Show that the set of the upper triangular matrices is a ring

  • #1
mathmari
Gold Member
MHB
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Hey! :eek:

A set $R$ with two operations $+$ und $\cdot$ is a ring, if the following properties are satisfied:
  1. $(R, +)$ is a commutative/abelian group
  2. Associativity : For all $a,b, c \in R$ it holds that $(a \cdot b)\cdot c = a \cdot (b \cdot c)$.
  3. Distributive property : For all $a,b, c \in R$ it holds that $(a+b)\cdot c=a\cdot c+b\cdot c$ und $a \cdot (b + c)=a\cdot b+a\cdot c$.

Let $\mathbb{K}$ be a field.
We have the set $T_n(\mathbb{K})=\{A\in \mathbb{K}^{n\times n} \mid a_{ij}=0 \text{ für alle } i,j\in \{1, \ldots , n\} \text{ mit } i>j\}$.

I want to show that $T_n(\mathbb{K})$ is a ring with the addition and multiplication of matrices. How can we show the first property? (Wondering)

Let $A,B,C\in T_n(\mathbb{K})$.
To show the other two properties do we have to find the form of the element of the the resulting matrix at the position $ij$ ? (Wondering)
$$[(A \cdot B)\cdot C]_{ij}=( \sum_{k=1}^ma_{ik}b_{kj})c_{ij}$$
To what is this equal? (Wondering)
 
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  • #2
mathmari said:
How can we show the first property? (Wondering)

Hi mathmari! (Smile)

Don't we already know that regular matrix multiplication over a field is a ring (with multiplicative identity)? (Wondering)
So wouldn't it suffice to show that $T_n(\mathbb K)$ is a sub ring?
If so then we only need to verify if it is closed for addition and multiplication, and if the additive identity is an element.
Let $A,B,C\in T_n(\mathbb{K})$.
To show the other two properties do we have to find the form of the element of the the resulting matrix at the position $ij$ ? (Wondering)
$$[(A \cdot B)\cdot C]_{ij}=( \sum_{k=1}^ma_{ik}b_{kj})c_{ij}$$
To what is this equal? (Wondering)

Shouldn't that be:
$$[(A \cdot B)\cdot C]_{ij}=\sum_{l=1}\left(\sum_{k=1}^ma_{ik}b_{kl}\right)c_{lj}$$
(Wondering)
 
  • #3
I like Serena said:
If so then we only need to verify if it is closed for addition and multiplication, and if the additive identity is an element.

Ah ok...

Let $A,B\in T_n(\mathbb{K})$, i.e., $A,B$ are two upper triangular matrices. We have that the addition and the mutiplication of two upper triangular matrices is a triangular matrix, or not? Do we have to prove it? If so how? (Wondering)
 
  • #4
mathmari said:
Ah ok...

Let $A,B\in T_n(\mathbb{K})$, i.e., $A,B$ are two upper triangular matrices. We have that the addition and the mutiplication of two upper triangular matrices is a triangular matrix, or not? Do we have to prove it? If so how? (Wondering)

For addition it is trivial. We might say something like:

For any $A, B \in T_n(\mathbb{K})$, let $C=A+B$.
$C$ is given by $c_{ij}=a_{ij}+b_{ij}$, which is $0$ for all $i>j$.
Therefore $C \in T_n(\mathbb{K})$ and thus $T_n(\mathbb{K})$ is closed for addition.

We can prove closure for multiplication as follows:

For any $A, B \in T_n(\mathbb{K})$, let $C=AB$.
For $i>j$ it follows that:
$$c_{ij} = \sum_{k=1}^n a_{ik}b_{kj}
=\sum_{k=1}^{i-1} a_{ik}b_{kj} + \sum_{k=i}^n a_{ik}b_{kj}
=\sum_{k=1}^{i-1} 0\cdot b_{kj} + \sum_{k=i}^n a_{ik}\cdot 0 = 0
$$
Therefore $C \in T_n(\mathbb{K})$ and thus $T_n(\mathbb{K})$ is closed for multiplication.

