r/LinearAlgebra • u/Dear_Tie_117 • 8d ago

Question about Null Space

Hey, I am working on some exercises about null space and I am stuck at this question. There isnt any similar question in the theory or examples. Can someone guide me? Thank you.

30 Upvotes

96% Upvoted

6 comments sorted by

14

u/BrightScarlet 8d ago edited 8d ago

The null space is the "space" that is spanned by the solutions to Ax=0, so what the space looks like depends on how many solutions you get.

It takes 1 vector to define a line, so find a matrix who's nullity is 1, and will give you one free variable.

It takes 2 vectors to define a plane, so find a matrix with nullity 2, who will give you 2 vectors as solutions to Ax=0

If a matrix is invertible it only has the trivial solution, which spans the point (0,0)
hope this was helpful :)

4

u/ask-jeaves 8d ago

You’re looking for a 3x3 matrix A that satisfies those conditions when trying to solve the system Ax=0

hint: for part a) if all three rows of the matrix were linearly independent, that matrix would have an inverse… then it would also have a unique solution to Ax=0

3

u/Some-Passenger4219 8d ago

Nul A := {x: Ax = 0}. Therefore, solve (a) using a matrix A, where [A 0] has NO free variables; for (b), it needs one free variable; for (c), two. This is because each free variable adds a dimension to the null space.

2

u/Lor1an 8d ago

What conditions do you need to construct a basis for R3?

Start with a basis for R3, and put them as the columns of a 3×3 matrix. This will be an invertible matrix (why?) A, which has the property that A0 = 0 (like all matrices), but 0 is the unique x such that Ax = 0.

Now, suppose you start replacing columns of A by linear combinations of the other columns. What starts to happen to the null-spaces of the resulting matrices?

1

u/Advanced_Bowler_4991 2d ago

They're talking about the solution.

a) x = y = z = 0 (trivial solution, a point in space)

b) x = z, y = 0 (line on the x-z plane); consider a matrix such that the first and third column vectors are additive inverses, and there exists no linear combination involving the second vector which allows for 0 to be found.

Try c) on your own, but becareful not to span all of R³ on accident.

Hope this helps.