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

[mathmari]

Hey!

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)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

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)

If so then we only need to verify if it is closed for addition and multiplication, and if the additive identity is an element.

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)

 

[I like Serena]

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!
Such a ring is also named rng instead to distinguish it.
See wikpedia.

 

[mathmari]

Because your ring doesn't have a multiplicative identity!
Such a ring is also named rng instead to distinguish it.
See wikpedia.

Why doesn't it contain the identity matrix? (Wondering)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

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.

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

There's a plus missing in the middle, next to $a_{kk}b_{kk}$, but other than that it looks fine! (Happy)

 

[mathmari]

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)