MHB Proving statements regarding invertible matrices

[Bruce Wayne1]

I've been watching the OCW Linear Algebra course and I have a textbook, and I came across something that I think is fascinating: The Invertible Matrix Theorem. I've seen some proofs and I find that a lot of the statements are linked in different orders and sometimes the author will site one tiny little theorem and justify 5 statements at once. I've been able to prove a few on my own. I'm looking at some and I don't understand why they work exactly. OK. I want to prove that the columns of matrix A span Rⁿ and that this is equivalent to the linear transformation x->Ax maps Rⁿ onto Rⁿ. I would like to prove the converse is also true.

So my thinking is:

Assume the columns of matrix A span Rⁿ. By the definition of Ax , for each b in Rⁿ , the equation Ax=b has a solution. This implies that T(x)=b has at least one solution.

That's when I get myself totally confused. I think I'm missing a few intermediary steps in the proof. How do I do this proof?

 

[Jameson]

Hi there. How many other parts of the invertible matrix theorem may we use while doing this proof? If we look at this list of components of this theorem, you are trying to prove that (h) implies (i) and the converse. Can we use (a)-(g)? If not you'll have to define and prove some other stuff first I believe.

 

[Bruce Wayne1]

Hi there. How many other parts of the invertible matrix theorem may we use while doing this proof? If we look at this list of components of this theorem, you are trying to prove that ( h) implies (i) and the converse. Can we use (a)-(g)? If not you'll have to define and prove some other stuff first I believe.

Correct, I am trying to prove ( h)<-->(i) from that list. I'm trying to do it directly between those two, without using the others. However, I do understand (g) can be used, as I tried, but how do I justify it? In other words, I need to have it justified. I'm not looking to just say "oh, statement blah is equivalent, done" because that's how the book does it and I'm trying to justify each step.

 

[I like Serena]

I've been watching the OCW Linear Algebra course and I have a textbook, and I came across something that I think is fascinating: The Invertible Matrix Theorem. I've seen some proofs and I find that a lot of the statements are linked in different orders and sometimes the author will site one tiny little theorem and justify 5 statements at once. I've been able to prove a few on my own. I'm looking at some and I don't understand why they work exactly. OK. I want to prove that the columns of matrix A span Rⁿ and that this is equivalent to the linear transformation x->Ax maps Rⁿ onto Rⁿ. I would like to prove the converse is also true.

So my thinking is:

Assume the columns of matrix A span Rⁿ. By the definition of Ax , for each b in Rⁿ , the equation Ax=b has a solution. This implies that T(x)=b has at least one solution.

That's when I get myself totally confused. I think I'm missing a few intermediary steps in the proof. How do I do this proof?

Hail, mighty B! (Bow)

In any proof you should always start from the definitions.

Definition from wiki:

the span of a set S may be defined as the set of all finite linear combinations of elements of S.​

In your case that means that any vector $\mathbf v \in \mathbb R^n$ can be written as
$\qquad \mathbf v = \lambda_1 \mathbf a_1 + \lambda_2 \mathbf a_2 + ... + \lambda_n \mathbf a_n$
where $\mathbf a_i$ is the set of column vectors of the matrix $A$, and $\lambda_i$ are the factors making up the linear combination with respect to these column vectors.The Definition from wiki about a function that maps onto:

a function f with domain X and codomain Y is surjective if for every y in Y there exists at least one x in X with f(x)=y.​

In your case that means that for any vector $\mathbf v \in \mathbb R^n$ there is at least one vector $\mathbf x \in \mathbb R^n$ such that $A \mathbf x = \mathbf v$.
So let's write your first half of the proof a bit neater.Lemma 1
If the columns of matrix $A$ span $\mathbb R^n$, then the linear transformation $\mathbf x \mapsto A \mathbf x$ maps $\mathbb R^n$ onto $\mathbb R^n$.

Proof
Assume the columns of matrix $A$ span $\mathbb R^n$.

Then any vector $\mathbf v \in \mathbb R^n$ can be written as
$\qquad \mathbf v = \lambda_1 \mathbf a_1 + \lambda_2 \mathbf a_2 + ... + \lambda_n \mathbf a_n$
where $\mathbf a_i$ is the set of column vectors of the matrix $A$, and $\lambda_i$ are the factors making up the linear combination with respect to these column vectors.

