Lecture 17 · Week 5· 15:57

The invertible matrix theorem

Elementary matrices, equivalent conditions for invertibility.

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In this lecture

Today: many questions, one answer
Elementary matrices
Row operations are multiplications
Undoing an elementary matrix
Why Gauss–Jordan inversion works
An example
The invertible matrix theorem
Proof: the main loop
Proof: the other statements
Using the theorem
Summary

Materials

Check yourself

Try it before you look

Write down the $3 \times 3$ elementary matrix $E$ of $R_{2} \to R_{2} + 4 R_{3}$ and its inverse, and check that $E E^{- 1} = I_{3}$ with Theorem 3.6.3.
Let $A = [2614]$ over $R$ . Find elementary matrices $E_{1}, E_{2}, E_{3}$ with $E_{3} E_{2} E_{1} A = I_{2}$ , and write both $A^{- 1}$ and $A$ as products of elementary matrices.
Let $A$ be $n \times n$ . True or false? (i) If $A x = e_{1}$ has two different solutions, then the columns of $A$ do not span $F^{n}$ . (ii) If $A x = e_{1}$ has a solution, then $A$ is invertible.

Work it out on paper first. Then open the hint if you are stuck, and the answer to check.

1Hint

Everything you need is in Section 3.6 of the course book. Open it at page 38 of the chapter PDF; any theorem the question names is stated there, with an example worked the same way.

2Answer

$E = [100010041]$ , and $E^{- 1} = [100010 0 - 4 1]$ is the matrix of the reverse operation $R_{2} \to R_{2} - 4 R_{3}$ (Theorem 3.6.6). By Theorem 3.6.3, $E E^{- 1}$ is $E^{- 1}$ after $R_{2} \to R_{2} + 4 R_{3}$ : row $2$ becomes $(0, 1, - 4) + 4 (0, 0, 1) = (0, 1, 0)$ , so $E E^{- 1} = I_{3}$ .
$R_{2} \to R_{2} - 3 R_{1}$ gives rows $(2, 1)$ , $(0, 1)$ ; $R_{1} \to R_{1} - R_{2}$ gives $(2, 0)$ , $(0, 1)$ ; $R_{1} \to \frac{1}{2} R_{1}$ gives $I_{2}$ . So $E_{1} = [1 - 3 01]$ , $E_{2} = [10 - 1 1]$ , $E_{3} = [\frac{1}{2} 0 01]$ , and by Theorem 3.6.8 $A^{- 1} = E_{3} E_{2} E_{1} = E_{3} [4 - 3 - 1 1] = [2 - 3 - \frac{1}{2} 1]$ (the $2 \times 2$ formula agrees: $a d - b c = 2$ ), while $A = E_{1}^{- 1} E_{2}^{- 1} E_{3}^{- 1} = [1301] [1011] [2001] = [1314] [2001] = [2614]$ .
(i) True. If $A u = A v = e_{1}$ with $u \neq = v$ , then $A (u - v) = 0$ with $u - v \neq = 0$ , so statement (3) of Theorem 3.6.10 fails, and so does (8). (ii) False. $A = [1000]$ has $A e_{1} = e_{1}$ , but its second column is zero, so its columns are dependent and $A$ is not invertible. Statement (7) asks for a solution for every $b$ , not just one.