Question

1. Let A be a symmetric matrix with eigenvalues $\lambda_1, \dots, \lambda_n$. It is known that there exists an orthogonal matrix C such that $A = CDC^\prime$, where $D = diag(\lambda_1, \dots, \lambda_n)$. Use this fact to prove the following: (a) $|A| = \lambda_1 \lambda_2 \dots \lambda_n$, where $|A|$ is the determinant of A. (b) $trace\{A\} = \lambda_1 + \lambda_2 + \dots + \lambda_n$.

          1. Let A be a symmetric matrix with eigenvalues $\lambda_1, \dots, \lambda_n$. It is known that there exists an orthogonal matrix C such that $A = CDC^\prime$, where $D = diag(\lambda_1, \dots, \lambda_n)$. Use this fact to prove the following:
(a) $|A| = \lambda_1 \lambda_2 \dots \lambda_n$, where $|A|$ is the determinant of A.
(b) $trace\{A\} = \lambda_1 + \lambda_2 + \dots + \lambda_n$.

$1. Let A be a symmetric matrix with eigenvalues λ1, …,. It is known that there exists an orthogonal matrix C such that A = CDC^', where D = diag(λ1, …, ). Use this fact to prove the following: (a) |A| = λ1 λ2 …, where |A| is the determinant of A. (b) trace{A} = λ1 + λ2 + … +.$

Added by Carolyn V.

Question

Please give Ace some feedback