Proof involving invertible matrices

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February 21st 2010, 11:36 AM
madmartigano

Proof involving invertible matrices

I need to prove that if A is an invertible matrix and B is row equivalent to A, then B is also invertible.

I understand the logic of that statement, but I'm not experienced enough to know how to write it into words that prove it.

Here's my thinking so far:

1.) B is row equivalent to A, therefore there exist a sequence of elementary row operations that can be performed on A to produce B:
(En*...*E2*E1)A=B

2.) A is invertible, therefore there exist a sequence of elementary row operations that can be performed on A to produce the identity matrix:
(En*...*E2*E1)A=I

3.) From 1 & 2, there exist a sequence of elementary row operations that can be performed on B to produce the identity matrix, therefore B is also invertible.

How can I write the above statements so that they flow in a step-by-step proof?
February 21st 2010, 11:40 AM
tonio

Quote:

Originally Posted by madmartigano

I need to prove that if A is an invertible matrix and B is row equivalent to A, then B is also invertible.

I understand the logic of that statement, but I'm not experienced enough to know how to write it into words that prove it.

Here's my thinking so far:

1.) B is row equivalent to A, therefore there exist a sequence of elementary row operations that can be performed on A to produce B:
(En*...*E2*E1)A=B

2.) A is invertible, therefore there exist a sequence of elementary row operations that can be performed on A to produce the identity matrix:
(En*...*E2*E1)A=I

Well, yes, but the first $E_i's$ do not HAVE to be the same as the second ones, do they? So call the elementary matrices representing the row operations in 2. as $F_1,F_2,...,F_m$ (watch the index!), and try again (you're close...!)

Tonio

3.) From 1 & 2, there exist a sequence of elementary row operations that can be performed on B to produce the identity matrix, therefore B is also invertible.

How can I write the above statements so that they flow in a step-by-step proof?

.
February 21st 2010, 12:11 PM
madmartigano

Is this a step in the right direction?

$\begin{array}{l} {E_n} \cdots {E_2}{E_1}A = I\,\,\, \Rightarrow \,\,\,\,A = {E_1}^{ - 1}{E_2}^{ - 1} \cdots {E_n}^{ - 1} \\ \\ {F_m} \cdots {F_2}{F_1}A = B\,\,\, \Rightarrow \,\,\,A = {F_1}^{ - 1}{F_2}^{ - 1} \cdots {F_m}^{ - 1}B \\ \\ {E_1}^{ - 1}{E_2}^{ - 1} \cdots {E_n}^{ - 1} = {F_1}^{ - 1}{F_2}^{ - 1} \cdots {F_m}^{ - 1}B\,\,\, \Rightarrow \,\,\,{E_n} \cdots {E_2}{E_1}B = {F_m} \cdots {F_2}{F_1} \\ \end{array} $

I can't figure out how to show that B will equal I after a sequence of elementary row operations.
February 21st 2010, 05:24 PM
patrick

If A is invertible, it has full rank and hence non-zero determinant. Since row-equivalence is achieved by multiplying by elementary (in particular, invertible) matrices, the determinant of B will be the product of the determinants of a bunch of matrices with non-zero determinants. Therefore, det(B) is non-zero, hence B has full rank and is invertible.
February 22nd 2010, 02:33 AM
HallsofIvy

I would be inclined to use the fact that every elementary matrix (except the ones constructed by multiplying a row by 0 which are not relvant here) is itself invertible.

If $E_1E_2\cdot\cdot\cdot E_n A= B$ then $B^{-1}= (E_1E_2\cdot\cdot\cdot E_n A)^{-1}$ $= A^{-1}E_n^{-1}\cdot\cdot\cdot\ E_2^{-1}E_1^{-1}$ so B is invertible.