MHB These are the only submodule

[mathmari]

Hey!

Let $R$ be a commutative ring with unit.
Let $T$ be a linear mapping from $\mathbb{R}^2$ to $\mathbb{R}^2$ that is given by the projection at the $y$-axis.
I want to show that the only $\mathbb{R}[x]$-submodule (as for the action that $T$ defines) of $\mathbb{R}^2$ are $\mathbb{R}^2$, $0$, $x$-axis and $y$-axis.

Could you give me some hints how we could show that? (Wondering)

 

[Deveno]

A $\Bbb R[x]$-submodule of $\Bbb R^2$ (under the action of $T$) is a $T$-invariant subspace.

Now clearly $\Bbb R^2$ is a $T$-invariant subspace (since $\text{im }T \subseteq \Bbb R^2$).

Just as clearly, the 0-subspace is $T$-invariant, since $T$ is linear.

The only other subspaces of $\Bbb R^2$ are of the form:

$W = \{t(x_0,y_0): t \in \Bbb R\}$, of dimension one, for any fixed non-zero vector $(x_0,y_0)$.

For such a subspace of $\Bbb R^2$ to be $T$-invariant, we must have $T(W) \subseteq W$.

This means that $T(t_0x_0,t_0y_0) = (0,t_0y_0) = (ax_0,ay_0)$ for some $a$ (which may depend on $t_0$).

If $x_0 = 0$, then we can pick $a = t_0$, so the $y$-axis is clearly $T$-invariant.

Otherwise, we are forced to choose $a = 0$, that is $(x_0,y_0) \in \text{ker }T$. Can you continue?

Alternatively, if $E_{\lambda}$ is an eigenspace of $T$, then if $v \in E_{\lambda}$, we have:

$x\cdot v = T(v) = \lambda v$, and thus:

$f(x) \cdot v = f(T)(v)$, if $f(x) = c_0 + c_1x +\cdots + c_nx^n$ then this means:

$f(x)\cdot v = (c_0I + c_1T + \cdots + c_nT^n)v = c_0v + \lambda v + \cdots + c_n\lambda^n v$

$= f(\lambda)v \in E_{\lambda}$, since subspaces are closed under scalar multiplication.

So what are the eigenspaces of $T$?

Can you show any one-dimensional $T$-invariant subspace (what we are looking for) *must* be generated by an eigenvector?

 

[mathmari]

Now clearly $\Bbb R^2$ is a $T$-invariant subspace (since $\text{im }T \subseteq \Bbb R^2$).

Just as clearly, the 0-subspace is $T$-invariant, since $T$ is linear.

The only other subspaces of $\Bbb R^2$ are of the form:

$W = \{t(x_0,y_0): t \in \Bbb R\}$, of dimension one, for any fixed non-zero vector $(x_0,y_0)$.

The subspaces of $\mathbb{R}^2$ can have the dimension $0$, $1$ or $2$, right? (Wondering)

The subspace with dimension $2$ is $\mathbb{R}^2$ and the subspace of dimension $0$ is the origin, or not? (Wondering)

Why is the subspace of dimension $1$ of the form $W = \{t(x_0,y_0): t \in \Bbb R\}$ ? Does this set represent a line? (Wondering)

For such a subspace of $\Bbb R^2$ to be $T$-invariant, we must have $T(W) \subseteq W$.

This means that $T(t_0x_0,t_0y_0) = (0,t_0y_0) = (ax_0,ay_0)$ for some $a$ (which may depend on $t_0$).

Why does it have to stand that $T(t_0x_0,t_0y_0) = (0,t_0y_0) = (ax_0,ay_0)$ for some $a$ ? (Wondering)

Otherwise, we are forced to choose $a = 0$, that is $(x_0,y_0) \in \text{ker }T$. Can you continue?

Why is $(x_0,y_0) \in \text{ker }T$ and not $(t_0x_0,t_0y_0) \in \text{ker }T$ ? (Wondering)

 

[Deveno]

The subspaces of $\mathbb{R}^2$ can have the dimension $0$, $1$ or $2$, right? (Wondering)

Yes.

The subspace with dimension $2$ is $\mathbb{R}^2$ and the subspace of dimension $0$ is the origin, or not? (Wondering)

Yes.

Why is the subspace of dimension $1$ of the form $W = \{t(x_0,y_0): t \in \Bbb R\}$ ? Does this set represent a line? (Wondering)

Yes, the line that passes through $(x_0,y_0)$ and the origin.

Why does it have to stand that $T(t_0x_0,t_0y_0) = (0,t_0y_0) = (ax_0,ay_0)$ for some $a$ ? (Wondering)

The line consists of all scalar multiples of $(x_0,y_0)$. If $T$ is to map an element of that line $(t_0x_0,t_0y_0)$ back to the line, it must be "some" scalar multiple of $(x_0,y_0)$.

But, according to the statement of your problem, $T(x,y) = (0,y)$.

