diff --git a/end of chapter exercises/Chapter 4 Exercises.ipynb b/end of chapter exercises/Chapter 4 Exercises.ipynb
index e8a5b22..aadeec4 100644
--- a/end of chapter exercises/Chapter 4 Exercises.ipynb	
+++ b/end of chapter exercises/Chapter 4 Exercises.ipynb	
@@ -246,7 +246,7 @@
     "## Question 11\n",
     "We need only show this in one direction, since we can replace $A$ with $A^{\\mathsf{T}}$ below, and the argument still hold true.\n",
     "\n",
-    "Suppose $\\lambda\\neq 0$ is an eigenvalue of $A^{\\mathsf{T}}A$. Then there is some vector $x\\neq 0$ such that $A^{\\mathsf{T}}Ax = \\lambda x$. Thus, $AA^{\\mathsf{T}}Ax = \\lambda Ax$. Therefore, $Ax$ (equivalently $\\lambda x$, or, indeed just $x$ itself!)~is an eigenvector of $AA^{\\mathsf{T}}$, with eigenvalue $\\lambda$.\n",
+    "Suppose $\\lambda\\neq 0$ is an eigenvalue of $A^{\\mathsf{T}}A$. Then there is some vector $x\\neq 0$ such that $A^{\\mathsf{T}}Ax = \\lambda x$. Thus, $AA^{\\mathsf{T}}Ax = \\lambda Ax$. Therefore, $Ax$ is an eigenvector of $AA^{\\mathsf{T}}$, with eigenvalue $\\lambda$. Please not that $x$ is in general not the eigenvector of $AA^{\\mathsf{T}}$, in fact it can't even be multiplied by it, because it is $n$-dimensional, and not $m$-dimensional.\n",
     "\n",
     "---"
    ]