1. Linear subspace!

Hi!

I have problem because I do not know how to prove this:

I have a set:
$\displaystyle T_n (\mathbb{R})=\big\{[a_i_j] \ : \ a_i_j \in \ \mathbb{R} \ , \ i,j \ = 1,...,n, \ a_i_j=0 \ for \ i>j\big\}$

and I must show that the $\displaystyle T_n (\mathbb{R})$ is linear subspace of space n x n matrices $\displaystyle M_n (\mathbb{R})$? Also I must to define the dimension of a space $\displaystyle T_n (\mathbb{R})$ and write 2 different bases of $\displaystyle T_3(\mathbb{R})$ space?

Anybody know how can I do that?

Tomi

2. Originally Posted by tom27
Hi!

I have problem because I do not know how to prove this:

I have a set:
$\displaystyle T_n (\mathbb{R})=\big\{[a_i_j] \ : \ a_i_j \in \ \mathbb{R} \ , \ i,j \ = 1,...,n, \ a_i_j=0 \ for \ i>j\big\}$

and I must show that the $\displaystyle T_n (\mathbb{R})$ is linear subspace of space n x n matrices $\displaystyle M_n (\mathbb{R})$? Also I must to define the dimension of a space $\displaystyle T_n (\mathbb{R})$ and write 2 different bases of $\displaystyle T_3(\mathbb{R})$ space?

Anybody know how can I do that?

Tomi

So $\displaystyle T_n(\mathbb{R})$ is simply the set of all nxn real upper triangular matrices. Show now that this set is closed

under sum of matrices and multiplication by scalar (piece of cake).

For its dimension: by what elements is any element of the above set uniquely and completely determined? In how many

ways can you choose lin. ind. real numbers for these elements? Well, this is the dimension, and to choose now

2 (or 100000) different basis for it is another piece of cake.

Tonio

3. Another help (for the dimension). Consider

$\displaystyle A_{11}=\begin{bmatrix} 1 & 0 & \ldots & 0\\ 0 & 0 & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & 0\end{bmatrix},\ldots,A_{nn}=\begin{bmatrix} 0 & 0 & \ldots & 0\\ 0 & 0 & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & 1\end{bmatrix}.$

$\displaystyle A_{12}=\begin{bmatrix} 0 & 1 & \ldots & 0\\ 0 & 0 & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & 0\end{bmatrix},\ldots,A_{1n}=\begin{bmatrix} 0 & 0 & \ldots & 1\\ 0 & 0 & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & 0\end{bmatrix},\ldots$

It is easy to prove that

$\displaystyle B=\{A_{11},\ldots,A_{nn},A_{12},\ldots,A_{n-1,n}\}$

is a basis of $\displaystyle T_n$.

Its dimension is:

$\displaystyle n+1+2+\ldots (n-1)=\dfrac{n(n+1)}{2}$

Fernando Revilla

4. Thank you both for reply!

So if I understand this:

$\displaystyle T_n(\mathbb{R})=\begin{bmatrix} a_1_1 & a_1_2 & \cdots & a_1_n \\ 0 & a_2_2 & \cdots & a_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & a_n_n \end{bmatrix}$

If I chose matrices:

$\displaystyle M_1_1(\mathbb{R})=\begin{bmatrix} a_1_1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix}$

and

$\displaystyle M_1_2(\mathbb{R})=\begin{bmatrix} 0 & a_1_2 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix}$

the sum of this two matrices is belong to set $\displaystyle T_n(\mathbb{R})$. Is this enough to prove that a set is "closed under addition"?

To prove that a set is "closed under multiplication" I must do:

$\displaystyle \left \begin{bmatrix} \alpha a_1_1 & \alpha a_1_2 & \cdots & \alpha a_1_n \\ 0 & \alpha a_2_2 & \cdots & \alpha a_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & \alpha a_n_n \end{bmatrix}\right =\alpha \begin{bmatrix} a_1_1 & a_1_2 & \cdots & a_1_n \\ 0 & a_2_2 & \cdots & a_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & a_n_n \end{bmatrix}$
Is that all?

So those two bases are ok?

$\displaystyle \left \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\right \ and \ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$.

Tomi

Every $\displaystyle A\in T_3(\mathbb{R})$ can be written in the form:

$\displaystyle A=\begin{bmatrix}{a_{11}}&{a_{12}}&{a_{13}}\\{0}&{ a_{22}}&{a_{23}}\\{0}&{0}&{a_{33}}\end{bmatrix}=a_ {11}\begin{bmatrix}{1}&{0}&{0}\\{0}&{0}&{0}\\0&0&0 \end{bmatrix}+a_{22}\begin{bmatrix}{0}&{0}&{0}\\{0 }&{1}&{0}\\0&0&0\end{bmatrix}+$

