Linear algebra: Prove that the set is a subspace

[gruba]

Homework Statement

Let $U$ is the set of all commuting matrices with matrix $A= \begin{bmatrix}<br /> 2 & 0 & 1 \\<br /> 0 & 1 & 1 \\<br /> 3 & 0 & 4 \\<br /> \end{bmatrix}$. Prove that $U$ is the subspace of $\mathbb{M_{3\times 3}}$ (space of matrices $3\times 3$). Check if it contains $span\{I,A,A^2,...\}$. Find the dimension and a basis for $U$ and $span\{I,A,A^2,...\}$.

Homework Equations

-Commuting matrices
-Subspaces
-Vector space span
Basis and dimension

The Attempt at a Solution

$U$ can be defined as $U=\{B\in\mathbb{M_{3\times 3}}: AB=BA\}$.

Letting $B=\begin{bmatrix}<br /> a & b & c \\<br /> d & e & f \\<br /> g & h & i \\<br /> \end{bmatrix}$ and solving the equation $AB=BA$ gives $B=\begin{bmatrix}<br /> i-\frac{2}{3}g & 0 & \frac{1}{3}g \\<br /> g-3f & i-3g & f \\<br /> g & 0 & i \\<br /> \end{bmatrix}$.

$U$ is a subspace of $\mathbb{M_{3\times3}}$ if $\forall u_1,u_2\in U\Rightarrow u_1+u_2\in U,\forall t\in\mathbb{R}\Rightarrow tu_1\in U$ which is correct.

It is easy to check that if $C\in span\{I,A,A^2,...\}\Rightarrow C\in U$:
Linear combination for C is
$C=c_0I+c_1A+c_2A^2+...\Rightarrow CA=AC\Rightarrow C\in U$

$U$ has the dimension $3$ and a basis are column vectors of identity matrix $3\times 3$.

How to find the dimension and a basis for $span\{I,A,A^2,...\}$?

 

[MostlyHarmless]

Homework Statement

Let $U$ is the set of all commuting matrices with matrix $A= \begin{bmatrix}<br /> 2 & 0 & 1 \\<br /> 0 & 1 & 1 \\<br /> 3 & 0 & 4 \\<br /> \end{bmatrix}$. Prove that $U$ is the subspace of $\mathbb{M_{3\times 3}}$ (space of matrices $3\times 3$). Check if it contains $span\{I,A,A^2,...\}$. Find the dimension and a basis for $U$ and $span\{I,A,A^2,...\}$.

Homework Equations

-Commuting matrices
-Subspaces
-Vector space span
Basis and dimension

The Attempt at a Solution

$U$ can be defined as $U=\{B\in\mathbb{M_{3\times 3}}: AB=BA\}$.

Letting $B=\begin{bmatrix}<br /> a & b & c \\<br /> d & e & f \\<br /> g & h & i \\<br /> \end{bmatrix}$ and solving the equation $AB=BA$ gives $B=\begin{bmatrix}<br /> i-\frac{2}{3}g & 0 & \frac{1}{3}g \\<br /> g-3f & i-3g & f \\<br /> g & 0 & i \\<br /> \end{bmatrix}$.

$U$ is a subspace of $\mathbb{M_{3\times3}}$ if $\forall u_1,u_2\in U\Rightarrow u_1+u_2\in U,\forall t\in\mathbb{R}\Rightarrow tu_1\in U$ which is correct.

It is easy to check that if $C\in span\{I,A,A^2,...\}\Rightarrow C\in U$:
Linear combination for C is
$C=c_0I+c_1A+c_2A^2+...\Rightarrow CA=AC\Rightarrow C\in U$

$U$ has the dimension $3$ and a basis are column vectors of identity matrix $3\times 3$.

How to find the dimension and a basis for $span\{I,A,A^2,...\}$?

What does it mean to be a basis?

I don't agree with your explanation of the dimension and basis of U. Column vectors (in this case) are $3x1$ matrices. You're claiming that you can generate a whole space of 3x3 matrices with linear combinations of 3x1 matrices. The basis that you are claiming is not even a subset of U.

Also, how are you getting B?

 

[gruba]

I made a mistake in the statement for the dimension and a basis of $U$. You could look at $9\times 1$ vectors.
But what is the way of algebraically finding a basis (dimension follows) of $U$ and $span\{I,A,A^2,...\}$?

Matrix $B$ can be reduced from $9$ to $3$ variables after solving the equation $AB=BA$ (too long to post).

 

[MostlyHarmless]

I made a mistake in the statement for the dimension and a basis of $U$. You could look at $9\times 1$ vectors.

Your basis has to be a set of 3x3 matrices.

But what is the way of algebraically finding a basis (dimension follows) of $U$ and $span\{I,A,A^2,...\}$?

Matrix $B$ can be reduced from $9$ to $3$ variables after solving the equation $AB=BA$ (too long to post).

That is fair, I'm just wondering the method you used.

You've (assuming your calculations are correct) found what matrices in the subspace U "look like". Can you write that matrix as a linear combination of other matrices? (I haven't verified, but it seems like the $span\{I,A,A^2, \dots\}$ would not have been given to you just to show that it was contained in U.

 

[gruba]

I think that it is not possible to form a linear combination on $span\{I,A,A^2,...\}$ because this set is not finite - defined (you could arbitrarily choose that $span\{I,A,A^2,A^3\}$ - this would be finite set).

 

[Samy_A]

I think that it is not possible to form a linear combination on $span\{I,A,A^2,...\}$ because this set is not finite - defined (you could arbitrarily choose that $span\{I,A,A^2,A^3\}$ - this would be finite set).

What can you say about the dimension of $span\{I,A,A^2,...\}$?

 

[MostlyHarmless]

I think that it is not possible to form a linear combination on $span\{I,A,A^2,...\}$ because this set is not finite - defined (you could arbitrarily choose that $span\{I,A,A^2,A^3\}$ - this would be finite set).

I don't follow what you are trying to say here.

I'm sorry, I must be missing something.

 

[pasmith]

I think that it is not possible to form a linear combination on $span\{I,A,A^2,...\}$ because this set is not finite

(a) The space spanned by an infinite set can nevertheless be finite-dimensional; as a trivial example, consider $\mathrm{span}\{ (x,0) : 0 < x \leq 1\}$ which is just the one-dimensional subspace of $\mathbb{R}^2$ spanned by (1,0).

(b) In this case, apply the Cayley-Hamilton theorem.

 

[HallsofIvy]

A linear combination on an infinite set is defined to be a linear combination on a finite subset so your objection in post 5 is not relevant.