[ps2] fix a few typos

1 files changed, 3 insertions, 3 deletions

diff --git a/ps2/main.tex b/ps2/main.tex
index e59a055..ecfba0b 100644
--- a/ps2/main.tex
+++ b/ps2/main.tex
@@ -111,7 +111,7 @@ Dividing by $x_{i^*}$ both sides (it is non-zero for the same reason as in
 where we again used the triangle inequality, the definition of $x_{i^*}$ and
 the fact that the sum of the entries of the rows of $M$ is one. Hence, the
 chain of inequalities is a chain of equalities. This implies that for all $j$
-such that $m_{i,j} \neq 0$, $x_j=x_{i^*}$. But we can then repeatedly apply the
+such that $m_{i^*,j} \neq 0$, $x_j=x_{i^*}$. But we can then repeatedly apply the
 same argument to all the neighbors of $i^*$, the neighbors of their neighbors,
 etc. By induction (over the distance to $i^*$), we see that for all vertices
 $j$ in the same connected component as $i^*$, $x_j = x_{i^*}>0$. This implies
@@ -186,7 +186,7 @@ index this path $i^* = i_1,\ldots, i_l = k$. Then:
     \sum_{(i,j)\in E} (x_i-x_j)^2 \geq  \sum_{s=1}^l (x_{i_{s+1}}
     - x_{i_s})^2
 \end{displaymath}
-where we restricted the sum on the path from $i^*$ to $k$. Applying the
+where we restricted the sum to the path from $i^*$ to $k$. Applying the
 Cauchy-Schwartz inequality:
 \begin{displaymath}
     \sum_{(i,j)\in E} (x_i-x_j)^2 \geq \frac{1}{l} \left(\sum_{s=1}^l x_{i_{s+1}}
@@ -197,7 +197,7 @@ Finally using $x_{i^*}^2 \geq \frac{1}{n}$ and $l\leq n$:
 \begin{displaymath}
     \sum_{(i,j)\in E} (x_i-x_j)^2 \geq \frac{1}{n^2}
 \end{displaymath}
-Taking the $\min$ over $x$, this implies using, the variational expression for
+Taking the $\min$ over $x$, this implies, using the variational expression for
 $\lambda_2$ (second largest eigenvalue) established at the beginning of the
 question: $\lambda_2\leq 1-\frac{1}{dn^2}$.