If ##A^3=0## then ##A-Z## is nonsingular

[Hall]

Homework Statement

Let $A$ be an $n \times$ square matrix and $Z$ denote the $n\times n$ identity matrix. If $A^3=0$ then prove that (or present a counter example) $A-Z$ is nonsingular.

Relevant Equations

A matrix is nonsingular if and only if its corresponding Linear transformation is invertible.

I'm really unable to have a start, because I cannot think of any matrix (other than $O$) such that its cube is the zero matrix. I tried to assume A = $\begin{bmatrix} a &c \\b &d \end{bmatrix}$ and computed $A^3$ and set it to $O$ to get an idea how the elements would look like, but the resulting equations will be non-linear.

 

[PeroK]

The first thing I thought of was to factorise $A^3 - I$.

 

[fresh_42]

Transform $A$ into the Jordan normal form. Also, $\exp(A) = 1+A+(1/2)A^2$ is regular, but I'm not sure whether this helps. To me $A$ is an element of the Heisenberg algebra, and $A+1$ an element of the Heisenberg group, but I'm not sure how to formalize that.

 

[PeroK]

Transform $A$ into the Jordan normal form. Also, $\exp(A) = 1+A+(1/2)A^2$ is regular, but I'm not sure whether this helps. To me $A$ is an element of the Heisenberg algebra, and $A+1$ an element of the Heisenberg group, but I'm not sure how to formalize that.

It's a bit simpler than that!

 

[FactChecker]

I'm really unable to have a start, because I cannot think of any matrix (other than $O$) such that its cube is the zero matrix.

Try $A = \begin{bmatrix} 0 &1 &0 \\ 0& 0 &1 \\ 0 &0 &0 \end{bmatrix}$
It shifts the first coordinate to the second and the second coordinate to the third and the third to zero. Three applications to any vector should give the zero vector.
And if $B$ is any invertable matrix, it is easy to show that $C = BAB^{-1}$ also satisfies $C^3=O$.

 

[PeroK]

In terms of 2x2 matrices, we have:
$$\begin{bmatrix}
1 \ \ \ \ \ \ b \\
{-\frac 1 b} \ \ {-1}
\end{bmatrix}^2 = 0
$$

 

[fresh_42]

Or simply consider eigenvectors of $A.$

 

[Hall]

The first thing I thought of was to factorise $A^3 - I$.

As those two matrices commute, we can have
$$
A^3 - I^3= (A+I)(A^2 - A + I)$$

 

[PeroK]

As those two matrices commute, we can have
$$
A^3 - I^3= (A+I)(A^2 - A + I)$$

That's not quite right.

Once you've fixed it, what do you know about $A^3$?

 

[PeroK]

In terms of 2x2 matrices, we have:
$$\begin{bmatrix}
1 \ \ \ \ \ \ b \\
{-\frac 1 b} \ \ {-1}
\end{bmatrix}^2 = 0
$$

There are actually no 2x2 matrices where $A^3 = 0$ but $A^2 \ne 0$.

 

[fresh_42]

As those two matrices commute, we can have
$$
A^3 - I^3= (A+I)(A^2 - A + I)$$

$A^3-I^3=(A-I)(A^2+A+I)$

 

[fresh_42]

It's a bit simpler than that!

My bad, I confused the forum. Sorry, @Hall!

 

[Hall]

That's not quite right.

Once you've fixed it, what do you know about $A^3$?

$$
A^3 -I^3 = (A-I) (A^2 +A +I)$$
$$
(A-I)(-A^2-A-I) = I$$
Hence, (A-I) is invertible because it has an inverse $(-A^2 -A -I)$.

 

[fresh_42]

$$
A^3 -I^3 = (A-I) (A^2 +A +I)$$
$$
(A-I)(-A^2-A-I) = I$$
Hence, (A-I) is invertible because it has an inverse $(-A^2 -A -I)$.

This is certainly a good start, and it would be sufficient in a multiplicative group.

But we do not have commutativity and automatic inverses, so strictly speaking you have to show that left and right inverse exist and are the same. Also, don't we need $A+I$ and not $A-I,$ and why can't the product of two singular matrices be a regular matrix, $I$ in that case.

I think you can assume a vector $v$ in the kernel of $A+I$ and show that it has to be the zero vector by this equation. Hint: Use $(-A^2-A-I)(A-I) = I$ instead and at the end apply $A$ once more.

 

[PeroK]

This is certainly a good start, and it would be sufficient in a multiplicative group.

But we do not have commutativity and automatic inverses, so strictly speaking you have to show that left and right inverse exist and are the same. Also, don't we need $A+I$ and not $A-I,$ and why can't the product of two singular matrices be a regular matrix, $I$ in that case.

I think you can assume a vector $v$ in the kernel of $A+I$ and show that it has to be the zero vector by this equation.

What the OP has posted is sufficient, given that he already proved in another thread that $AB = I$ is sufficient for matrices $A$ and $B$ to be inverses of each other.

In any case, we could just change the order of the factors:
$$
A^3 -I^3 = (A-I) (A^2 +A +I) = (A^2 + A + I)(A-I)$$
Hence:$$(A-I)B = B(A-I) = I$$ with $B = -A^2 -A -I$

 

[Hall]

This is certainly a good start, and it would be sufficient in a multiplicative group.

But we do not have commutativity and automatic inverses, so strictly speaking you have to show that left and right inverse exist and are the same. Also, don't we need $A+I$ and not $A-I,$ and why can't the product of two singular matrices be a regular matrix, $I$ in that case.

