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Determinant as expression

Petrus

Well-known member
Feb 21, 2013
739
Matrice \(\displaystyle c_n\) is a 2nxn2 matrice and is given by \(\displaystyle (\delta_{ij}+2\delta_{i,2n-j+1})_{ij}\) determine the expression of \(\displaystyle |c_n|\)
progress:
I have been drawing the matrice \(\displaystyle n=1,2,3\) and calculate the determinant.

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

Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Matrice \(\displaystyle c_n\) is a 2x2 matrice and is given by \(\displaystyle (\delta_{ij}+2\delta_{i,2n-j+1})_{ij}\) determine the expression of \(\displaystyle |c_n|\)
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

Well-known member
Feb 21, 2013
739
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 \(\displaystyle c_n\) 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 alot when I do it.. I mean what I am suposed to see?

\(\displaystyle |\pi\rangle\)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Thanks for pointing My misstake!
I am trying with \(\displaystyle c_n\) 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 alot when I do it.. I mean what I am suposed to see?

\(\displaystyle |\pi\rangle\)
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

Well-known member
Feb 21, 2013
739
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
\(\displaystyle c_n= 1•(n-1)-2(n-1)\) what I am suposed to see? I am just having hard to understand. Or is this wrong?

Regards,
\(\displaystyle |\pi\rangle\)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Sorry to bother you but after doing it once I got
\(\displaystyle c_n= 1•(n-1)-2(n-1)\) what I am suposed to see? I am just having hard to understand. Or is this wrong?

Regards,
\(\displaystyle |\pi\rangle\)
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

Well-known member
Feb 21, 2013
739
$\det(C_2) = 1*\det(C_1) - 4*\det(C_1) = -3\det(C_1).$
Is that -4 suposed to be -2? I think I start to understand

Regards,
\(\displaystyle |\pi\rangle\)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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

Well-known member
Feb 21, 2013
739
Hello!
After taking a brake from this problem and finally got back to it things start to make sense!
so we got for \(\displaystyle |c_6|=-3|c_4|\) and \(\displaystyle |c_4=9=-3•-3\) Ahh we see a pattern!
\(\displaystyle |c_n|=-3^n\) 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,
\(\displaystyle |\pi\rangle\)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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

Well-known member
Feb 21, 2013
739
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 \(\displaystyle n-1\) I mean normaly when I proof it is \(\displaystyle n+1\)
and thank you both for taking your time!
Regrds,
\(\displaystyle |\pi\rangle\)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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.