MHB What is the expression for |c_n| in terms of the given matrice?

[Petrus]

Matrice $$c_n$$ is a 2nxn2 matrice and is given by $$(\delta_{ij}+2\delta_{i,2n-j+1})_{ij}$$ determine the expression of $$|c_n|$$
progress:
I have been drawing the matrice $$n=1,2,3$$ and calculate the determinant.

As you see there is a pattern with the matrice and I would like to have tips how i write $$|c_n|$$ as a expression cause I got no idé

Regards,
$$|\pi\rangle$$

 

[Opalg]

Matrice $$c_n$$ is a 2x2 matrice and is given by $$(\delta_{ij}+2\delta_{i,2n-j+1})_{ij}$$ determine the expression of $$|c_n|$$

First, I assume that $C_n$ is meant to be a $2n\times2n$ matrix (not $2\times2$). Next, you have made a mistake with the matrix $C_1.$ It should be $\begin{bmatrix}1&2 \\ 2&1\end{bmatrix}$, with determinant $-3.$ So $C_1,\,C_2,\,C_3$ have determinants $-3,\,9,\,-27.$ That ought to suggest a pattern.

To evaluate $C_n$ in general, expand along the top row, and I think that you will be able to deduce a recurrence formula relating $C_n$ to $C_{n-1}$.

 

[Petrus]

First, I assume that $C_n$ is meant to be a $2n\times2n$ matrix (not $2\times2$). Next, you have made a mistake with the matrix $C_1.$ It should be $\begin{bmatrix}1&2 \\ 2&1\end{bmatrix}$, with determinant $-3.$ So $C_1,\,C_2,\,C_3$ have determinants $-3,\,9,\,-27.$ That ought to suggest a pattern.

To evaluate $C_n$ in general, expand along the top row, and I think that you will be able to deduce a recurrence formula relating $C_n$ to $C_{n-1}$.

Thanks for pointing My misstake!
I am trying with $$c_n$$ I just Dont get the point with expand along the top row.. It is n big then n-1, n-2..which Dont tell me a lot when I do it.. I mean what I am suposed to see?

$$|\pi\rangle$$

 

[Opalg]

Thanks for pointing My misstake!
I am trying with $$c_n$$ I just Dont get the point with expand along the top row.. It is n big then n-1, n-2..which Dont tell me a lot when I do it.. I mean what I am suposed to see?

$$|\pi\rangle$$

Most of the entries in the top row are zero. The only ones you have to worry about are the first and last entries. For each of them, the corresponding cofactor has a column with only one nonzero element; and the cofactor of that element is a copy of $C_{n-1}.$

 

[Petrus]

Most of the entries in the top row are zero. The only ones you have to worry about are the first and last entries. For each of them, the corresponding cofactor has a column with only one nonzero element; and the cofactor of that element is a copy of $C_{n-1}.$

Sorry to bother you but after doing it once I got
$$c_n= 1•(n-1)-2(n-1)$$ what I am suposed to see? I am just having hard to understand. Or is this wrong?

Regards,
$$|\pi\rangle$$

 

[Opalg]

Sorry to bother you but after doing it once I got
$$c_n= 1•(n-1)-2(n-1)$$ what I am suposed to see? I am just having hard to understand. Or is this wrong?

Regards,
$$|\pi\rangle$$

When you expand the matrix $A = (a_{ij})$ along the top row, the formula is $\det(A) = a_{11}B_{11} - a_{12}B_{12} +\ldots + (-1)^na_{1n}B_{1n},$ where $B_{ij}$ is the cofactor of $a_{ij}.$ What that means is that $B_{ij}$ is the determinant of the $(n-1)\times(n-1)$-matrix obtained by deleting row $i$ and column $j$ from $A$. In the case of your matrix $C_2$, that tells you that $$\begin{vmatrix}1&0&0&2 \\ 0&1&2&0 \\ 0&2&1&0 \\ 2&0&0&1 \end{vmatrix} = 1*\begin{vmatrix}1&2&0 \\ 2&1&0 \\ 0&0&1 \end{vmatrix} - 2*\begin{vmatrix}0&1&2 \\ 0&2&1 \\ 2&0&0 \end{vmatrix}.$$ If you expand each of those $3\times3$-determinants along the bottom row, you see that $\det(C_2) = 1*\det(C_1) - 4*\det(C_1) = -3\det(C_1).$