I think you can assume a vector $v$ in the kernel of $A+I$ and show that it has to be the zero vector by this equation.

Can we make this argument: If a matrix has a left inverse then it also has a right inverse and that right inverse is same as the left one?
$$
(A-I)(-A^2-A -I) = I $$
The left inverse of $(-A^2 -A-I)$ is $(A-I)$, so we have: $$
(-A^2 -A -I) (A-I) =I$$
And therefore, the left inverse of $A-I$ exists.

 

[PeroK]

Can we make this argument: If a matrix has a left inverse then it also has a right inverse and that right inverse is same as the left one?
$$
(A-I)(-A^2-A -I) = I $$
The left inverse of $(-A^2 -A-I)$ is $(A-I)$, so we have: $$
(-A^2 -A -I) (A-I) =I$$
And therefore, the left inverse of $A-I$ exists.

See:

https://www.physicsforums.com/threads/prove-b-is-invertible-if-ab-i.1012582/

 

[fresh_42]

Can we make this argument: If a matrix has a left inverse then it also has a right inverse and that right inverse is same as the left one?
$$
(A-I)(-A^2-A -I) = I $$
The left inverse of $(-A^2 -A-I)$ is $(A-I)$, so we have: $$
(-A^2 -A -I) (A-I) =I$$
And therefore, the left inverse of $A-I$ exists.

Yes, in this case where the inverse is a polynomial of $A.$ Honestly? I do not have the group theory proof in mind to say which group axioms besides associativity are actually necessary to prove left inverse equals right inverse. It is not automatically the case.

And there is still the problem with $A+I$ instead of $A-I.$ I guess we may assume characteristic zero.

 

[PeroK]

And there is still the problem with $A+I$ instead of $A-I.$

Why do we need to do this one as well? I don't see it in the OP. In any case:

If $A^3 = 0$, then $(-A)^3 = 0$ and $-A - I = -(A + I)$ is invertible; hence, $A + I$ is invertible. QED

 

[fresh_42]

Why do we need to do this one as well? I don't see it in the OP. In any case:

If $A^3 = 0$, then $(-A)^3 = 0$ and $-A - I = -(A + I)$ is invertible; hence, $A + I$ is invertible. QED

Where have you shown the invertibility of $A+I$ or $-A-I$?

\begin{align*}
v\in \operatorname{ker}(A+I) &\Longrightarrow Av=-v\\
&\Longrightarrow (-A^2-A-I)(A-I)v=(-A^2-A-I)(A.v-Iv)\\
&\ldots
\end{align*}

 

[PeroK]

Where have you shown the invertibility of $A+I$?

It's a corollary of the invertibility of $A - I$, given that $A^3 = 0$.

 

[Hall]

Where have you shown the invertibility of $A+I$?

Post #8.

$$A^3 + I^3= (A+I)(A^2 - A +I)$$
$$ I = (A+I)(A^2-A+I)$$

 

[fresh_42]

Post #8.

There was a mistake in your formula in post #8. Look at the $I\cdot I$ term!

It's a corollary of the invertibility of $A - I$, given that $A^3 = 0$.

And where have you or some magician shown it?

Edit: Now it is in post #22. But nowhere before.

 

[fresh_42]

Why do we need to do this one as well? I don't see it in the OP.

I had $A+Z$ in mind. Bad hair day ...

 

[Office_Shredder]

This whole thread seems stupidly complicated.

Suppose $A-Z$ is singular. Then there exists $v\neq 0$ such that $(A-Z)v=0$ -> $Av=v$.

Apply $A$ twice more and be done with it. No factorization required.

 

[PeroK]

This whole thread seems stupidly complicated.

Sorry to be so stupid, but I'm just trying to help the OP and factorisation was the first thing I thought of. I can't see it's particularly more complicated than considering an eigenvector.

Most of the muddle stemmed from the thread being derailed somewhat.

 

[Hall]

Sorry to be so stupid, but I'm just trying to help the OP and factorisation was the first thing I thought of. I can't see it's particularly more complicated than considering an eigenvector.

Most of the muddle stemmed from the thread being derailed somewhat.

I think the question maker had exactly the factorization method in mind, because this question was there in chapter before the chapters on Determinants and Eigenvectors.

 

[Mark44]

Most of the muddle stemmed from the thread being derailed somewhat.

And not just somewhat. Given that $A^3 = 0$ it's easy to see that
$-I = A^3 - I = (A - I)(A^2 + A + I) \Rightarrow I = A^3 - I = (A - I)(-A^2 - A - I)$
Further, one can show that $(-A^2 - A - I)(A - 1) = -A^3 + I = I$
Clearly $A - I$ is invertible, which was shown in several other posts in this thread, and was the point that @PeroK was making. Any discussion of Jordan normal form, Heisenberg algebra, or eigenvectors was completely unnecessary.

 

[Mark44]

This whole thread seems stupidly complicated.

Suppose $A-Z$ is singular. Then there exists $v\neq 0$ such that $(A-Z)v=0$ -> $Av=v$.

Apply $A$ twice more and be done with it. No factorization required.

In a post following the one I quoted, the OP stated that determinants and eigenvalues/eigenvectors haven't yet been presented (see below). The proof outline shown above is a good one, but with no knowledge yet of eigenvectors, about the only technique remaining was by using factorization.

this question was there in chapter before the chapters on Determinants and Eigenvectors.

The thread seems to have run its course, so I'm closing it now.