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Let $B$ be an all-ones matrix of dimension $n \times n$, and $I$ be the identity matrix of dimension $n \times n$. Define $A = \frac{1}{n}B - I$. For $n=3$, $A$ would yield

\begin{equation*} \begin{bmatrix} -2/3 & 1/3 & 1/3 \\ 1/3 & -2/3 & 1/3 \\ 1/3 & 1/3 & -2/3 \end{bmatrix}. \end{equation*}

How would one compute rank$(A)$ for any $n$?

I have noted that any row $r_i$ of $A$ can be written as a linear combination of the other $n-1$ rows. Specifically, $r_i = - \sum_{j\neq i} r_j$. This implies rank$(A) < n$, but I don’t know how to compute the exact rank of $A$.

Any hint would be appreciated.

Eva
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3 Answers3

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Notice the matrix rank is subadditive. i.e. for any $n\times n$ matrix $P, Q$, we have

$${\rm rank}(P+Q) \le {\rm rank}(P) + {\rm rank}(Q)$$

Since ${\rm rank}(B) = 1$, we have

$$n = {\rm rank}(-I) = {\rm rank}\left(A - \frac{B}{n}\right) \le {\rm rank}(A) + {\rm rank}\left(-\frac{B}{n}\right) = {\rm rank}(A) + 1$$ This implies ${\rm rank}(A) \ge n - 1$. Together with the inequality you have ${\rm rank}(A) < n$, you can deduce ${\rm rank}(A) = n -1$.

achille hui
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It is easy to compute $ker(A)$:

$ker(A)= span \{(1,1,1,...,1)^T\}$.

Hence $\dim ker(A)=1$ and so $rank(A)=n-1$.

Fred
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Suppose we are in characteristic zero. Then $B$ is symmetric over the reals and is diagonalisable.

The eigenvalues of $B$ are $n$ (with multiplicity one) and $0$ (with multiplicity $n-1$). Therefore the eigenvalues of $rB+sI$ are $rn+s$ (with multiplicity one) and $s$ (with multiplicity $n-1$). When, say $rn+s=0$ and $s\ne0$, then $rB+sI$ has zero as an eigenvalue with multiplicity one, so its rank is $n-1$.

Angina Seng
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