From a3dd791a3c87e58dbd9aad98008ab7279346281e Mon Sep 17 00:00:00 2001
From: Cminorwhy <112066107+Cminorwhy@users.noreply.github.com>
Date: Sun, 5 Nov 2023 16:27:54 +0000
Subject: [PATCH] motify

---
 .../3_Algebraic_and_geometric_multiplicaties_of_eigenvalues.md  | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/study/Imperial_mathematics/year_2/Linear_Algebra_and_Numerical_Analysis/Part_I/3_Algebraic_and_geometric_multiplicaties_of_eigenvalues.md b/study/Imperial_mathematics/year_2/Linear_Algebra_and_Numerical_Analysis/Part_I/3_Algebraic_and_geometric_multiplicaties_of_eigenvalues.md
index db44f0fe6..f8872e65d 100644
--- a/study/Imperial_mathematics/year_2/Linear_Algebra_and_Numerical_Analysis/Part_I/3_Algebraic_and_geometric_multiplicaties_of_eigenvalues.md
+++ b/study/Imperial_mathematics/year_2/Linear_Algebra_and_Numerical_Analysis/Part_I/3_Algebraic_and_geometric_multiplicaties_of_eigenvalues.md
@@ -131,7 +131,7 @@ $$
 
             `Claim`: $$B$$ is a basis of $$V$$.
 
-            `Proof of Claim`: Since $$|B| = n = \dim V$$, it is enough to show that $$B$$ is linearly independent. Suppose there is a linear relation on the vector in $$B$$, and we write it as
+            `Proof of Claim`: Since $$|B| = n = \dime V$$, it is enough to show that $$B$$ is linearly independent. Suppose there is a linear relation on the vector in $$B$$, and we write it as
 
             $$
             \sum_{a \in B_1} \alpha_0 a + \sum_{a \in B_2} \alpha_1 b + \dots + \sum_{a \in B_r} \alpha_{r-1} z = 0