What is the order of the matrix m

[YungCalculator]

What is the order of the matrix m :=

\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1\\ 0 & 0 & 1 \end{pmatrix}

Multiplying the matrix by itself never yields the identity matrix, so in my opinion the order of the matrix m :=must be infinite. Is this correct?

 

[topsquark]

But you need to be able to prove it. What does the matrix [math]\left ( \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} \right )^n[/math] look like? What happens each time we raise the power by 1?

-Dan

 

[pka]

What is the order of the matrix m :=\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1\\ 0 & 0 & 1 \end{pmatrix}

SEE HERE

 

[Steven G]

Induction??

 

[YungCalculator]

But you need to be able to prove it. What does the matrix [math]\left ( \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} \right )^n[/math] look like? What happens each time we raise the power by 1?

-Dan

the elements in the top right corner and center right will be equal to [math]n[/math] for instance [math]\left ( \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} \right )^5 = \left ( \begin{matrix} 1 & 0 & 5 \\ 0 & 1 & 5 \\ 0 & 0 & 1 \end{matrix} \right )[/math].

but to be honest I'm not exactly sure how this helps me because in my opinion (besides taking the [math]m^0[/math]) no multiple of the matrix gives me the identity matrix. So I wouldn't know what an finite order must be then, so that's why I guess that the order is infinite.

 

[lex]

Induction??

Yes - the nice thing about induction is all you need to do initially is 'guess' the answer, then prove it.
If you do a couple of examples, [MATH]A^2[/MATH], [MATH]A^3[/MATH] it would appear that:

[MATH]\left ( \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} \right )^n[/MATH]
is

[MATH]\left ( \begin{matrix} 1 & 0 & n \\ 0 & 1 & n \\ 0 & 0 & 1 \end{matrix} \right )[/MATH]
which is very easily proven by induction.

 

[Deleted member 4993]

What is the order of the matrix m :=

\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1\\ 0 & 0 & 1 \end{pmatrix}

Multiplying the matrix by itself never yields the identity matrix, so in my opinion the order of the matrix m :=must be infinite. Is this correct?

Please define:

The order of a matrix

- According to YOUR textbook.

 

[lex]

Please define:
The order of a matrix

Yes a good point; the context is important. As a matrix, the order is 3 [MATH]\times[/MATH] 3. It would seem it is being considered as an element of a group however.

 

[YungCalculator]

Yes a good point; the context is important. As a matrix, the order is 3 [MATH]\times[/MATH] 3. It would seem it is being considered as an element of a group however.

Apologies, you are absolutely right, I should have made it clear that this is a question in the part group theory of my abstract algebra course

 

[Deleted member 4993]

Apologies, you are absolutely right, I should have made it clear that this is a question in the part group theory of my abstract algebra course

According to that class & TEXTBOOK -

what is the definition of "order" of the "matrix m"?​

 

[lex]

You can see the result directly too. As the first two columns of A are the first two columns of the identity matrix, multiplying a matrix by A keeps the first two columns the same and then it adds the three columns of the matrix to get the final column:

(a b c) A = (a b (a+b+c)) (*)

using (*)
[MATH]A^2[/MATH]=(a b (a+b+c))[MATH] \hspace28ex \boldsymbol{a}=\left ( \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right )\,\boldsymbol{b}=\left ( \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right )\,\boldsymbol{c}=\left ( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right )[/MATH][MATH]A^3[/MATH]=(a b 2(a+b)+c)
[MATH]A^4[/MATH]=(a b 3(a+b)+c)
[MATH]A^n[/MATH]=(a b (n-1)(a+b)+c)

[MATH]=\left ( \begin{matrix} 1 & 0 & n \\ 0 & 1 & n \\ 0 & 0 & 1 \end{matrix} \right )[/MATH]
which can then be proved by induction.

 

[Steven G]

the elements in the top right corner and center right will be equal to [math]n[/math] for instance [math]\left ( \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} \right )^5 = \left ( \begin{matrix} 1 & 0 & 5 \\ 0 & 1 & 5 \\ 0 & 0 & 1 \end{matrix} \right )[/math].

but to be honest I'm not exactly sure how this helps me because in my opinion (besides taking the [math]m^0[/math]) no multiple of the matrix gives me the identity matrix. So I wouldn't know what an finite order must be then, so that's why I guess that the order is infinite.

You are correct that the order of this element, that is matrix, is infinite. But as topsquark stated, you need to prove this. I suggested that you prove this by induction as you seem to know the pattern of this matrix.

You just can't say that the order is infinite or even 17--you must prove it!