Dimension of the solution space for Ax=0

[eyehategod]

Find the dimension of the solution space of Ax=0, where
A=1 2 5
-1 3 1

is the rank(A)=2
so the nullity(A)=2?
is this correct?

 

[HallsofIvy]

If A is a linear transformation from vector space U to vector space V, then range(A)+ nullity(A)= dimension of U. Here, A is from R³ to R². Yes, the rank of A is 2. No, the nullity of A is not 2 also.

You could just as well do this directly: if x+ 2y+ 5z= 0 and -x+ 2y+ z= 0, then, adding the equations, you must also have 4y+ 6z= 0 or z= (2/3)y. You can pick y to be anything you like and then calculate both x and z. What does that tell you about the nullity?

 

[eyehategod]

so the nullity(A)=1?

 

[Office_Shredder]

Read HallsofIvy's post again... if rank(A)=2, and rank(A)+nullity(A)=2, what does nullity(A) equal?

 

[eyehategod]

so your saying n=2? i thought it was 3. n is always the number of rows?

 

[eyehategod]

//the last blury part is "Furthermore, rank of the matrix is 2,"

This is an example from my book. When I reduce A i get
1 0 13/5
0 1 6/5so according to the example in the book
my rank would be 2
and nullity would be 1
with n=3
Is this wrong?

 

[HallsofIvy]

Yes, that is correct.

 

[eyehategod]

just to make sure, what Office_Shredder said was wrong?

 

[HallsofIvy]

Yes, I believe he confused the dimension of the domain and range spaces.

At any rate, as I also pointed out, you could find the nullity directly:
" if x+ 2y+ 5z= 0 and -x+ 2y+ z= 0, then, adding the equations, you must also have 4y+ 6z= 0 or z= (2/3)y. You can pick y to be anything you like and then calculate both x and z. "

Since you can pick one number arbitrarily, the kernel has dimension 1.

 

[eyehategod]

great thanks [solved]

 

[Office_Shredder]

I apologize, I can't read :/