Why is $(x_0,y_0) \in \text{ker }T$ and not $(t_0x_0,t_0y_0) \in \text{ker }T$ ? (Wondering)

If $v \in \text{ker }T$ for any linear $T$, so is $\alpha v$ for any scalar $\alpha$, because the kernel of a linear map is a subspace of the domain.

 

[mathmari]

This means that $T(t_0x_0,t_0y_0) = (0,t_0y_0) = (ax_0,ay_0)$ for some $a$ (which may depend on $t_0$).

If $x_0 = 0$, then we can pick $a = t_0$, so the $y$-axis is clearly $T$-invariant.

Can we not say also that if $y_0 = 0$, then we can pick $a = t_0$, so the $x$-axis is $T$-invariant ? (Wondering)

If $v \in \text{ker }T$ for any linear $T$, so is $\alpha v$ for any scalar $\alpha$, because the kernel of a linear map is a subspace of the domain.

What information do we get by having that $(x_0,y_0)\in \ker T$ ? (Wondering)

 

[Deveno]

Can we not say also that if $y_0 = 0$, then we can pick $a = t_0$, so the $x$-axis is $T$-invariant ? (Wondering)

Yes, it is not hard to show the $x$-axis is $T$-invariant. It's a bit harder to show that ONLY the $x$-axis and $y$ axis are 1-dimensional $T$-invariant subspaces.

What information do we get by having that $(x_0,y_0)\in \ker T$ ? (Wondering)

Calculate the kernel of $T$-what do you find?

 

[mathmari]

It's a bit harder to show that ONLY the $x$-axis and $y$ axis are 1-dimensional $T$-invariant subspaces.

So, to show that only these are the 1-dimensional $T$-invariant subspaces, do we use the kernel? (Wondering)

Calculate the kernel of $T$-what do you find?

We have that $\ker T=\{(t_0x_0, t_0y_0) : (ax_0, ay_0)=(0,0)\}$.
When $a=0$, then $(t_0x_0, t_0y_0)\in \ker T, \forall (x_0, y_0)$.
Otherwise, it must be $x_0=y_0=0$, so $(0,0)\in \ker T$.

Is this correct? (Wondering)

 

[Deveno]

So, to show that only these are the 1-dimensional $T$-invariant subspaces, do we use the kernel? (Wondering)

Suppose that $T$ is *any* linear transformation $T: V \to V$, and we seek a one-dimensional $T$-invariant subspace $W = \langle v_0\rangle$.

The condition that $W$ is $T$-invariant, means that $Tv_0 = \alpha v_0$ for some scalar $\alpha$, that is, $v_0$ is an eigenvector with eigenvalue $\alpha$ (note that $v_0 \neq 0$ since we have a one-dimensional subspace of $V$, not the $0$-dimensional one)$.

We have that $\ker T=\{(t_0x_0, t_0y_0) : (ax_0, ay_0)=(0,0)\}$.
When $a=0$, then $(t_0x_0, t_0y_0)\in \ker T, \forall (x_0, y_0)$.
Otherwise, it must be $x_0=y_0=0$, so $(0,0)\in \ker T$.

Is this correct? (Wondering)

No, the kernel of $T$ is $(a,b) \in \Bbb R^2$ such that $T(a,b) = (0,0)$.

Since $T$ is (by your stated definition) the map $(x,y) \mapsto (0,y)$, we have $T(a,b) = (0,b)$.

This equals $(0,0)$ if and only if $b = 0$, that is, if $(a,b)$ is of the form $(a,0)$ (this works for any real number $a$).

So the kernel of $T$ is the $x$-axis.

Note that the kernel of a linear map is the eigenspace corresponding to the eigenvalue $0$ (if a linear transformation $T: V \to V$ does not have $0$ as an eigenvalue, this subspace is trivial, and is usually not called an eigenspace because it has no eigenvectors, which of necessity must be non-zero vectors).

In fact, the characteristic (and minimal) polynomial of your specific $T$ is: $x^2 - x$, since we clearly have $T^2 = T$, and $T$ is neither the 0-map nor the identity (the only (irreducible monic) polynomial factors of $x^2 - x = x(x - 1)$ are $x$ and $x - 1$ and $T$ satisfies neither of these).

So, $T$ has two eigenvalues, $0$ and $1$, and the one-dimensional $T$-invariant subspaces correspond to these one-dimensional eigenspaces.

What makes this problem *easy* is that $\Bbb R^2$ is only two-dimensional, which severely limits the dimensionality of its subspaces. With a three-dimensional (or higher) vector space, we would have a greater possible variety of $T$-invariant subspaces (for example, if $T$ had an eigenspace of dimension two, any one-dimensional subspace of this eigenspace would also be $T$-invariant).

At the extremes, with $T = I$, or $T = 0$, we have that *every* subspace is $T$-invariant. In other words, these maps reveal nothing of the internal structure of the vector space they act upon.

 

[mathmari]

Suppose that $T$ is *any* linear transformation $T: V \to V$, and we seek a one-dimensional $T$-invariant subspace $W = \langle v_0\rangle$.

