Elementary matrices (1)

Posted: January 28, 2011 in Elementary Algebra; Problems & Solutions, Linear Algebra

Let k be a field and let M_n(k) be the set of all n \times n matrices with entries from k. Let I \in M_n(k) be the identity matrix and let e_{ij} \in M_n(k) be the matrix whose (i,j)-entry is 1 and all other entries zero. We will denote by \text{diag} \{x_1, x_2, \cdots , x_n \} the diagonal matrix whose (i,i)-entry is x_i.

Definition. For every 1 \leq i \neq j \leq n and \alpha \in k define E_{ij}(\alpha)=I + \alpha e_{ij}. We will call E_{ij}(\alpha) an elementary matrix.

Remark. If A \in M_n(k), then clearly multiplying A on the left by E_{ij}(\alpha) takes A and adds on \alpha times the j-th row of A to the i-th row of A. Similarly multiplying A on the right by E_{ij}(\alpha) takes A and adds on \alpha times the i-th column of A to the j-th column of A.

Problem 1. Prove that

1) \det E_{ij}(\alpha)=1.

2) (E_{ij}(\alpha))^{-1}=E_{ij}(-\alpha).

Solution. The first part is obvious because E_{ij}(\alpha) is a triangular matrix and so its determinant is the product of its diagonal entries, which are all 1. To prove the second part of the problem, note that since i \neq j we have e_{ij}^2=0 and thus E_{ij}(\alpha) E_{ij}(-\alpha)=(I + \alpha e_{ij})(I - \alpha e_{ij})=I. \Box

Problem 2. Let A \in M_n(k) be a diagonal matrix with \det A = 1. Prove that A is a product of 5(n-1) of elementary matrices.

Solution. Let A=\text{diag} \{a_1, a_2, \cdots , a_n \}. Since \det A = 1, we have

a_1a_2 \cdots a_n = 1. \ \ \ \ \ \ \ \ \ \ (*)

In partcular, all a_i are non-zero and hence invertible in k. See that

E_{n,n-1}(1-a_n^{-1})E_{n-1,n}(-1)E_{n,n-1}(1)E_{n,n-1}(-a_n)E_{n-1,n}(a_n^{-1})A = \text{diag} \{a_1, a_2 , \cdots , a_{n-1}a_n, 1 \}.

As you see we have multiplied A on the left by 5 elementary matrices and we got a diagonal matrix whose last diagonal entry is 1. If we continue by induction, we will get a matrix B which is the product of 5(n-1) elementary matrices and

BA = \text{diag} \{a_1a_2 \cdots a_n, 1, \cdots , 1 \} = I,

by (*). Thus A=B^{-1} and the result follows from the second part of Problem 1. \Box


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