 

[Petrus]

$\det(C_2) = 1*\det(C_1) - 4*\det(C_1) = -3\det(C_1).$

Is that -4 suposed to be -2? I think I start to understand

Regards,
$$|\pi\rangle$$

 

[Deveno]

No, it's supposed to be a -4. We have:

$\begin{vmatrix}1&2&0\\2&1&0\\0&0&1 \end{vmatrix} = 0 \ast (-1)^{3+1}\ast \begin{vmatrix}2&0\\1&0 \end{vmatrix} + 0 \ast (-1)^{3+2} \ast \begin{vmatrix}1&0\\2&0 \end{vmatrix} + 1 \ast (-1)^{3+3} \ast \begin{vmatrix}1&2\\2&1 \end{vmatrix}$

$= 0 + 0 + 1 \ast \det(C_1)$

While:

$\begin{vmatrix}0&1&2\\0&2&1\\2&0&0 \end{vmatrix} = 2 \ast (-1)^{3+1}\ast \begin{vmatrix}1&2\\2&1 \end{vmatrix} + 0 \ast (-1)^{3+2} \ast \begin{vmatrix}0&2\\0&1 \end{vmatrix} + 0 \ast (-1)^{3+3} \ast \begin{vmatrix}0&1\\0&2 \end{vmatrix}$

$= 2 \ast 1 \ast \det(C_1) + 0 + 0$

So that:

$1 \ast \begin{vmatrix}1&2&0\\2&1&0\\0&0&1 \end{vmatrix} - 2 \ast \begin{vmatrix}0&1&2\\0&2&1\\2&0&0 \end{vmatrix}$

$= \det(C_1) - 2 \ast 2 \ast \det(C_1) = -3\det(C_1) = (-3)(-3) = 9$

 

[Petrus]

Hello!
After taking a brake from this problem and finally got back to it things start to make sense!
so we got for $$|c_6|=-3|c_4|$$ and $$|c_4=9=-3•-3$$ Ahh we see a pattern!
$$|c_n|=-3^n$$ ok now we have to do a induction proof for this and does anyone got tips well I pretty much Dont know how this induction proof Will work for this

Regards,
$$|\pi\rangle$$

 

[Deveno]

First, we must establish the base case for $n = 1$. Although this is typically very easy, it is most important.

Now:

$|C_1| = \begin{vmatrix}1&2\\2&1 \end{vmatrix} = (1)(1) - (2)(2) = 1 - 4 = -3 = (-3)^1$

Assume that $|C_{n-1}| = (-3)^{n-1}$ (this is our inductive step, we seek to prove given this that $|C_n| = (-3)^n$).

Now to do this, we are going to have to go back "two sizes" (from $2n \times 2n$ to $2(n-1) \times 2(n-1)$).

We have (expanding by minors along the first row):

$|C_n| = 1 \ast |C_{1|1}| - 2 \ast |C_{1|2n}|$

(by $C_{i|j}$ I mean the matrix formed by deleting row i and column j from $C_n$).

since all the other cofactors are 0.

Now in $C_{1|1}$, if we expand THIS determinant by minors along its LAST row (which is all 0's except for the final 1), we get:

$|C_{1|1}| = (-1)^{(2n-1)+(2n-1)} \ast |C_{n-1}| = |C_{n-1}|$.

Similarly, if we expand $|C_{1|2n}|$ by minors along its last row (which is all 0's except for the initial 2), we get:

$|C_{1|2n}| = (-1)^{1+(2n-1)} \ast 2 \ast |C_{n-1}| = 2 \ast |C_{n-1}|$.