Using the definition of matrix multiplication, we can rewrite this as:
$$\mathbf v = A \begin{bmatrix}\lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n\end{bmatrix}$$
In other words, for any vector $\mathbf v \in \mathbb R^n$ there is at least one vector $\mathbf x \in \mathbb R^n$ such that $A \mathbf x = \mathbf v$.

Therefore the linear transformation $\mathbf x \mapsto A \mathbf x$ maps $\mathbb R^n$ onto $\mathbb R^n$. $\qquad \blacksquare$
Perhaps you can do the second (converse) half of the proof?

 

[Bruce Wayne1]

Hail, mighty B! (Bow)

In any proof you should always start from the definitions.

Definition from wiki:

the span of a set S may be defined as the set of all finite linear combinations of elements of S.​

In your case that means that any vector $\mathbf v \in \mathbb R^n$ can be written as
$\qquad \mathbf v = \lambda_1 \mathbf a_1 + \lambda_2 \mathbf a_2 + ... + \lambda_n \mathbf a_n$
where $\mathbf a_i$ is the set of column vectors of the matrix $A$, and $\lambda_i$ are the factors making up the linear combination with respect to these column vectors.The Definition from wiki about a function that maps onto:

a function f with domain X and codomain Y is surjective if for every y in Y there exists at least one x in X with f(x)=y.​

In your case that means that for any vector $\mathbf v \in \mathbb R^n$ there is at least one vector $\mathbf x \in \mathbb R^n$ such that $A \mathbf x = \mathbf v$.
So let's write your first half of the proof a bit neater.Lemma 1
If the columns of matrix $A$ span $\mathbb R^n$, then the linear transformation $\mathbf x \mapsto A \mathbf x$ maps $\mathbb R^n$ onto $\mathbb R^n$.

Proof
Assume the columns of matrix $A$ span $\mathbb R^n$.

Then any vector $\mathbf v \in \mathbb R^n$ can be written as
$\qquad \mathbf v = \lambda_1 \mathbf a_1 + \lambda_2 \mathbf a_2 + ... + \lambda_n \mathbf a_n$
where $\mathbf a_i$ is the set of column vectors of the matrix $A$, and $\lambda_i$ are the factors making up the linear combination with respect to these column vectors.

Using the definition of matrix multiplication, we can rewrite this as:
$$\mathbf v = A \begin{bmatrix}\lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n\end{bmatrix}$$
In other words, for any vector $\mathbf v \in \mathbb R^n$ there is at least one vector $\mathbf x \in \mathbb R^n$ such that $A \mathbf x = \mathbf v$.

Therefore the linear transformation $\mathbf x \mapsto A \mathbf x$ maps $\mathbb R^n$ onto $\mathbb R^n$. $\qquad \blacksquare$
Perhaps you can do the second (converse) half of the proof?

Thanks. Here's my attempt:

Assume the linear transformation $\mathbf x \mapsto A \mathbf x$ maps $\mathbb R^n$ onto $\mathbb R^n$.

For all $\mathbf v \in \mathbb R^n$ there is at least one vector $\mathbf x \in \mathbb R^n$ such that $A \mathbf x = \mathbf v$.

$A \mathbf x = \mathbf v$ can be rewritten as $$\mathbf v = A \begin{bmatrix}\lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n\end{bmatrix}$$

This can further be rewritten as $\qquad \mathbf v = \lambda_1 \mathbf a_1 + \lambda_2 \mathbf a_2 + ... + \lambda_n \mathbf a_n$

Therefore the columns of matrix $A$ span $\mathbb R^n$. $\qquad \blacksquare$

 

[I like Serena]

Thanks. Here's my attempt:

Assume the linear transformation $\mathbf x \mapsto A \mathbf x$ maps $\mathbb R^n$ onto $\mathbb R^n$.

For all $\mathbf v \in \mathbb R^n$ there is at least one vector $\mathbf x \in \mathbb R^n$ such that $A \mathbf x = \mathbf v$.

$A \mathbf x = \mathbf v$ can be rewritten as $$\mathbf v = A \begin{bmatrix}\lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n\end{bmatrix}$$

This can further be rewritten as $\qquad \mathbf v = \lambda_1 \mathbf a_1 + \lambda_2 \mathbf a_2 + ... + \lambda_n \mathbf a_n$

Therefore the columns of matrix $A$ span $\mathbb R^n$. $\qquad \blacksquare$

Looks just fine! ;)

 

[Bruce Wayne1]

Thanks! It was a lot simpler than I had anticipated. (Smile)