Proving Non-Equivalence of Matrices Using Trace in Linear Algebra

[Ted123]

Homework Statement

[PLAIN]http://img152.imageshack.us/img152/3162/linal.jpg

Homework Equations

The Attempt at a Solution

How do I do part (i) and follow the hint?

 

[micromass]

What does the sentence "U is a subspace of V" mean? what do you need to check for this?

 

[Ted123]

What does the sentence "U is a subspace of V" mean? what do you need to check for this?

(i) U non-empty
(ii) U closed under vector addition
(iii) U closed under scalar multiplication

 

[micromass]

Fine, so which of these three properties is bothering you? and why?

 

[Ted123]

Fine, so which of these three properties is bothering you? and why?

Well can I show the properties like so?:

(i) {\bf 0}\;\text{matrix}\in U so U\neq \emptyset

For a,b,c,d,e,f,g \in \mathbb{R},

(ii) Let \begin{bmatrix} 2a & b & c \\ d & -a & e \\ f & g & -a \end{bmatrix} \in U

Then \begin{bmatrix} 2a & b & c \\ d & -a & e \\ f & g & -a \end{bmatrix} + \begin{bmatrix} 2a & b & c \\ d & -a & e \\ f & g & -a \end{bmatrix} = \begin{bmatrix} 4a & 2b & 2c \\ 2d & -2a & 2e \\ 2f & 2g & -2a \end{bmatrix} \in U

(iii) \alpha\begin{bmatrix} 2a & b & c \\ d & -a & e \\ f & g & -a \end{bmatrix} = \begin{bmatrix} 2\alpha a & \alpha b & \alpha c \\ \alpha d & -\alpha a & \alpha e \\ \alpha f & \alpha g & -\alpha a \end{bmatrix}\in U

\therefore U is a subspace of V=M_3(\mathbb{R})

But what is the hint saying? (and how to find the dimension in (ii)?)

 

[Mark44]

For ii, you need to take two arbitrary matrices, say M and N, that are in U, and show that M + N is in U. For the two matrices M and N, the only entries that are important are the ones on the diagonal. I suspect that's where the hint will be successful.

You made some unwarranted assumptions in your matrix in part ii; namely, that the entries on the diagonal were 2a, -a, and -a. If the entries on the diagonal for M are a, b, and c, what makes it so that M is in U?

 

[Ted123]

For ii, you need to take two arbitrary matrices, say M and N, that are in U, and show that M + N is in U. For the two matrices M and N, the only entries that are important are the ones on the diagonal. I suspect that's where the hint will be successful.

You made some unwarranted assumptions in your matrix in part ii; namely, that the entries on the diagonal were 2a, -a, and -a. If the entries on the diagonal for M are a, b, and c, what makes it so that M is in U?

Does this prove that U is a subspace of V?:

{\bf 0}\;\text{matrix} \in U so U\neq\emptyset .

For For a,a',b,b',c,c',d,d',e,e',f,f',g,g',h,h',i,i'\in\mathbb{R} ,

Let A=\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\in U

and B=\begin{bmatrix} a' & b' & c' \\ d' & e' & f' \\ g' & h' & i' \end{bmatrix}\in U

so \text{trace}(A) = a+e+i=0 and \text{trace}({B}) = a'+e'+i'=0

\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} + \begin{bmatrix} a' & b' & c' \\ d' & e' & f' \\ g' & h' & i' \end{bmatrix} = \begin{bmatrix} a+a' & b+b' & c+c' \\ d+d' & e+e' & f+f' \\ g+g' & h+h' & i+i' \end{bmatrix}\in U

since (a+a')+(e+e')+(i+i')=(a+e+i)+(a'+e'+i')= \text{trace}(A) + \text{trace}(B) = 0

\therefore U is closed under vector addition.

\alpha A=\begin{bmatrix} \alpha a & \alpha b & \alpha c \\ \alpha d & \alpha e & \alpha f \\ \alpha g & \alpha h & \alpha i \end{bmatrix}\in U

since \text{trace}(\alpha A}) = \alpha a +\alpha e +\alpha i = \alpha (a+e+i) = \alpha \text{trace}(A) = 0

\therefore U is closed under scalar multiplication.

\therefore U is a subspace of V=M_3(\mathbb{R})

 

[micromass]

Yes, this is a good proof.

 

[Ted123]

Yes, this is a good proof.

For part (ii) (finding the dimension) why is the following correct?

Letting M = (m_{ij}), we need that m_{11} + m_{22} + m_{33} = 0 .

Thus, \text{dim}(U) = \text{dim}(M_3(\mathbb{R})) - 1 = 3^2 - 1 = 8 .

And how do I do part (iii)?

 

[micromass]

Well the trace is a linear map tr:M_{3\times 3}\rightarrow \mathbb{R} and thus satisfies

\dim(ker(tr))+\dim(im(tr))=M_{3\times 3}

You will need to find the dimension of ker(tr). So you just need to shom that the dimension of im(tr) is 1, and that the dimension of M_{3\times 3} is 9.


If you didnt see the dimension theorem, then you could also proceed by explicitly finding a base of U...

 

[Ted123]

Well the trace is a linear map tr:M_{3\times 3}\rightarrow \mathbb{R} and thus satisfies

\dim(ker(tr))+\dim(im(tr))=M_{3\times 3}

You will need to find the dimension of ker(tr). So you just need to shom that the dimension of im(tr) is 1, and that the dimension of M_{3\times 3} is 9.