The additive identity is the zero matrix, which is an element of $T_n(\mathbb{K})$.

Thus the proof is complete and $T_n(\mathbb{K})$ is a sub rng of $\mathbb{K}^{n\times n}$, and therefore a rng. (Nerd)
 
  • #5
I like Serena said:
For addition it is trivial. We might say something like:

For any $A, B \in T_n(\mathbb{K})$, let $C=A+B$.
$C$ is given by $c_{ij}=a_{ij}+b_{ij}$, which is $0$ for all $i>j$.
Therefore $C \in T_n(\mathbb{K})$ and thus $T_n(\mathbb{K})$ is closed for addition.

We can prove closure for multiplication as follows:

For any $A, B \in T_n(\mathbb{K})$, let $C=AB$.
For $i>j$ it follows that:
$$c_{ij} = \sum_{k=1}^n a_{ik}b_{kj}
=\sum_{k=1}^{i-1} a_{ik}b_{kj} + \sum_{k=i}^n a_{ik}b_{kj}
=\sum_{k=1}^{i-1} 0\cdot b_{kj} + \sum_{k=i}^n a_{ik}\cdot 0 = 0
$$
Therefore $C \in T_n(\mathbb{K})$ and thus $T_n(\mathbb{K})$ is closed for multiplication.

I understand! (Happy) (Yes)
I like Serena said:
If so then we only need to verify if it is closed for addition and multiplication, and if the additive identity is an element.

I like Serena said:
The additive identity is the zero matrix, which is an element of $T_n(\mathbb{K})$.

Why do we have to check only if the additive identity of the matrix ring is an element of $T_n(\mathbb{K})$ and not also the multiplicative identity? (Wondering)
 
  • #6
mathmari said:
Why do we have to check only if the additive identity of the matrix ring is an element of $T_n(\mathbb{K})$ and not also the multiplicative identity? (Wondering)

Because your ring doesn't have a multiplicative identity! :eek:
Such a ring is also named rng instead to distinguish it.
See wikpedia.
 
  • #7
I like Serena said:
Because your ring doesn't have a multiplicative identity! :eek:
Such a ring is also named rng instead to distinguish it.
See wikpedia.

Why doesn't it contain the identity matrix? (Wondering)
 
  • #8
mathmari said:
Why doesn't it contain the identity matrix? (Wondering)

Oh it contains the identity matrix all right.
It's just that it is not required according to the definition of a ring that you gave. (Smirk)
 
  • #9
In general the matrix ring contains the multiplicative identity, the idenity matrix, or not? (Wondering)
But we don't have to show that $T_n(\mathbb{K})$ contains it because the definition of a ring doesn't require it, only the definition of a unit ring, right? (Wondering)
 
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  • #10
mathmari said:
In general the matrix ring contains the multiplicative identity, the idenity matrix, or not? (Wondering)
But we don't have to show that $T_n(\mathbb{K})$ contains it because the definition of a ring doesn't require it, only the definition of a unit ring, right? (Wondering)

Yep. (Nod)

Consider for instance $T'_n(\mathbb{K})=\{A\in \mathbb{K}^{n\times n} \mid a_{ij}=0 \text{ für alle } i,j\in \{1, \ldots , n\} \text{ mit } {\color{green} i\ge j}\}$.
I believe it's still a rng, but it doesn't have the identity matrix.
 
  • #11
I like Serena said:
Yep. (Nod)

Consider for instance $T'_n(\mathbb{K})=\{A\in \mathbb{K}^{n\times n} \mid a_{ij}=0 \text{ für alle } i,j\in \{1, \ldots , n\} \text{ mit } {\color{green} i\ge j}\}$.
I believe it's still a rng, but it doesn't have the identity matrix.

