From 883939892ec1d1a7a514344496bdbe970d0d6c17 Mon Sep 17 00:00:00 2001 From: Lei Mao Date: Sun, 13 Aug 2023 10:26:54 -0700 Subject: [PATCH] Fix a Typo --- src/diagonalization.xml | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/diagonalization.xml b/src/diagonalization.xml index a924a26..8a8d58e 100644 --- a/src/diagonalization.xml +++ b/src/diagonalization.xml @@ -1286,7 +1286,7 @@ The eigenvectors v_1,v_2,v_3 are linearly independent: v_1,v_2 for Now suppose that the sum of the geometric multiplicities equals n. As above, this forces the sum of the algebraic multiplicities to equal n as well. As the algebraic multiplicities are all greater than or equal to the geometric multiplicities in any case, this implies that they are in fact equal.

- Finally, suppose that the third condition is satisfied. Then the sum of the geometric multiplicities equals n. Suppose that the distinct eigenvectors are \lambda_1,\lambda_2,\ldots,\lambda_k, and that \cB_i is a basis for the \lambda_i-eigenspace, which we call V_i. We claim that the collection \cB = \{v_1,v_2,\ldots,v_n\} of all vectors in all of the eigenspace bases \cB_i is linearly independent. Consider the vector equation + Finally, suppose that the third condition is satisfied. Then the sum of the geometric multiplicities equals n. Suppose that the distinct eigenvalues are \lambda_1,\lambda_2,\ldots,\lambda_k, and that \cB_i is a basis for the \lambda_i-eigenspace, which we call V_i. We claim that the collection \cB = \{v_1,v_2,\ldots,v_n\} of all vectors in all of the eigenspace bases \cB_i is linearly independent. Consider the vector equation 0 = c_1v_1 + c_2v_2 + \cdots + c_nv_n.