If you didnt see the dimension theorem, then you could also proceed by explicitly finding a base of U...

What is ker and I am ? I haven't seen that before but I'd like to learn it... (is ker kernel?) how do you find them?

But I'm more interested at the moment in how to do (iii)?

 

[Ted123]

For part (iii), in answer to the hint:

A \sim B\iff A-B \in U \iff \text{trace}(A-B)=0\iff \text{trace}(A)=\text{trace}(B)

but how does knowing this help me solve the question (show that the equivalence classes can be labelled by elements of \mathbb{R})?

 

[micromass]

Well, take an equivalence class [A] (so all elements of this class have the same trace as A), to what number would you assign [A]?

 

[Ted123]

Well, take an equivalence class [A] (so all elements of this class have the same trace as A), to what number would you assign [A]?

I really don't know. Why do I need to assign a number?

 

[micromass]

That's what the exercise is... You need to assign a real number to every equivalence class such that every real number corresponds to a unique equivalence class.

Now, every matrix in an equivalence class has the same trace. So it's obvious you need to do something with traces. What would then be the real number you assign to an equivalence class?

 

[Ted123]

That's what the exercise is... You need to assign a real number to every equivalence class such that every real number corresponds to a unique equivalence class.

Now, every matrix in an equivalence class has the same trace. So it's obvious you need to do something with traces. What would then be the real number you assign to an equivalence class?

The trace of the matrices in the equivalance class?

 

[micromass]

Yes! To an equivalence class [A] just assign trace(A).

You'll need to check two things now:
- if [A]=, then the assigned number is also the same
- if [A] and are different classes, then the assigned number is different.

 

[Ted123]

Yes! To an equivalence class [A] just assign trace(A).

You'll need to check two things now:
- if [A]=, then the assigned number is also the same
- if [A] and are different classes, then the assigned number is different.

How?

 

[micromass]

Come on, that's not that difficult.

Take [A]=, then A and B are in the same equivalence class. It's easy to see that they have thesame trace.
The other thing is as easy...

 

[Ted123]

Come on, that's not that difficult.

Take [A]=, then A and B are in the same equivalence class. It's easy to see that they have thesame trace.
The other thing is as easy...

Well it probably isn't but I'm not that familiar with equivalence classes and such.

You've said 'matrices in the same equivalence class have the same trace'. Can I take this as a fact?

Why are you using a small 'b' in and a capital B?

What's part (iv) asking me to do, and how do I go about it?

 

[micromass]

You've basically proven it youreself.

If A and B are in the same equivalence class, then (by definition) A and B are equivalent. And in page 1, you have shown that A\sim B implies tr(A)=tr(B). So you've already proven it...

small b is a typo, it's supposed to be B, sorry

 

[Ted123]

You've basically proven it youreself.

If A and B are in the same equivalence class, then (by definition) A and B are equivalent. And in page 1, you have shown that A\sim B implies tr(A)=tr(B). So you've already proven it...

small b is a typo, it's supposed to be B, sorry

So how do I, \forall \alpha\in \mathbb{R} write down an explicit representative of the corresponding equivalence class?

 

[micromass]

Take a in R. You'll need to find a matrix A such that [A] is labelled by a. Remember that labelling is given by the trace...

 

[Ted123]

Take a in R. You'll need to find a matrix A such that [A] is labelled by a. Remember that labelling is given by the trace...

Well can't I just use A=\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} ? \alpha=\text{trace}(A)=0

 

[micromass]

Yes, that matrix would be labelled by 0 (since the trace of the matrix is 0).
Can you find out what matrix is labelled by 2?
More generally, what matrix is labelled by a?

 

[Ted123]

Yes, that matrix would be labelled by 0 (since the trace of the matrix is 0).
Can you find out what matrix is labelled by 2?
More generally, what matrix is labelled by a?

A=\begin{bmatrix} 0 & 0 & 0\\ 0 & \alpha & 0 \\ 0 & 0 & 0 \end{bmatrix} is labelled by \alpha

 

[micromass]

Yes, can you now find in general what matrix is labelled by a?

 

[Ted123]

Yes, can you now find in general what matrix is labelled by a?

A=\begin{bmatrix} 0 & 0 & 0\\ 0 & \alpha & 0 \\ 0 & 0 & 0 \end{bmatrix} is labelled by \alpha

 

[micromass]

Correct! Thats the solution to this question.

 

[Ted123]

Correct! Thats the solution to this question.

Thanks. Just to clarify part (iii) and the checking of those 2 things:

Is this correct?

[A]=<b> \iff A \sim B\iff A-B \in U \iff \text{trace}(A-B)=0\iff \text{trace}(A)=\text{trace}(B)</b>

[A]\neq<b> \not\iff A \sim B\not\iff A-B \in U \not\iff \text{trace}(A-B)=0\not\iff \text{trace}(A)=\text{trace}(B)</b>

 

[micromass]

Yes, that seems to do it! It looks like you've got it!

 

[Ted123]

Yes, that seems to do it! It looks like you've got it!

Or it may be better to write the second one as

[A]\neq<b> \iff A \not\sim B \iff A-B \not\in U \iff \text{trace}(A-B)\neq 0 \iff \text{trace}(A)\neq\text{trace}(B)</b>