# A real parameter guaranteeing subspace invariance

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1. Nov 4, 2016

### TheSodesa

1. The problem statement, all variables and given/known data
Let $A$ and $B$ be square matrices, such that $AB = \alpha BA$. Investigate, with which value of $\alpha \in \mathbb{R}$ the subspace $N(B)$ is $A$-invariant.

2. Relevant equations

If $S$ is a subspace and $A \in \mathbb{C}^{n \times n}$, we define multiplying $S$ by $A$ as follows:

AS = \{ A \vec{x}: \vec{x} \in S \}

The subspace $S$ is said to be $A$-invariant, if

AS \subset S

The null space of a matrix $A \in \mathbb{C}^{n \times n}$ is

N(A) = \{ \vec{x} \in \mathbb{C}^{n}: A \vec{x} = \vec{0} \}

3. The attempt at a solution

Alright, I have no idea what I'm doing and I came close to failing (or failed, the exam results haven't come back yet) my matrix algebra course, but here goes:

Let us assume that $N(B)$ is indeed $A$-invariant, and that $AB = \alpha BA$. Then according to $(2)$:
$$A N(B) \subset N(B)$$

Also, solving for $\alpha$ in the given equation gives us (assuming $A$ and $B$ are non-singular):
\begin{align*}
&& AB &= \alpha BA\\
&&\iff\\
&& ABA^{-1} &=\alpha B\\
& &\iff\\
&& AB A^{-1} B^{-1} &= \alpha I,
\end{align*}

Looking at the middle equation, $\alpha B$ seems to similar with $B$. I wonder if this is something I can use to my advantage. Regardless, I'm not sure how to progress from here on out. Like I said, my understanding of the theory is on a very shaky foundation, so I really could use some help.

2. Nov 5, 2016

### PeroK

Let $x \in N(B)$ and consider $Ax$. Is this in $N(B)$ or not?

3. Nov 5, 2016

### TheSodesa

Well, since $N(B)$ is $A$-similar, yes it is, since $AN(B)$ is a subset of $N(B)$. All of the vectors in $AN(B)$ can be found in $N(B)$.

Hmm. Then, since the null space is a subspace, any linear combination of vectors in $N(B)$ is also in $N(B)$, so $\alpha \vec{x}$ is also in $N(B)$. $\alpha A \vec{x}$ should also be in $N(B)$, because of $N(B)$'s $A$-invariance. Am I going in the right direction?

I'm not sure how I can make the connection between these ideas and the equation I'm supposed to solve. I can't plug an $\vec{x}$ or an $\alpha \vec{x}$ in there anywhere, can I?

4. Nov 5, 2016

### PeroK

That's assuming what you are trying to show. Why not start with what you were given:

$AB = \alpha BA$

Then, as I suggested above:

5. Nov 5, 2016

### TheSodesa

So finding what $\alpha$ is is equivalent to showing, that all of the vectors $A\vec{x} \in N(B)$?

If I multiply the given equation with $\vec{x} \in N(B)$ from the right, I get:
\begin{align*}
AB\vec{x} &= \alpha BA \vec{x}\\
\vec{0} &= \alpha BA \vec{x}
\end{align*}
Now assuming $B$, $A$ and $\vec{x}$ are all non-zero matrices/vectors, the only way the right side can be equal to zero is if $\alpha = 0$ or $A\vec{x} = \vec{0}$ (or both, if you want to get pedantic about this). If $A\vec{x} = \vec{0}$, that would imply that $A\vec{x}$ is indeed in $N(B)$, and $N(B)$ is $A$-invariant?

6. Nov 5, 2016

### PeroK

You've got the basis of the argument, but you need to work on your logic. Also, there is no reason to assume that "$B$, $A$ and $\vec{x}$ are all non-zero matrices/vectors". In fact, you cannot make any of those assumptions at all.

To try to tighten up your logic, let's go back to:

$0 = A(Bx) = (AB)x = (\alpha BA)x = \alpha B(Ax)$

What can you deduce from that?

7. Nov 5, 2016

### TheSodesa

I'm a bit hesitant to say this, but looking at $$A(B\vec{x}) = \alpha B(A\vec{x})$$ almost makes it look like $$A = \alpha B$$

8. Nov 5, 2016

### PeroK

First, assuming two matrices satisfy $AB = \alpha BA$ is not going to lead to $A = \alpha B$ or anything like that. Also, that's not the sort of thing we are trying to show!

So, let's go back to:

$0 = \alpha B(Ax)$

Let me do the next step:

Either $\alpha = 0$

or $B(Ax) = 0$

What next?