Therefore, $|C_n| = 1 \ast |C_{n-1}| - 2 \ast 2 \ast |C_{n-1}| = |C_{n-1}| - 4 \ast |C_{n-1}| = -3 \ast |C_{n-1}|$.

Since by our induction hypothesis:

$|C_{n-1}| = (-3)^{n-1}$,

$|C_n| = (-3)(-3)^{n-1} = (-3)^{1+(n-1)} = (-3)^n$ QED.

 

[Petrus]

First, we must establish the base case for $n = 1$. Although this is typically very easy, it is most important.

Now:

$|C_1| = \begin{vmatrix}1&2\\2&1 \end{vmatrix} = (1)(1) - (2)(2) = 1 - 4 = -3 = (-3)^1$

Assume that $|C_{n-1}| = (-3)^{n-1}$ (this is our inductive step, we seek to prove given this that $|C_n| = (-3)^n$).

Now to do this, we are going to have to go back "two sizes" (from $2n \times 2n$ to $2(n-1) \times 2(n-1)$).

We have (expanding by minors along the first row):

$|C_n| = 1 \ast |C_{1|1}| - 2 \ast |C_{1|2n}|$

(by $C_{i|j}$ I mean the matrix formed by deleting row i and column j from $C_n$).

since all the other cofactors are 0.

Now in $C_{1|1}$, if we expand THIS determinant by minors along its LAST row (which is all 0's except for the final 1), we get:

$|C_{1|1}| = (-1)^{(2n-1)+(2n-1)} \ast |C_{n-1}| = |C_{n-1}|$.

Similarly, if we expand $|C_{1|2n}|$ by minors along its last row (which is all 0's except for the initial 2), we get:

$|C_{1|2n}| = (-1)^{1+(2n-1)} \ast 2 \ast |C_{n-1}| = 2 \ast |C_{n-1}|$.

Therefore, $|C_n| = 1 \ast |C_{n-1}| - 2 \ast 2 \ast |C_{n-1}| = |C_{n-1}| - 4 \ast |C_{n-1}| = -3 \ast |C_{n-1}|$.

Since by our induction hypothesis:

$|C_{n-1}| = (-3)^{n-1}$,

$|C_n| = (-3)(-3)^{n-1} = (-3)^{1+(n-1)} = (-3)^n$ QED.

Hello,
I understand all of the part without one( I want Also to say your explaining Was Really good and it is all logic) why $$n-1$$ I mean normaly when I proof it is $$n+1$$
and thank you both for taking your time!
Regrds,
$$|\pi\rangle$$

 

[Deveno]

You can do an inductive proof either way, it's largely a matter of style. Technically, it's more logically defensible to prove:

P(n₀) and [P(n) --> P(n+1)].

However, for:

P(k₀) and [P(k-1) --> P(k)]

to work out the same, all you have to do is take an initial value for "k" of k₀+1 (it doesn't really matter "which" natural number you start an inductive proof with, as long as it's true for SOME natural number, although if you start with P(17), for example, you may have to prove more "bases cases", or qualify your result).

In other words, the "first" n in my proof is 2, and I prove the base case for n - 1 = 1.

(it doesn't really make any difference if the "dummy variable" in an inductive proof is labelled "n" or "k" or "n-1" its just a LABEL...what IS important is that we prove that given it's true for some (variable) natural number, that in and of itself proves its true for the NEXT natural number).

Or, to put it another way, the sets:

{n in Z: n is in Z⁺}

{k-1 in Z: k-1 is in Z⁺}

are both the same sets (namely, N, the natural numbers (starting with 1)) even though n runs from 1 to infinity in the first set, and k runs from 2 to infinity in the second set.

If you wish to amend my proof replacing the "n-1" with "n" (and some of the consequent changes involving various subscripts and exponents) it's fine with me.