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Let $x,y \in K$ and $n \in \mathbb{N}$. Determine the rank of the following $n \times n$ Matrix: $xI_n$ + $y\cdot \sum_{i\neq j}E_{ij}$ in dependence of $x$ and $y$.

$I_n$ is the identity matrix and $E_{ij}$ is the standard matrix. The matrix has $x$ on its diagonal and the rest looks like a triangular matrix with $y$. So we are trying to find the number of linearly independent rows or columns. The standard procedure would be to find the echelon form and count the non-zero rows.

My first idea was to subtract the first row from the others, when I subtract it from a $3\times 3$ matrix I get a rank of $2$, when we take a $4\times 4$ matrix we get a rank of $3$, so maybe the rank of our matrix is $n$?

Lovecraft
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MassU
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  • It doesn't look like a triangular matrix: ALL the off-diagonal entries are $y$, not just those in the upper triangle (nor just those in the lower triangle). – John Hughes Jan 11 '17 at 19:51

1 Answers1

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Hint (quick approach): Write the matrix as $$ A = (x-y)I_n + y \sum_{i,j=1}^n E_{ij} $$ Note that $B = \sum_{i,j=1}^n E_{ij}$ is a rank $1$ matrix. Once we have its eigenvalues, it's easy to find the eigenvalues of $(x-y)I + yB$. Since $A$ is symmetric (and therefore diagonalizable), the rank of $A$ is simply $n$ minus the algebraic multiplicity of the eigenvalue $0$.

Elementary Approach: In the case that $x = y = 0$, the rank is of course $0$. In the case that $x = y$ but $x \neq 0$, the rank is $1$, since each row is a multiple of the first.

The next special case is when $x = (1-n)y$. In this case, we row-reduce to the form $$ \pmatrix{ 1&0&\cdots&&0&-1\\ &1&0&\cdots&0&-1\\ &&\ddots &\ddots &\vdots&\vdots\\ &&&1&0&-1\\ &&&&1&-1\\ &&&&0&0 } $$ Try this in the $3 \times 3$ case, and see if you can explain why the pattern persists.

Finally, in any other case, we will find that the matrix has full rank, which is to say that we can row-reduce all the way to the identity matrix.

Ben Grossmann
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