diff --git a/_posts/2023-03-06-a-favourite-proof-of-mine.md b/_posts/2023-03-06-a-favourite-proof-of-mine.md
index 5674a26..adb09c8 100644
--- a/_posts/2023-03-06-a-favourite-proof-of-mine.md
+++ b/_posts/2023-03-06-a-favourite-proof-of-mine.md
@@ -108,7 +108,7 @@ Now we can solve $(S-\lambda I_2) \hat v = 0$ for $\lambda_0$ and $\lambda_1$, a
 
 $$ \hat v_0 = \begin{bmatrix} -2 \\ \sqrt 5 - 1 \end{bmatrix},\; \hat v_1 = \begin{bmatrix} 2 \\ \sqrt 5 + 1 \end{bmatrix} $$
 
-These vectors are indeed linearly dependent, and we can use them as basis vectors for our diagonal matrix. We will now to write $S = BDB^{-1}$ where 
+These vectors are indeed linearly independent, and we can use them as basis vectors for our diagonal matrix. We will now to write $S = BDB^{-1}$ where 
 $$ B = \begin{bmatrix} -2 & 2 \\ \sqrt 5 - 1 & \sqrt 5 + 1 \end{bmatrix}$$ 
 $$ D = \begin{bmatrix} \frac{1 - \sqrt 5}{2} & 0 \\ 0 & \frac{1 + \sqrt 5}{2} \end{bmatrix}$$