9. Nov 5, 2016

### TheSodesa

I assume we are not allowed to assume non-singularity for either $A$ nor $B$, so I can't just say that $$B^{-1} B (A\vec{x}) = A\vec{x} = \vec{0}$$, which would imply, that either $\alpha = 0$ or $A\vec{x} \in N(B)$?

If we are allowed to assume non-singularity, then $A \alpha \vec{x}$ is also in $N(B)$, and $\alpha$ can be any real number.

10. Nov 5, 2016

### PeroK

Nothing like that. We already know that $\alpha = 0$ is one option. For the second case, what if we let $y = Ax$, then we have:

$By = 0$

What does that tell you about $y$?

11. Nov 5, 2016

### TheSodesa

That $\vec{y} = A \vec{x}$ is also in $N(B)$, which implies that $AN(B) \subset N(B)$. Now since $N(B)$ is a subspace, I could take any non-zero real number $\alpha$, so that $\alpha A \vec{x} \in N(B)$, right?

12. Nov 5, 2016

### PeroK

You were right up to here. Then you went off the point!

Try to summarise what we've done so far. It's spread over a number of posts, so try to pull it all together into a logical argument that starts with:

Let $AB = \alpha BA$

And ends with:

$\alpha = 0$ or $N(B)$ is $A$-invariant.

13. Nov 5, 2016

### TheSodesa

Ahh, wait, I think I see it now. Since we are not allowed to make any other assumptions other than the given $AB = \alpha BA$, the only thing we can say for certain is that if we want $N(B)$ to be $A$-invariant, $\alpha$ must not be equal to zero, but it can be any other real number.

Is this correct?

14. Nov 5, 2016

### PeroK

Pretty much, but again your logic is loose. In purely mathematical language, I'd say we've proved:

$AB = \alpha BA \ \Rightarrow \ \alpha = 0 \$ or $\ N(B)$ is $A$-invariant.

What you can do with this is say:

$\alpha \ne 0$ and $AB = \alpha BA \ \Rightarrow \ N(B)$ is $A$-invariant. (1)

What we haven't done is said anything further about the case $\alpha = 0$. In other words:

$AB = 0 \ \Rightarrow \ ?$

We haven't shown anything one way or the other in this case about the $A$-invariance of $N(B)$

In any case, I think you should try to do a clear, logical proof of statement (1), based on this thread.

15. Nov 5, 2016

### TheSodesa

Alright.

Let $AB = \alpha BA$ and $\vec{x} \in N(B)$. Now
\begin{align*}
AB &= \alpha BA &\iff\\
A \overbrace{B \vec{x}}^{ \vec{0} \text{ by definition} } &= \alpha BA \vec{x} &\iff\\
\vec{0} &= \alpha BA \vec{x}\\
\end{align*}

This is true if $\alpha = 0$ (inclusive) or $BA\vec{x} = \vec{0}$. Now to make things clearer, let us denote $A\vec{x} = \vec{y}$. If we now assume, that $\alpha \neq 0$, $BA \vec{x} = B \vec{y} = 0 \iff \vec{y} \in N(B)$.

Therefore $\alpha \neq 0 \Rightarrow AN(B) \subset N(B)$.

Next I should probably try to show that $\alpha = 0 \Rightarrow AN(B) \not\subset N(B)$.

16. Nov 5, 2016

### PeroK

Yes, I think that's good.

On the last point, perhaps you might have both cases. Can you find $A_1, B_1$ where

$A_1B_1 = 0$ and $N(B_1)$ is $A_1$-invariant.

And, find $A_2, B_2$ where:

$A_2B_2 = 0$ and $N(B_2)$ is not $A_2$-invariant?

Hint: try some simple 2x2 matrices.

17. Nov 5, 2016

### TheSodesa

Pretty much every single simple 2x2 matrix I tried in Matlab row-reduces into the identity matrix $I_2$, as does their product. Also, every random 2x2 matrix generated seems to produce the identity matrix when row-reduced.

The null space of the identity matrix is just the zero vector: $N(I_n) = span\{\vec{0}\}$. This is supposed to be a huge red flag, isn't it?

18. Nov 5, 2016

### PeroK

The identity matrix isn't much good, but perhaps the $0$ matrix is something to try?

Also, can you find two non-zero matrices whose product is the $0$ matrix?

19. Nov 5, 2016

### TheSodesa

What know is that if I wanted the product $AB=0$, all of the column vectors of $B$ would have to be in $N(A)$.

20. Nov 5, 2016

### PeroK

Do you understand the concept of an example or counterexample?

If I said: all matrices commute, how would you disprove that?