Why is $W$ of the form $\langle v_0\rangle$ ? I got stuck right now... (Wondering)

 

[Deveno]

A one-dimensional subspace has a basis consisting of one non-zero vector $v_0$. Every vector in this subspace is a scalar multiple of the basis vector.

 

[mathmari]

In fact, the characteristic (and minimal) polynomial of your specific $T$ is: $x^2 - x$, since we clearly have $T^2 = T$, and $T$ is neither the 0-map nor the identity (the only (irreducible monic) polynomial factors of $x^2 - x = x(x - 1)$ are $x$ and $x - 1$ and $T$ satisfies neither of these).

So, $T$ has two eigenvalues, $0$ and $1$, and the one-dimensional $T$-invariant subspaces correspond to these one-dimensional eigenspaces.

How do we know that $T^2 = T$ ? (Wondering)

 

[mathmari]

Could we say the following for $1$-dimensional subspaces? (Wondering)

If the subspace $W$ is of dimenion $1$, then it is a line.
So, it is of the form $\{t(x_0, y_0) : t\in \mathbb{R}\}=\mathbb{R}(x_0, y_0)$.
So that it is $T$-invariant, i.e., $T(W)\subseteq W$, we have that $T(t_0x_0, t_0y_0)=(ax_0, ay_0) \Rightarrow (0, t_0y_0)=(ax_0, ay_0)$.
If $y_0=0$ then the above relation holds for $a=0$.
So, $W$ is the $x$-axis.
Therefore, the $x$-axis is a $\mathbb{R}[x]$-submodule of $\mathbb{R}^2$.
If $y_0\neq 0$ then we choose $a=t_0$ and it must be $x_0=0$.
So, $W$ is the $y$-axis.
Therefore, the $y$-axis is a $\mathbb{R}[x]$-submodule of $\mathbb{R}^2$.

Have we shown in that way that the only $1$-dimensional subspaces are the $x$-axis and the $y$-axis? (Wondering)

 

[Deveno]

How do we know that $T^2 = T$ ? (Wondering)

$T^2(x,y) = T(T(x,y)) = T(0,y) = T(x,y)$ for all $(x,y) \in \Bbb R^2$.

Could we say the following for $1$-dimensional subspaces? (Wondering)

If the subspace $W$ is of dimenion $1$, then it is a line.
So, it is of the form $\{t(x_0, y_0) : t\in \mathbb{R}\}=\mathbb{R}(x_0, y_0)$.
So that it is $T$-invariant, i.e., $T(W)\subseteq W$, we have that $T(t_0x_0, t_0y_0)=(ax_0, ay_0) \Rightarrow (0, t_0y_0)=(ax_0, ay_0)$.
If $y_0=0$ then the above relation holds for $a=0$.
So, $W$ is the $x$-axis.
Therefore, the $x$-axis is a $\mathbb{R}[x]$-submodule of $\mathbb{R}^2$.
If $y_0\neq 0$ then we choose $a=t_0$ and it must be $x_0=0$.
So, $W$ is the $y$-axis.
Therefore, the $y$-axis is a $\mathbb{R}[x]$-submodule of $\mathbb{R}^2$.

Have we shown in that way that the only $1$-dimensional subspaces are the $x$-axis and the $y$-axis? (Wondering)

The important thing to realize here is that if a line is non-degenerate, it contains some non-zero point $(x_0,y_0)$.

Now we know that for any element $t(x_0,y_0) = (tx_0,ty_0)$, that for $T(W) \subseteq W$ we must have:

$T(tx_0,ty_0) = (ax_0,ay_0)$ for some $a \in \Bbb R$.

If $y_0 = 0$, this is ALWAYS TRUE, since:

$T(tx_0,t0) = t(T(x_0,0)) = t(0,0) = (0,0)$, and we can take for ANY $t$, a value of $a = 0$.

This is because the $x$-axis is the kernel of $T$, so $T$ sends the entire $x$-axis to $(0,0)$ and $(0,0)$ is an element of EVERY subspace of $\Bbb R^2$.

This shows the $x$-axis is ONE $T$-invariant subspace.

On the other hand if $(x_0,y_0) \neq (0,0)$ and $y_0 \neq 0$, let's say it's $b$ for now, we need to show that $x_0$ MUST BE $0$.

Now, by the DEFINITION of $T$, we have $T(tx_0,ty_0) = t(T(x_0,y_0)) = t(0,y_0) = (0,ty_0)$.

So for $y_0 = b$, for example, $T(tx_0,tb) = (0,tb)$.

Now if we have a (one-dimensional) $T$-invariant subspace, $(0,tb) = a(x_0,b)$, that is:

$ax_0 = 0$
$tb = ab$.

Note this has to be true for ANY $t$. So let's use $t = 1$. This gives:

$ax_0 = 0$
$b = ab$

and we can solve the second equation by dividing by $b$ (since $b \neq 0$), to get $a = 1$. Then the first equation is:

$x_0 = 0$.

So what you wrote is correct, but be sure you understand WHY.