Prove a $n times n $ matrix has rank 3












11












$begingroup$


I have been examining a problem dealing with finding the rank of a $n times n $ matrix $M$ as follows:



begin{bmatrix}
0&1&4&9&16&cdots &(n-1)^2\
1&0&1&4&9&cdots&(n-2)^2\
4&1&0&1&4&cdots&(n-3)^2\
9&4&1&0&1&cdots&(n-4)^2\
16&9&4&1&0&cdots&(n-5)^2\
vdots&vdots&vdots&vdots&vdots&ddots&vdots\
(n-1)^2&(n-2)^2&cdots&(n-4)^2&cdots&cdots&0
end{bmatrix}



That is, the matrix with $k^2$ on its $k^{th}$ super and subdiagonals.
One of my colleagues claims the matrix is of rank 3.
I have tried to factor this matrix somehow, but have not really gotten very far.



Can we develop a proof of the claim that the rank of $M$ is indeed 3 (assume that $ngeq3$).






linear-algebra matrices







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    If you down vote, at least indicate why. What do you find objectionable about the question? That way I can fix it.
    $endgroup$
    – Vladhagen
    Feb 14 '14 at 20:46
















11












$begingroup$


I have been examining a problem dealing with finding the rank of a $n times n $ matrix $M$ as follows:



begin{bmatrix}
0&1&4&9&16&cdots &(n-1)^2\
1&0&1&4&9&cdots&(n-2)^2\
4&1&0&1&4&cdots&(n-3)^2\
9&4&1&0&1&cdots&(n-4)^2\
16&9&4&1&0&cdots&(n-5)^2\
vdots&vdots&vdots&vdots&vdots&ddots&vdots\
(n-1)^2&(n-2)^2&cdots&(n-4)^2&cdots&cdots&0
end{bmatrix}



That is, the matrix with $k^2$ on its $k^{th}$ super and subdiagonals.
One of my colleagues claims the matrix is of rank 3.
I have tried to factor this matrix somehow, but have not really gotten very far.



Can we develop a proof of the claim that the rank of $M$ is indeed 3 (assume that $ngeq3$).






linear-algebra matrices







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    If you down vote, at least indicate why. What do you find objectionable about the question? That way I can fix it.
    $endgroup$
    – Vladhagen
    Feb 14 '14 at 20:46














11












11








11


2



$begingroup$


I have been examining a problem dealing with finding the rank of a $n times n $ matrix $M$ as follows:



begin{bmatrix}
0&1&4&9&16&cdots &(n-1)^2\
1&0&1&4&9&cdots&(n-2)^2\
4&1&0&1&4&cdots&(n-3)^2\
9&4&1&0&1&cdots&(n-4)^2\
16&9&4&1&0&cdots&(n-5)^2\
vdots&vdots&vdots&vdots&vdots&ddots&vdots\
(n-1)^2&(n-2)^2&cdots&(n-4)^2&cdots&cdots&0
end{bmatrix}



That is, the matrix with $k^2$ on its $k^{th}$ super and subdiagonals.
One of my colleagues claims the matrix is of rank 3.
I have tried to factor this matrix somehow, but have not really gotten very far.



Can we develop a proof of the claim that the rank of $M$ is indeed 3 (assume that $ngeq3$).






linear-algebra matrices







share|cite|improve this question









$endgroup$




I have been examining a problem dealing with finding the rank of a $n times n $ matrix $M$ as follows:



begin{bmatrix}
0&1&4&9&16&cdots &(n-1)^2\
1&0&1&4&9&cdots&(n-2)^2\
4&1&0&1&4&cdots&(n-3)^2\
9&4&1&0&1&cdots&(n-4)^2\
16&9&4&1&0&cdots&(n-5)^2\
vdots&vdots&vdots&vdots&vdots&ddots&vdots\
(n-1)^2&(n-2)^2&cdots&(n-4)^2&cdots&cdots&0
end{bmatrix}



That is, the matrix with $k^2$ on its $k^{th}$ super and subdiagonals.
One of my colleagues claims the matrix is of rank 3.
I have tried to factor this matrix somehow, but have not really gotten very far.



