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