Prove Matrix Rank of A*A = Rank of A

[Maybe_Memorie]

Homework Statement

Show that for all nxn matricies A with real entries we have
rk(A*A) = rk(A) where A* is the transpose of A.

Homework Equations

The Attempt at a Solution

I'm working over a vector space V.

Im(A) = {A(v) | vEV}

Im(A*A) = {A*(A(v)) | vEV}

So Im(A*A) is a subset of Im(A)
So rk(A*A) =< rk(A)

Ker(A) = {A(w) = 0 | wEV}

Ker(A*A) = {A*(A(w)) = 0 | wEV}

so Ker(A*A) is a subset of Ker(A)
so dimKer(A*A) =< dimKer(A)

dimKer(A) = dimV - rk(A)

so -rk(A*A) =< -rk(A)
so rk(A*A) >= rk(A)

Combining the results yields rk(A*A) = rk(A)


Is this correct??

 

[micromass]

Im(A*A) = {A*(A(v)) | vEV}

So Im(A*A) is a subset of Im(A)

Ker(A*A) = {A*(A(w)) = 0 | wEV}

so Ker(A*A) is a subset of Ker(A)

Why are these two things true?? They don't exactly seem trivial to me...

 

[I like Serena]

Ker(A) = {A(w) = 0 | wEV}

Note that you have reversed your definition of Ker.
It should be:
Ker(A) = {wEV | A(w) = 0}

 

[Maybe_Memorie]

Im(A*A) = {A*(A(v)) | vEV}

A maps the vector v to another vector v1.
Then A* maps the vector v1 to another vector v2, but the set of all v1s can either be less than or equal to the set of original vectors, then A* has either the original amount of vectors or a smaller amount to map to v2.
So Im(A*A) is a subset of Im(A)

Similar logic can be applied to the Kernel.

 

[micromass]

Im(A*A) = {A*(A(v)) | vEV}

A maps the vector v to another vector v1.
Then A* maps the vector v1 to another vector v2, but the set of all v1s can either be less than or equal to the set of original vectors, then A* has either the original amount of vectors or a smaller amount to map to v2.

That only means that Im(A*A) has less elements than Im(A), which is true. This doesn't imply that it's a subset. You didn't prove that every element in Im(A*A) is in Im(A).

 

[Maybe_Memorie]

That only means that Im(A*A) has less elements than Im(A), which is true. This doesn't imply that it's a subset. You didn't prove that every element in Im(A*A) is in Im(A).

Ah I see. Any points in the right direction?

 

[micromass]

Ah I see. Any points in the right direction?

Yes. The trick is to prove that

$$Ker(A^*A)=Ker(A)$$

Now, I claim that $Ker(A)\subseteq Ker(A^*A)$ (why??)
To see the other inclusion, assume that A*Ax=0. What happens if you multiply both sides with x*?

 

[Maybe_Memorie]

Yes. The trick is to prove that

$$Ker(A^*A)=Ker(A)$$

Now, I claim that $Ker(A)\subseteq Ker(A^*A)$ (why??)
To see the other inclusion, assume that A*Ax=0. What happens if you multiply both sides with x*?

A*Ax = 0, then A*Axx* = 0
But if x is a nx1 matrix and x* a 1xn matrix xx* is a 1x1 matrix?
But the only way this can be zero is if either A*A is 0?

As for the kernel fact, I'm a bit confused..

 

[micromass]

A*Ax = 0, then A*Axx* = 0
But if x is a nx1 matrix and x* a 1xn matrix xx* is a 1x1 matrix?
But the only way this can be zero is if either A*A is 0?

Multiply the other side by x*

And use that in general z*z=|z| for vectors z.

As for the kernel fact, I'm a bit confused..

Let x be in Ker(A). You know that A(x)=0. Can you prove that A*A(x)=0?

 

[Maybe_Memorie]

Multiply the other side by x*

And use that in general z*z=|z| for vectors z.



Let x be in Ker(A). You know that A(x)=0. Can you prove that A*A(x)=0?

Well if Ax = 0 then A*A(x) = A*(0) = 0, as under a linear operator 0 is always mapped to 0.

 

[micromass]

Well if Ax = 0 then A*A(x) = A*(0) = 0, as under a linear operator 0 is always mapped to 0.

Yes, that's ok!

 

[Maybe_Memorie]

I'm afraid I don't see how that shows

$Ker(A)\subseteq Ker(A^*A)$

 

[I like Serena]

I'm afraid I don't see how that shows

$Ker(A)\subseteq Ker(A^*A)$

In words it says: if x is in the kernel of A, then it is also in the kernel of A^*A.

Rephrasing: for every x in V we have: if Ax = 0, then A^*Ax = 0.

 

[Maybe_Memorie]

Right I understand now!

So if $Ker(A)\subseteq Ker(A^*A)$

Then by taking the dimension of each side and applying the rank-nullity theorem and applying my original result then we get the answer?

 

[I like Serena]

Nope.
Previously you had the subset relationship the wrong way around.
You'll see if you try it again.
Sorry.