix}$, ..., $v_{n} = \begin{pmatrix} 0 \\ 0 \\ 0\\ \cdot \\ \cdot \\ \cdot \\ 0 \\1 \\-1\\ \end{pmatrix}$.

Now, it is easier to see that $v_{1}, ..., v_{n}$ are linearly independent and they are eigenvectors of $A$ with $v_{1}$ associated with the eigenvalue $\lambda =n$ and $v_{j}$ for $j \geq 2$ associated with $\lambda = 0$. From the linearly independence of these eigenvectors we see $A$ is diagonalizable.