Can we develop a proof of the claim that the rank of $M$ is indeed 3 (assume that $ngeq3$).






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Feb 14 '14 at 19:39









VladhagenVladhagen

3,27252134




3,27252134








  • 5




    $begingroup$
    If you down vote, at least indicate why. What do you find objectionable about the question? That way I can fix it.
    $endgroup$
    – Vladhagen
    Feb 14 '14 at 20:46














  • 5




    $begingroup$
    If you down vote, at least indicate why. What do you find objectionable about the question? That way I can fix it.
    $endgroup$
    – Vladhagen
    Feb 14 '14 at 20:46








5




5




$begingroup$
If you down vote, at least indicate why. What do you find objectionable about the question? That way I can fix it.
$endgroup$
– Vladhagen
Feb 14 '14 at 20:46




$begingroup$
If you down vote, at least indicate why. What do you find objectionable about the question? That way I can fix it.
$endgroup$
– Vladhagen
Feb 14 '14 at 20:46










1 Answer
1






active

oldest

votes


















7












$begingroup$

When $nge4$, let $P=pmatrix{1&-1\ &ddots&ddots\ &&ddots&-1\ &&&1}^3
=pmatrix{1&-3&3&-1\
&ddots&ddots&ddots&ddots\
&&ddots&ddots&ddots&-1\
&&&ddots&ddots&3\
&&&&ddots&-3\
&&&&&1}$. Then $P^TMP=pmatrix{0&1&1\ 1&-6&1\ 1&1&0}oplus 0_{(n-3)times(n-3)}$. Hence $M$ has rank 3.





Remark.

Knowing that $M$ has rank $3$ allows us to find the eigenvalues of $M$ relatively easily. One can readily verify that $(1-n,3-n,ldots,n-3,n-1)^T$ is an eigenvector of $M$ corresponding to the eigenvalue $-frac{n(n^2-1)}{6}$. Since $M$ has zero trace, if we also know $operatorname{tr}(M^2)$, the other two nonzero eigenvalues can be found. It turns out that
$$
operatorname{tr}(M^2)=2sum_{i=1}^{n-1} i(n-i)^4
=frac1{30}n^2(n^2-1)(2n^2-3)
$$
(see also OEIS A101089) and the other two nonzero eigenvalues of $M$ are
$$
frac{n(n^2-1)}{12}pmfrac{n}{2}sqrt{frac{(n^2-1)(3n^2-7)}{60}}.
$$






share|cite|improve this answer











$endgroup$









  • 5




    $begingroup$
    How on Earth did you come up with this? (+1 BTW)
    $endgroup$
    – John
    Feb 14 '14 at 20:50








  • 4




    $begingroup$
    @John The answer was obtained by observing that the third-order Newton difference of the sequence $0,1,4,9,16,ldots$ is zero.
    $endgroup$
    – user1551
    Feb 14 '14 at 20:54










  • $begingroup$
    A very elegant solution.
    $endgroup$
    – John
    Feb 14 '14 at 21:02






  • 3




    $begingroup$
    I see no way that someone could top this. Most excellent.
    $endgroup$
    – Vladhagen
    Feb 14 '14 at 21:18











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

When $nge4$, let $P=pmatrix{1&-1\ &ddots&ddots\ &&ddots&-1\ &&&1}^3
=pmatrix{1&-3&3&-1\
&ddots&ddots&ddots&ddots\
&&ddots&ddots&ddots&-1\
&&&ddots&ddots&3\
&&&&ddots&-3\
&&&&&1}$. Then $P^TMP=pmatrix{0&1&1\ 1&-6&1\ 1&1&0}oplus 0_{(n-3)times(n-3)}$. Hence $M$ has rank 3.





Remark.