$\displaystyle a_{33}\begin{bmatrix}{0}&{0}&{0}\\{0}&{0}&{0}\\0&0 &1\end{bmatrix}+a_{12}\begin{bmatrix}{0}&{1}&{0}\\ {0}&{0}&{0}\\0&0&0\end{bmatrix}+a_{13}\begin{bmatr ix}{0}&{0}&{1}\\{0}&{0}&{0}\\0&0&0\end{bmatrix}+a_ {23}\begin{bmatrix}{0}&{0}&{0}\\{0}&{0}&{1}\\0&0&0 \end{bmatrix}$

so, those six matrices generate $\displaystyle T_3(\mathbb{R})$ . On the other hand, you can easily prove that also are linearly independent i.e. those six matrices form a basis for $\displaystyle T_3(\mathbb{R})$. Note that $\displaystyle 3(3+1)/2=6$

You have to choose two generic matrices of $\displaystyle T_n(\mathbb{R})$ and prove that the sum belongs to $\displaystyle T_n(\mathbb{R})$.

Fernando Revilla

6. Originally Posted by tom27

So if I understand this:

$\displaystyle T_n(\mathbb{R})=\begin{bmatrix} a_1_1 & a_1_2 & \cdots & a_1_n \\ 0 & a_2_2 & \cdots & a_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & a_n_n \end{bmatrix}$

If I chose matrices:

$\displaystyle M_1_1(\mathbb{R})=\begin{bmatrix} a_1_1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix}$

and

$\displaystyle M_1_2(\mathbb{R})=\begin{bmatrix} 0 & a_1_2 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix}$

the sum of this two matrices is belong to set $\displaystyle T_n(\mathbb{R})$. Is this enough to prove that a set is "closed under addition"?

To prove that a set is "closed under multiplication" I must do:

$\displaystyle \left \begin{bmatrix} \alpha a_1_1 & \alpha a_1_2 & \cdots & \alpha a_1_n \\ 0 & \alpha a_2_2 & \cdots & \alpha a_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & \alpha a_n_n \end{bmatrix}\right =\alpha \begin{bmatrix} a_1_1 & a_1_2 & \cdots & a_1_n \\ 0 & a_2_2 & \cdots & a_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & a_n_n \end{bmatrix}$
Is that all?

So those two bases are ok?

$\displaystyle \left \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\right \ and \ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$.

Tomi
First, those aren't "two bases". Those are two matrices and a "basis" here is a set of matrices.

Second, FernandoRevilla told you that the set of upper triangular n by n matrices has dimension n(n+1)/2. In the case that n= 2, that is equal to 2(3)/2= 3. A basis must consist of three matrices.

7. Thank you all!

a) To prove that subspace is linear I do that:

I choose generic matrices (A and B in this case).

$\displaystyle \left A=\begin{bmatrix} a_1_1 & a_1_2 & \cdots & a_1_n \\ 0 & a_2_2 & \cdots & a_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & a_n_n \end{bmatrix} B=\begin{bmatrix} b_1_1 & b_1_2 & \cdots & b_1_n \\ 0 & b_2_2 & \cdots & b_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & b_n_n \end{bmatrix}$

1) Try to prove $\displaystyle T_n(A+B)=T_n(A)+T_n(B)$

$\displaystyle T_n(A+B)=\begin{bmatrix} a_1_1+b_1_1 & a_1_2+b_1_2 & \cdots & a_1_n+b_1_n \\ 0 & a_2_2+b_2_2 & \cdots & a_2_n+b_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & a_n_n+b_n_n \end{bmatrix}=$

$\displaystyle \left T_n\begin{bmatrix} a_1_1 & a_1_2 & \cdots & a_1_n \\ 0 & a_2_2 & \cdots & a_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & a_n_n \end{bmatrix} + T_n\begin{bmatrix} b_1_1 & b_1_2 & \cdots & b_1_n \\ 0 & b_2_2 & \cdots & b_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & b_n_n \end{bmatrix}=T_n(A)+T_n(B)$

2) Try to prove $\displaystyle T_n(\alpha A)= \alpha T_n(A)$
$\displaystyle \left T_n(\alpha A)=T_n \begin{bmatrix} \alpha a_1_1 & \alpha a_1_2 & \cdots & \alpha a_1_n \\ 0 & \alpha a_2_2 & \cdots & \alpha a_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & \alpha a_n_n \end{bmatrix}\right =\alpha T_n \begin{bmatrix} a_1_1 & a_1_2 & \cdots & a_1_n \\ 0 & a_2_2 & \cdots & a_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & a_n_n \end{bmatrix}= \alpha T_n(A)$

Because the both are true (1. and 2.) this is a linear subspace. Is that right?

b) So the basis in uper case (post 5) is all six matrices. The second case can be:

$\displaystyle \left \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\right \ \ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \right \ \ \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Are those three matrices a basis of$\displaystyle T_3(\mathbb{R})$?

Tomi

8. Originally Posted by tom27
Thank you all!

a) To prove that subspace is linear I do that:

I choose generic matrices (A and B in this case).

