Null Space of a matrix

  • Thread starter sarumman
  • Start date
  • #1
2
1

Homework Statement


given
upload_2018-12-19_23-51-14.png
I am required to proove or disprove:[/B]
lTxizl2.jpg


Homework Equations


rank
dim
null space

The Attempt at a Solution


I tried to base my answer based on the fact that null A and null A^2 is Contained in F (n)
and
dim N(A)+rank(A)=N
same goes for A^2.
 

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Answers and Replies

  • #2
member 587159
Why don't you just use the definition?

##x \in Null(A) \implies Ax = 0 ##

If the statement is true, you have to prove that ##A^2 x = 0##. Can you show that?
 
  • #3
2
1
Why don't you just use the definition?

##x \in Null(A) \implies Ax = 0 ##

If the statement is true, you have to prove that ##A^2 x = 0##. Can you show that?
thank you! you mean like so:
upload_2018-12-20_0-5-50.png
 

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  • #4
member 587159
Yes, the idea is certainly correct. The proof exposition can be better though. Here is how I would write it:

We want to prove that ##Null(A) \subseteq Null(A^2)##, so let's take an arbitrary element ##x \in Null(A)##. By definition, this means that ##Ax = 0##. Since ##A^2x = (AA)x = A(Ax) = A0 = 0## (here we used associativity of matrix multiplication/function composition), it follows that ##x \in Null(A^2)##, and we are done.
 
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