Knowing that $M$ has rank $3$ allows us to find the eigenvalues of $M$ relatively easily. One can readily verify that $(1-n,3-n,ldots,n-3,n-1)^T$ is an eigenvector of $M$ corresponding to the eigenvalue $-frac{n(n^2-1)}{6}$. Since $M$ has zero trace, if we also know $operatorname{tr}(M^2)$, the other two nonzero eigenvalues can be found. It turns out that
$$
operatorname{tr}(M^2)=2sum_{i=1}^{n-1} i(n-i)^4
=frac1{30}n^2(n^2-1)(2n^2-3)
$$
(see also OEIS A101089) and the other two nonzero eigenvalues of $M$ are
$$
frac{n(n^2-1)}{12}pmfrac{n}{2}sqrt{frac{(n^2-1)(3n^2-7)}{60}}.
$$






share|cite|improve this answer











$endgroup$









  • 5




    $begingroup$
    How on Earth did you come up with this? (+1 BTW)
    $endgroup$
    – John
    Feb 14 '14 at 20:50








  • 4




    $begingroup$
    @John The answer was obtained by observing that the third-order Newton difference of the sequence $0,1,4,9,16,ldots$ is zero.
    $endgroup$
    – user1551
    Feb 14 '14 at 20:54










  • $begingroup$
    A very elegant solution.
    $endgroup$
    – John
    Feb 14 '14 at 21:02






  • 3




    $begingroup$
    I see no way that someone could top this. Most excellent.
    $endgroup$
    – Vladhagen
    Feb 14 '14 at 21:18
















7












$begingroup$

When $nge4$, let $P=pmatrix{1&-1\ &ddots&ddots\ &&ddots&-1\ &&&1}^3
=pmatrix{1&-3&3&-1\
&ddots&ddots&ddots&ddots\
&&ddots&ddots&ddots&-1\
&&&ddots&ddots&3\
&&&&ddots&-3\
&&&&&1}$. Then $P^TMP=pmatrix{0&1&1\ 1&-6&1\ 1&1&0}oplus 0_{(n-3)times(n-3)}$. Hence $M$ has rank 3.





Remark.

Knowing that $M$ has rank $3$ allows us to find the eigenvalues of $M$ relatively easily. One can readily verify that $(1-n,3-n,ldots,n-3,n-1)^T$ is an eigenvector of $M$ corresponding to the eigenvalue $-frac{n(n^2-1)}{6}$. Since $M$ has zero trace, if we also know $operatorname{tr}(M^2)$, the other two nonzero eigenvalues can be found. It turns out that
$$
operatorname{tr}(M^2)=2sum_{i=1}^{n-1} i(n-i)^4
=frac1{30}n^2(n^2-1)(2n^2-3)
$$
(see also OEIS A101089) and the other two nonzero eigenvalues of $M$ are
$$
frac{n(n^2-1)}{12}pmfrac{n}{2}sqrt{frac{(n^2-1)(3n^2-7)}{60}}.
$$






share|cite|improve this answer











$endgroup$









  • 5




    $begingroup$
    How on Earth did you come up with this? (+1 BTW)
    $endgroup$
    – John
    Feb 14 '14 at 20:50








  • 4




    $begingroup$
    @John The answer was obtained by observing that the third-order Newton difference of the sequence $0,1,4,9,16,ldots$ is zero.
    $endgroup$
    – user1551
    Feb 14 '14 at 20:54










  • $begingroup$
    A very elegant solution.
    $endgroup$
    – John
    Feb 14 '14 at 21:02






  • 3




    $begingroup$
    I see no way that someone could top this. Most excellent.
    $endgroup$
    – Vladhagen
    Feb 14 '14 at 21:18














7












7








7





$begingroup$

When $nge4$, let $P=pmatrix{1&-1\ &ddots&ddots\ &&ddots&-1\ &&&1}^3
=pmatrix{1&-3&3&-1\
&ddots&ddots&ddots&ddots\
&&ddots&ddots&ddots&-1\
&&&ddots&ddots&3\
&&&&ddots&-3\
&&&&&1}$. Then $P^TMP=pmatrix{0&1&1\ 1&-6&1\ 1&1&0}oplus 0_{(n-3)times(n-3)}$. Hence $M$ has rank 3.