$\displaystyle \left A=\begin{bmatrix} a_1_1 & a_1_2 & \cdots & a_1_n \\ 0 & a_2_2 & \cdots & a_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & a_n_n \end{bmatrix} B=\begin{bmatrix} b_1_1 & b_1_2 & \cdots & b_1_n \\ 0 & b_2_2 & \cdots & b_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & b_n_n \end{bmatrix}$

1) Try to prove $\displaystyle T_n(A+B)=T_n(A)+T_n(B)$

$\displaystyle T_n(A+B)=\begin{bmatrix} a_1_1+b_1_1 & a_1_2+b_1_2 & \cdots & a_1_n+b_1_n \\ 0 & a_2_2+b_2_2 & \cdots & a_2_n+b_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & a_n_n+b_n_n \end{bmatrix}=$

$\displaystyle \left T_n\begin{bmatrix} a_1_1 & a_1_2 & \cdots & a_1_n \\ 0 & a_2_2 & \cdots & a_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & a_n_n \end{bmatrix} + T_n\begin{bmatrix} b_1_1 & b_1_2 & \cdots & b_1_n \\ 0 & b_2_2 & \cdots & b_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & b_n_n \end{bmatrix}=T_n(A)+T_n(B)$

2) Try to prove $\displaystyle T_n(\alpha A)= \alpha T_n(A)$
$\displaystyle \left T_n(\alpha A)=T_n \begin{bmatrix} \alpha a_1_1 & \alpha a_1_2 & \cdots & \alpha a_1_n \\ 0 & \alpha a_2_2 & \cdots & \alpha a_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & \alpha a_n_n \end{bmatrix}\right =\alpha T_n \begin{bmatrix} a_1_1 & a_1_2 & \cdots & a_1_n \\ 0 & a_2_2 & \cdots & a_2_n \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & a_n_n \end{bmatrix}= \alpha T_n(A)$

Because the both are true (1. and 2.) this is a linear subspace. Is that right?

b) So the basis in uper case (post 5) is all six matrices. The second case can be:

$\displaystyle \left \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\right \ \ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \right \ \ \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Are those three matrices a basis of$\displaystyle T_3(\mathbb{R})$?

Tomi

What part of "the dimension of $\displaystyle T_n(\mathbb{R})$ equals $\displaystyle \frac{n(n+1)}{2}$" haven't you yet understood after

two different people already told you this? You must really be more careful...

Tonio

I hope that now I understand this problem. Because we have space $\displaystyle T_3(\mathbb{R}) (n=3)$ and we know that dimension of upper triangular nxn matrices is $\displaystyle \frac{n(n+1)}{2}$, we get the dimension for our case $\displaystyle \frac{3(3+1)}{2}=6$. That mean the solution (basis) is set of six independent matrices. So the other basis looks like basis in post 5, only the numbers (in post 5 is 1) are different . Is that right?

$\displaystyle \left \begin{bmatrix} 3 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\right \ \ \begin{bmatrix} 0 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\right \ \ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 4 \\ 0 & 0 & 0 \end{bmatrix}\right \ \ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

10. Originally Posted by tom27

I hope that now I understand this problem. Because we have space $\displaystyle T_3(\mathbb{R}) (n=3)$ and we know that dimension of upper triangular nxn matrices is $\displaystyle \frac{n(n+1)}{2}$, we get the dimension for our case $\displaystyle \frac{3(3+1)}{2}=6$. That mean the solution (basis) is set of six independent matrices. So the other basis looks like basis in post 5, only the numbers (in post 5 is 1) are different . Is that right?

$\displaystyle \left \begin{bmatrix} 3 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\right \ \ \begin{bmatrix} 0 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\right \ \ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 4 \\ 0 & 0 & 0 \end{bmatrix}\right \ \ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

This time you got it, but I don't understand why all those numbers: 4,8,...why not all the nonzero entries in all the matrices 1 ? Ain't that simpler?

Tonio

11. I must write the two different basis of space $\displaystyle T_3(\mathbb{R})$ so the one is with 1 on nonzero entries and the other must be different.

12. Okay, yes, that is a basis and the same thing with "1" in place of those non-zero numbers is another basis. Or you could create a new basis by adding some of the matrices in the "standard" basis. For example
$\displaystyle \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$, $\displaystyle \begin{bmatrix}1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$, $\displaystyle \begin{bmatrix}1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$
etc. will give a basis.

13. So the one possibility for basis in this case is:
$\displaystyle \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix},\begin{bmatrix}1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix},\begin{bmatrix}1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix},\begin{bmatrix}1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix},\begin{bmatrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{bmatrix},\begin{bmatrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$.

That mean that in all six matrices the upper triangular element must be included at least once and the set of matrices represent basic?

14. I think the basis is ok, because I do not see any linear dependence between those matrices. Can anybody confirmed this?

Enrico

15. Yes. The last matrix has a "1" in a place no other does so it is independent of the first five. The fifth matrix has a "1" in a place none of the previous four does so it is independent of them, etc.

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