Question

Prove
If A has independent columns, then Ax=0 has only a trivial solution x=0

Answer 1

it depends on the space

I will assume you mean \(n\) independent columns in some \(n\) dimensional space.

Suppose \(Ax=0\)

Since \(A\) has full rank, it must have an inverse. Multiply both sides (on the left) by \(A^{-1}\)

Answer 2

A could be rectangular.
"Independent columns" only implies a full column rank, right ?

Answer 3

full rank \(\iff\) invertable

Answer 4

full column rank need not imply full rank ?

Answer 5

what?

Answer 6

again, the question, as it is stated, can not be answered. So I made some assumptions.

Answer 7

We can assume a general matrix, A is a mxn matrix with n independent column vectors

Answer 8

if the only thing in the kernel is the 0 vector then the matrix has full column rank and thus full row rank

Answer 9

check out 'invertable matrix theorem'

Answer 10

That is true only if A is a square matrix

Answer 11

oh hahahah you are the person that posted the question...

Answer 12

We cannot define a regular inverse for rectangular matrices

Since the question doesn't specifically state the type of matrix, I think it is reasonable assume that the matrix could be rectangular

Answer 13

ok sorry forget what I said. I was making assumptions because I was not clear from your question.

Answer 14

OK suppose you have \(n\) linearly independent columns and

\(Ax=0\), then

\(x_1v_1+x_2v_2+...+x_nv_n=0\), correct?

Answer 15

Example matrix equation

\[\begin{bmatrix}1&2\\2&3\\1&1\end{bmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix} \]

Answer 16

yes yes, I am with you now. I didn't know you were the guy that asked the qeustion.

Answer 17

are you with me above?

Answer 18

Yeah... I think we need to show
\(Ax = x_1v_1+x_2v_2+...+x_nv_n=0 \implies x_1=x_2=\cdots = x_n=0 \)

Answer 19

How could you add things up to get 0 unless, at some point, you are subtracting some quantity from itself.

Answer 20

hint, A+B+C+D=0 implies A=-B-C-D

Answer 21

Since it is not the case that all of the \(v_i\) are 0....

Answer 22

\(Ax = x_1v_1+x_2v_2+...+x_nv_n=0\)


\(x_1v_1=-x_2v_2-...-x_nv_n\)

Answer 23

and for sure both sides are not 0 right?

Answer 24

this shows they are not lin ind

Answer 25

contradiction...

Answer 26

None of the \(v_i\)'s are 0
but how do we know that the right hand is not 0 ?

Answer 27

How do we know that some combination of linearly independent vectors doesn't evaluate to 0 ?

Answer 28

well, we know the \(v_i\) are non zero and there is some \(x_i\) that is not zero.

So \(x_iv_i=-x_1v_1-x_2v_2-...-x_{i-1}v_{i-1}-x_{i+1}v_{i+1}-...-x_nv_n\)

Answer 29

now, since the left is not zero, the right is not zero

Answer 30

The \(v_i\) are not zero because they are lin ind, there is some \(x_i\) that is not zero because \(x\ne 0\).

Answer 31

Oh oh I think I see what you're saying... one moment

Answer 32

If the right hand side evaluates to the 0 vector, then the coefficient xi on left hand side must be the number 0. Beautiful ! Thank you xD

Answer 33

Something looks wrong,
right hand side can evaluate to some other nonzero vector right ?

Answer 34

A + B + C = 0

A = -B - C

A = 2
B = -1
C = -1

Answer 35

the left is non zero, so the right is not zero, so we have non zero multiples of \(v_i\) equal to scalar multiples of the other vectors

Answer 36

so the v_i are not lin ind as assumed.

Answer 37

2,-1,-1 are not lin ind vectors

Answer 38

what is the problem

Answer 39

?

Answer 40

Ohk that proves the contrapositive :

if a nontrivial solution exists for Ax=0, then the columns of A are not independent.

Nice nice :)

Answer 41

correct

Answer 42

It seems to me that more work is being done here than necessary (though admittedly, I've only invested a little over two minutes looking at the comments above).

To be brief: If the columns of \(\mathbf{A}_{m\times n}\) are all mutually independent, then the columns span all of \(\mathbb{R}^n\) and \(\mathrm{rank}(\mathbf{A})=n\) (i.e. full column rank if \(m\neq n\), or simply full rank if \(m=n\)). The rank-nullity theorem (number of columns = rank + nullity) tells you the dimension of the nullspace of \(\mathbf{A}\) must be \(0\) and so only contains the zero vector.

Answer 43

Alternatively, if you're not familiar with the result of that theorem, you can think of it this way: given \(\mathbf{A}_{m\times n}\) with \(n\) independent columns, you know that \(\mathbf{A}\) can be row reduced with exactly \(n\) pivots, i.e.
\[\mathrm{rref}(\mathbf{A})=\begin{bmatrix}p_1&\cdots&\cdots&\cdots\\
0&p_2&\cdots&\cdots\\
\vdots&\vdots&\ddots&\vdots\\
0&0&\cdots&p_n\\
0&0&\cdots&0\end{bmatrix}\]With \(n\) pivot variables, there is no room for any free variables, which means there are no "special" vectors in the nullspace.

Answer 44

Sure that works but you use way more "stuff" than we did here. The above got a little wacky but the following is the proof using definitions only.

Suppose \(A\) is made of \(n\) independent vectors and suppose b.w.o.c. \(x\ne 0\) and \(Ax=0\) . \(\exists \ x_i \) s.t. \(x_i\ne 0\). Then \(x_1v_1+x_1v_2+...+x_nv_n=0\) so \(x_iv_i=-x_1v_1-x_2v_2+...+x_{i-1}v_{i-1}+x_{i+1}v_{i+1}+...+x_nv_n\). Since the \(v_i\ne0 \ \forall i\in[n]\), we have the left side is non zero and thus the right is also non zero. So \(v_i\) is a linear combination of the other vectors and thus a contradiction.

i.e. This is basic work a 6th grader can understand, which I think is cleaner and easier than the proposed proof.

Answer 45

Fair enough, but the OP's use of "rank" is suggestive of knowing something about the R-N theorem. Ultimately up to him/her to decide which approach is easier.

Answer 46

obviously