Ah ok... I see! (Smile) I want to show also that $\phi_k : T_n(\mathbb{K})\rightarrow \mathbb{K}, (a_{ij})\mapsto a_{kk}, \ k\in \{1, \ldots , n\}$ is a ring homomorphim, so we have to show that for elements $(a_{ij}), (b_{ij})\in T_n(\mathbb{K})$ the following holds:
$$\phi ((a+b)_{ij})=\phi (a_{ij})+\phi (b_{ij}) \\ \phi (a\cdot b)_{ij})=\phi (a_{ij})\cdot \phi (b_{ij})$$

We have the following:
$$\phi ((a +b)_{ij})=((a+b)_{kk})=(a_{kk})+(b_{kk})=\phi (a_{ij})+\phi (b_{ij}) \\ \phi ((a\cdot b)_{ij}))=((a\cdot b)_{kk})=(a_{kk})\cdot (b_{kk})=\phi (a_{ij})\cdot \phi (b_{ij})$$

Is this correct? Could I improve something? (Wondering)
 
  • #12
mathmari said:
I want to show also that $\phi_k : T_n(\mathbb{K})\rightarrow \mathbb{K}, (a_{ij})\mapsto a_{kk}, \ k\in \{1, \ldots , n\}$ is a ring homomorphim, so we have to show that for elements $(a_{ij}), (b_{ij})\in T_n(\mathbb{K})$ the following holds:
$$\phi ((a+b)_{ij})=\phi (a_{ij})+\phi (b_{ij}) \\ \phi (a\cdot b)_{ij})=\phi (a_{ij})\cdot \phi (b_{ij})$$

We have the following:
$$\phi ((a +b)_{ij})=((a+b)_{kk})=(a_{kk})+(b_{kk})=\phi (a_{ij})+\phi (b_{ij}) \\ \phi ((a\cdot b)_{ij}))=((a\cdot b)_{kk})=(a_{kk})\cdot (b_{kk})=\phi (a_{ij})\cdot \phi (b_{ij})$$

Is this correct? Could I improve something? (Wondering)

I think we need $\phi_k$ everywhere instead of $\phi$.

And generally for matrices $((a\cdot b)_{kk})\ne(a_{kk})\cdot (b_{kk})$.
Instead we have:
$$(a\cdot b)_{kk} = \sum_{l=1}^n a_{kl}b_{lk}$$
(Thinking)
 
  • #13
I like Serena said:
I think we need $\phi_k$ everywhere instead of $\phi$.

And generally for matrices $((a\cdot b)_{kk})\ne(a_{kk})\cdot (b_{kk})$.
Instead we have:
$$(a\cdot b)_{kk} = \sum_{l=1}^n a_{kl}b_{lk}$$
(Thinking)

Ah ok...

$$\phi_k ((a +b)_{ij})=(a+b)_{kk}=a_{kk}+b_{kk}=\phi_k (a_{ij})+\phi_k (b_{ij})$$
This property is correct, isn't it? (Wondering)

$$\phi_k ((a\cdot b)_{ij}))=(a\cdot b)_{kk}=\sum_{l=1}^n a_{kl}b_{lk}=\sum_{l=1}^{k-1} a_{kl}b_{lk}+a_{kk}b_{kk}\sum_{l=k+1}^n a_{kl}b_{lk}=\sum_{l=1}^{k-1} 0b_{lk}+a_{kk}b_{kk}+\sum_{l=k}^n a_{kl}0=a_{kk}b_{kk}=\phi_k (a_{ij})\cdot \phi_k (b_{ij})$$

Is this correct now? (Wondering)
 
  • #14
There's a plus missing in the middle, next to $a_{kk}b_{kk}$, but other than that it looks fine! (Happy)
 
  • #15
I like Serena said:
There's a plus missing in the middle, next to $a_{kk}b_{kk}$, but other than that it looks fine! (Happy)

Oh yes... (Blush)

Thank you very much! (Happy)
 

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