- #1

- 2

- 1

## Homework Statement

given

## 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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- Thread starter sarumman
- Start date

- #1

- 2

- 1

given

rank

dim

null space

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.

- #2

member 587159

##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

thank you! you mean like so:

##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?

- #4

member 587159

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.

- #5

PeterDonis

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