Proving the Independence and Span of Matrix Columns | Theorem Help

[stunner5000pt]

Let A be a mxn matrix with columns C1,...Cn. If rank A = n, show taht
${A^T C_{1},...,A^T C_{N}}$is a basis of Rn

since Rank A = n, then the columns are linearly independant

so does that automatically mean that any multiple,, like A transpose for example, will keep the independance of the Columns?

A theorem also tells us that if the Rank A = n, then the column span Rn. So the columns span Rn in this case

is this adequate for a proof?

 

[0rthodontist]

No, it is not. Multiplication by A is not just multiplication by a scalar. It could also rotate the vectors.

You need to show that the set {A^T C1, ... A^T Cn} is linearly independent, which implies it is a basis in Rn since it has n elements. Assume otherwise. Then you have a1A^T C1 + ... + anA^T Cn = 0 for a1 ... an not all zero. Can you manipulate this to get a contradiction?

 

[AKG]

The answer is below. Highlight the first white line if you need a hint. Every time you need a hint, highlight the next line. For your own good, highlight as few lines as possible, i.e. try to do most of it yourself.

{A^TC₁, ..., A^TC_n} is a basis of Rⁿ

iff {A^TC₁, ..., A^TC_n} is linearly independent

iff the columns of the matrix (A^TC₁ ... A^TC_n) are linearly independent

iff det[(A^TC₁ ... A^TC_n)] is non-zero

iff det[A^T(C₁ ... C_n)] is non-zero

iff det[A^TA] is non-zero

iff det[A^T]det[A] is non-zero

iff det[A]det[A] is non-zero

iff det[A] is non-zero

iff rank[A] = n

 

[nocturnal]

The answer is below. Highlight the first white line if you need a hint. Every time you need a hint, highlight the next line. For your own good, highlight as few lines as possible, i.e. try to do most of it yourself.

{A^TC₁, ..., A^TC_n} is a basis of Rⁿ

iff {A^TC₁, ..., A^TC_n} is linearly independent

iff the columns of the matrix (A^TC₁ ... A^TC_n) are linearly independent

iff det[(A^TC₁ ... A^TC_n)] is non-zero

iff det[A^T(C₁ ... C_n)] is non-zero

iff det[A^TA] is non-zero

iff det[A^T]det[A] is non-zero

iff det[A]det[A] is non-zero

iff det[A] is non-zero

iff rank[A] = n

This would be true if A were a nxn matrix, but the question specifies that A is a mxn matrix.

 

[stunner5000pt]

No, it is not. Multiplication by A is not just multiplication by a scalar. It could also rotate the vectors.

You need to show that the set {A^T C1, ... A^T Cn} is linearly independent, which implies it is a basis in Rn since it has n elements. Assume otherwise. Then you have a1A^T C1 + ... + anA^T Cn = 0 for a1 ... an not all zero. Can you manipulate this to get a contradiction?

ok supose that
$$a_{1}A^T C_{1} + ... + a_{n}A^T C_{n} = 0$$
where not all ai are zero

we know that A is not a zero matrix because the rank A = n <= m
not A transpose is not zero
then the Columns must be all zero
But again A is not zero. Thus by contradiction we get that all the ai must be zero

IS that good?

 

[AKG]

This would be true if A were a nxn matrix, but the question specifies that A is a mxn matrix.

Oops, I overlooked that.

 

[AKG]

ok supose that
$$a_{1}A^T C_{1} + ... + a_{n}A^T C_{n} = 0$$
where not all ai are zero

we know that A is not a zero matrix because the rank A = n <= m
not A transpose is not zero
then the Columns must be all zero
But again A is not zero. Thus by contradiction we get that all the ai must be zero

IS that good?

No. Why is the underlined stuff true? Try this:

{A^TC₁, ..., A^TC_n} is a basis
iff {A^TC₁, ..., A^TC_n} is linearly independent
iff c₁A^TC₁ + ... + c_nA^TC_n = 0 implies c₁ = ... = c_n = 0
iff A^T(c₁C₁ + ... + c_nC_n) = 0 implies c₁ = ... = c_n = 0
iff A^T(c₁C₁ + ... + c_nC_n) = 0 implies c₁C₁ + ... + c_nC_n = 0 [since the C_i are linearly independent since Rank[A] = n, so $\sum c_iC_i = 0 \Leftrightarrow \forall i,\, c_i=0$]
iff A^T(X) = 0 implies X = 0

But rank[A^T] = rank[A] = n. Is there some theorem which says that, given this, that italicized line must be true?

 

[0rthodontist]

A^T has all independent columns since it has rank n. Therefore, A^T(X) = 0 can only be true if X is 0. Otherwise you would have a linear combination of the columns of A^T, with not all nonzero coefficients, yielding 0, which we know cannot happen.

 

[Hurkyl]

A^T has all independent columns since it has rank n

You mean rows.


Anyways, I think everyone's overlooking something important!

$$a_{1}A^T C_{1} + ... + a_{n}A^T C_{n} = 0$$

This means that the vector $a_1 C_1 + \cdots + a_n C_n$ is an element of the null space of $A^T$. People seem to be arguing that the null space is trivial, and therefore the vector must be zero.

But that's wrong!

$A^T$ is a rank n nxm matrix. Since it's operating on an m-dimensional space, it must have a null space of dimension (m-n).


You have to use the fact that these m-long column vectors are special -- each of the $C_n$ is the transpose of one of the rows of $A^T$.


In fact, if we place the $A^T C_i$ into a matrix, we get:

$$[ A^T C_1 \mid A^T C_2 \mid \cdots \mid A^T C_n ] = A^T A$$

There's actually a 1-line proof that this (square) matrix is nonsingular, but I think the approach you guys are using ought to work, as long as you start using the fact that the $C_i$ are special m-long vectors, and not arbitrary m-long vectors.

 

[0rthodontist]

You mean rows.

Oh yeah... I was picturing it wrong.

 

[AKG]

There's actually a 1-line proof that this (square) matrix is nonsingular

When m=n, this is easy, and that's what I did in my first post because I assumed we were dealing with square matrices. If m is not n, what is the 1-line proof?

 

[Hurkyl]

$$A^TA \vec{v} = 0 \implies \vec{v}^T A^T A \vec{v} = 0 \implies (A\vec{v})^T (A \vec{v}) = 0 \implies A \vec{v} = 0 \implies \vec{v} = 0$$
Where the last one follows because A has rank n and is operating on R^n.

 

[AKG]

Very nice, thanks Hurkyl.