Question

How would I find the kernel of the linear transformation and express the answer as the span of a set of vectors? For A=[2 1 -1; 1 -2 1; 1 -7 4; 3 4 -3]

Answer 1

I reduced it to row echelon form and got two independent variables [1 -7 4; 0 1 -3/5; 0 0 0; 0 0 0]. Then I got stuck from there. I don't know how to set the equations up for this matrix.

Answer 2

Oh my bad the ref I got was [1 0 -1/5; 0 1 -3/5; 0 0 0; 0 0 0]

Answer 3

when you row reduce the the matrix, you identify the pivot columns and the free variable columns. Then you write the pivot variables in terms of the free variables, and factor out the free variables to get the vectors that span the Null Space (or Kernel).

Answer 4

Why is it sometimes the last column is not a variable column?

Answer 5

I get confused when that applies

Answer 6

you mean, not a free variable column? if the last column isnt free, and they are all pivot columns, that means those 3 columns are linearly independent, and the Null Space (Kernel) will just be the 0 vector.

Answer 7

Umm I don't know if we're talking about the same thing like let's say it's [1 0 5; 0 1 4; 0 0 0] sometimes the answer becomes x_1 = 5 and x_2 =4?

Answer 8

And why is it for the answer there's only 3 variables? If there's two rows of zeroes aren't there 2 free variables?

Answer 9

in that example, when we turn that matrix back into a system of a equations it would look like this:
\[x_1+5x_3= 0 \]
\[x_2+4x_3= 0\]
\[0 = 0\]
Its a 3x3 matrix, so thats 3 equations in 3 unknowns.
The pivot variables are x1 and x2, the free variable is x3.

Answer 10

The reason for the zeros on the right side of the equation comes from the fact we are looking for the kernel, which is finding the solutions to Ax = 0

Answer 11

So to complete the problem, we solve for the pivot variables:
\[x_1+5x_3 = 0 \Rightarrow x_1 = -5x_3\]
\[x_2+4x_3=0\Rightarrow x_2 = -4x_3\]
x3 is free.
so the vector looks like:
-5x3
-4x3
x3
and when you factor out the x3 you get the vector:
-5
-4
1

Answer 12

Oh okay that make sense thank you, I forgot about the Ax=0. But last question so even if you have two rows of zeroes that would count as 1 free variable.

Answer 13

When would you have two sets of basis for the kernel T?

Answer 14

yes, that would still just be one free variable. it doesnt matter how many zero rows you have.

I think you are asking when would there be two vectors that span the Kernel of a matrix? That would be the case if there were 2 free variables. The number of vectors you get for your basis of the Kernel = number of free variables.

Answer 15

Oh okay thank you! That makes a lot more sense

Answer 16

Actually sorry here's what I'm confused about like for this example in my book it says A=[1 -8 -7 -4; 2 -3 -1 5; 3 2 5 14] and when reduced it becomes [1 0 1 4 -8; 0 1 1 1 -2; 0 0 0 0]

Answer 17

x_1 = -8 s - 4t x_2 = -2 -s -t x_3 = s x_4 =t

Answer 18

Any ideas why the last column isn't a free variable column? Oh and this question isn't asking for a the kernel for whether b is the column space of A

Answer 19

So is it only if it's asking for the kernel the last column is also a free variable column?

Answer 20

i dont like your books notation =/ anywhos, When we are looking for the Column space of a matrix, we are only looking for the pivot columns. So we row reduce a matrix, see which columns have pivots, then look back at the original problem to see which columns those were, and thats a basis for the column space. Those columns span the columns space.
When looking for the Kernel, thats when we are interested in the free variables instead.
Im looking at the problem you posted, writing out a solution on paper.

Answer 21

Oh my bad I get what they did

Answer 22

They gave a b vector too b=[8 -10 -28]

Answer 23

b=[8; -10; -28]

Answer 24

Oh your explanation actually makes a lot more sense than what it states in the textbook!! Thanks a lot man, you really know your linear algebra material haha

Answer 25

Oh I see so it only works if all solutions exists

Answer 26

Thanks for your help!

Answer 27

no problem :)