# Linear Algebra Vector Spaces: Prove equivalence

## Homework Statement

Prove that the following are equivalent:
1. N(A)=0
2. A is nonsingular
3. Ax=b has a unique solution for each b that exists in R^n.

## The Attempt at a Solution

I think you prove this by showing that 1 implies 2, 2 implies 3, & 3 implies 1.
But after that I don't know how to prove that.

## Answers and Replies

Related Calculus and Beyond Homework Help News on Phys.org
Start by trying one of the three implications you must show.
For example, for 1 implies 2, what does N(A) = 0 tell you about the equation Ax=0? Can you imply this to determine that A is invertible?

if N(A)=0 then in the equation Ax=0, x equals 0. But then how do you show that A is invertible from that?

What does Ax=0 tell you about the columns of A? Let a1, ...., an be some vector in A, and the components of x are x1,..., xn. Then from Ax=0, you get equations of the form a1x1 + ..... + anxn = 0, and from N(A)=0 you get that all the xi's are 0. What can you conclude from this?

Ok I think I figured the first part out:
The columns of A are linearly independent.
So, A is row equivalent to the identity matrix, and Ax=0 and Ix=0 have only the solution, x=0. So, the A=E1E2...Ek, which says that the product of invertiable matrices is invertiable and E is invertiable, so A is invertable.

So, then How do you imply that 2 equals 3, and 3 equals 1?

Give them a try, pick one of the two. What have you done so far?

For example, 2 to 3:
A is non singular, So A-1 exists. Then multiply Ax=b by A-1 . What does this imply? Is this unique?

Last edited:
And the other directions are probably similar. It's all a matter of shuffling between invertibility, linear dependence, etc.

ok, i think that for 2 implies 3. A being nonsingular says that A inverse exists. So, then you can show that x= (inverseA)*B which proves #3. Is this right?

Now, the one I am stuck on is 3 implies 1. Any hints to get me started?

Yes, you are correct and it is easy to understand that this is in fact a unique solution (since for all other solutions you can simply left multiply by A-1 to get the same conclusion.

I have not yet thought about 3 to 1. Think about it. Maybe there is some trick you can use from what you already proved in this problem.

You are assuming that Ax=b has a unique solution. Then b=0 also has a unique solution, and you know what it is.

I still have no idea on the last one. I have looked all through my book...

vela
Staff Emeritus
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}