## Elementary matrices (1)

Posted: January 28, 2011 in Elementary Algebra; Problems & Solutions, Linear Algebra
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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. $\Box$