Prove B is invertible if AB = I

[PeroK]

I think I can show $B$ is onto but using pivot point. Since $B$ has pivot point on each row, $B$ is onto.

How to show it without using pivot?

Thanks

I must admit I've never heard of "pivot point" in linear algebra. You can show that $B$ is onto if it is one-to-one by using a basis for the vector space, $V$, where $B: V \to V$ is a linear mapping.

Let $\{ e_1, e_2 \dots e_n\}$ be a basis for V and consider $\{ Be_1, Be_2 \dots Be_n\}$. By the linearity of $B$ we have:
$$a_1Be_1 + a_2Be_2 + \dots a_nBa_n = 0 \ \Rightarrow \ B(a_1e_1 + \dots + a_ne_n) = 0$$(And, as $B$ is one-to-one and the $e_i$ are linearly independent):
$$\Rightarrow \ a_1e_1 + \dots + a_ne_n = 0 \ \Rightarrow \ a_1 = a_2 \dots = a_n = 0$$This shows that the vectors $Be_i$ are linearly independent and hence a basis for $V$.

Finally, if $v \in V$, then for some scalars $v_i$:
$$v = v_1Be_1 + \dots + v_nBe_n = B(v_1e_1 + \dots + v_ne_n)$$And, as $v$ was arbitrary, we see that $B$ is onto.

 

[fresh_42]

This shows that the vectors $Be_i$ are linearly independent and hence a basis for $V$.

Finally, if $v \in V$, then for some scalars $v_i$:
$$v = v_1Be_1 + \dots + v_nBe_n = B(v_1e_1 + \dots + v_ne_n)$$And, as $v$ was arbitrary, we see that $B$ is onto.

And how does this differ from the rank-nullity theorem the OP already correctly used? You use surjectivity to prove surjectivity. But I don't want to confuse the OP more than he already is. His solution might not have been perfectly phrased, nevertheless, it was correct (post #18).

The idea with the Pivots is ok. In the end, we do have not enough information on what he may use according to his book, and what lies ahead.

 

[PeroK]

And how does this differ from the rank-nullity theorem the OP already correctly used?

The OP asked how to prove it without using pivots, so I showed him.

You use surjectivity to prove surjectivity.

I used injectivity to prove surjectivity. Which seems like an important concept in linear algebra and a neat solution.

 

[fresh_42]

The OP asked how to prove it without using pivots, so I showed him.I used injectivity to prove surjectivity. Which seems like an important concept in linear algebra and a neat solution.

Yes, and it is called the rank-nullity theorem.

 

[songoku]

I must admit I've never heard of "pivot point" in linear algebra.

Maybe the term that you are familiar with is "pivot" or "pivot element" or "pivot position"

Thank you very much for all the help and explanation fresh_42, PeroK, Office_Shredder, Addez123, WWGD

 

[Hall]

It seems to me that the thread went a little out of course. A fine approach to this problem would be to consider the matrices as connected with the Linear transformations.

If A (n x n dimension) is the coefficient matrix for T:V_n -> V_n , T is a Linear transformation and the columns of A represent the coefficients of the basis elements of range V_n when T is applied to one of the basis elements of domain V_n.

Prove that A is invertible if and only if T is invertible. And then prove that the inverse of A is the coefficient matrix of $T^{-1}$.

 

[nuuskur]

$\det AB\neq 0$ implies $\det B\neq 0$. Therefore, as was already pointed out, $B=BAB$ and therefore $BA$ is the identity.

 

[JFerreira]

Hi everyone,
I founded this discussion very interesting, so a search if someone did it before.

"A is linear transformation from a finite dimensional vector space to itself. AB=I amounts to saying A is surjective, hence it is bijective and has a left inverse C, so that CA=I. Of course, B=(CA)B=C(AB)=C, i.e. BA=I."

I founded this in the thread https://math.stackexchange.com/questions/152668/what-'s-the-short-proof-that-for-square-matrices-$ab-=-i$-implies-$ba-=-i$?
(searching for "\(AB = I\) for square matrices " on SearchOnMath) that can give some other contribuitions.

 

[Prof B]

You say that you learned about ranks. Did you learn the following?

1. rank(A) $\geq$ rank(AB)?
2. If A is nxn and rank(A) = n then A is invertible.

You could prove that A is invertible using 1 and 2.

 

[Delta2]

Using determinants and that $det(AB)=det(A)det(B)$ for square matrices+ A square matrix is invertible if its determinant is not zero, this problem is very easy. But OP says at post #4 that they haven't covered determinants yet.

 

[nuuskur]

Since $AB=I$, it means $A$ is surjective and $B$ is injective. Thus, $A,B$ are both bijective due to finite dimension.