1. ## Linear subspace!

Hi!

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

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

and I must show that the $T_n (\mathbb{R})$ is linear subspace of space n x n matrices $M_n (\mathbb{R})$? Also I must to define the dimension of a space $T_n (\mathbb{R})$ and write 2 different bases of $T_3(\mathbb{R})$ 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:
$T_n (\mathbb{R})=\big\{[a_i_j] \ : \ a_i_j \in \ \mathbb{R} \ , \ i,j \ = 1,...,n, \ a_i_j=0 \ for \ i>j\big\}$

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

Anybody know how can I do that?

Tomi

So $T_n(\mathbb{R})$ 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

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

$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

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

is a basis of $T_n$.

Its dimension is:

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

Fernando Revilla

4. Thank you both for reply!

So if I understand this:

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

If I chose matrices:

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

and

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

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

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

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

Is that all?

So those two bases are ok?

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

Tomi

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

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

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

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

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

Fernando Revilla

6. Originally Posted by tom27

So if I understand this:

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

If I chose matrices:

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

and

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

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

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

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

Is that all?

So those two bases are ok?

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

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).

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

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

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

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

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

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:

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

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

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).

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

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

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

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

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

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:

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

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

Tomi

What part of "the dimension of $T_n(\mathbb{R})$ equals $\frac{n(n+1)}{2}$" 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 $T_3(\mathbb{R}) (n=3)$ and we know that dimension of upper triangular nxn matrices is $\frac{n(n+1)}{2}$, we get the dimension for our case $\frac{3(3+1)}{2}=6$. 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?

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

10. Originally Posted by tom27

I hope that now I understand this problem. Because we have space $T_3(\mathbb{R}) (n=3)$ and we know that dimension of upper triangular nxn matrices is $\frac{n(n+1)}{2}$, we get the dimension for our case $\frac{3(3+1)}{2}=6$. 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?

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

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 $T_3(\mathbb{R})$ 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
$\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$, $\begin{bmatrix}1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$, $\begin{bmatrix}1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$
etc. will give a basis.

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

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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