<p>Prove that the identity matrix cannot be the product of 3 row exchanges (or five). It can be the product of 2 exchanges (or 4).</p>
<p>Now, to start, we try to count the number of rows that are different from the identity matrix. For the first row exchange, it’s 2. Second, it becomes either 0, 3, or 4 [both can change to identity, two more might be different, or a different row may switch with a same row - resulting in one extra different row]. Third, it can go to 1,2,3,4,6. Four, 0,1,2,3,4,5,6,8… But I think I’m doing this incorrectly. Can anyone help?</p>
<p>Let A be n by n identity matrix. Let e1, … , ek be elementary row-swtich matrices; e^(-1), …, ek^(-1) be their inverses.</p>
<p>A is invertible. Therefore</p>
<p>ek * e(k-1) * … * e1 * A = I
A = e1^(-1) * … * ek^(-1) * I</p>
<p>Because A = I,
e1^(-1) * … * ek^(-1) = I</p>
<p>Therefore, ek(-1) * e(k-1)^(-1) * … * e1^(-1) = I . Therefore one e^(-1) must be an inverse of another e^(-1). (ie e^(-2) is an inverse of e^(-1), so that e1^(-1) * e2^(-2) = I
e3^(-1) to e4^(-1) so that e3^(-1) * e4^(-1) = I).</p>
<p>For all that to happen for all e^(-1)'s, k must be even, ie there must be even number of e^(-1)matrices. </p>
<p>Therefore, A n by n identity matrix can’t be a product of odd number of elementary row-swtich matrices.</p>