Remark.

Knowing that $M$ has rank $3$ allows us to find the eigenvalues of $M$ relatively easily. One can readily verify that $(1-n,3-n,ldots,n-3,n-1)^T$ is an eigenvector of $M$ corresponding to the eigenvalue $-frac{n(n^2-1)}{6}$. Since $M$ has zero trace, if we also know $operatorname{tr}(M^2)$, the other two nonzero eigenvalues can be found. It turns out that
$$
operatorname{tr}(M^2)=2sum_{i=1}^{n-1} i(n-i)^4
=frac1{30}n^2(n^2-1)(2n^2-3)
$$
(see also OEIS A101089) and the other two nonzero eigenvalues of $M$ are
$$
frac{n(n^2-1)}{12}pmfrac{n}{2}sqrt{frac{(n^2-1)(3n^2-7)}{60}}.
$$






share|cite|improve this answer











$endgroup$



When $nge4$, let $P=pmatrix{1&-1\ &ddots&ddots\ &&ddots&-1\ &&&1}^3
=pmatrix{1&-3&3&-1\
&ddots&ddots&ddots&ddots\
&&ddots&ddots&ddots&-1\
&&&ddots&ddots&3\
&&&&ddots&-3\
&&&&&1}$. Then $P^TMP=pmatrix{0&1&1\ 1&-6&1\ 1&1&0}oplus 0_{(n-3)times(n-3)}$. Hence $M$ has rank 3.





Remark.

Knowing that $M$ has rank $3$ allows us to find the eigenvalues of $M$ relatively easily. One can readily verify that $(1-n,3-n,ldots,n-3,n-1)^T$ is an eigenvector of $M$ corresponding to the eigenvalue $-frac{n(n^2-1)}{6}$. Since $M$ has zero trace, if we also know $operatorname{tr}(M^2)$, the other two nonzero eigenvalues can be found. It turns out that
$$
operatorname{tr}(M^2)=2sum_{i=1}^{n-1} i(n-i)^4
=frac1{30}n^2(n^2-1)(2n^2-3)
$$
(see also OEIS A101089) and the other two nonzero eigenvalues of $M$ are
$$
frac{n(n^2-1)}{12}pmfrac{n}{2}sqrt{frac{(n^2-1)(3n^2-7)}{60}}.
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 22 '18 at 8:14

























answered Feb 14 '14 at 20:16









user1551user1551

72.8k566127




72.8k566127








  • 5




    $begingroup$
    How on Earth did you come up with this? (+1 BTW)
    $endgroup$
    – John
    Feb 14 '14 at 20:50








  • 4




    $begingroup$
    @John The answer was obtained by observing that the third-order Newton difference of the sequence $0,1,4,9,16,ldots$ is zero.
    $endgroup$
    – user1551
    Feb 14 '14 at 20:54










  • $begingroup$
    A very elegant solution.
    $endgroup$
    – John
    Feb 14 '14 at 21:02






  • 3




    $begingroup$
    I see no way that someone could top this. Most excellent.
    $endgroup$
    – Vladhagen
    Feb 14 '14 at 21:18














  • 5




    $begingroup$
    How on Earth did you come up with this? (+1 BTW)
    $endgroup$
    – John
    Feb 14 '14 at 20:50








  • 4




    $begingroup$
    @John The answer was obtained by observing that the third-order Newton difference of the sequence $0,1,4,9,16,ldots$ is zero.
    $endgroup$
    – user1551
    Feb 14 '14 at 20:54










  • $begingroup$
    A very elegant solution.
    $endgroup$
    – John
    Feb 14 '14 at 21:02






  • 3




    $begingroup$
    I see no way that someone could top this. Most excellent.
    $endgroup$
    – Vladhagen
    Feb 14 '14 at 21:18








5




5




$begingroup$
How on Earth did you come up with this? (+1 BTW)
$endgroup$
– John
Feb 14 '14 at 20:50






$begingroup$
How on Earth did you come up with this? (+1 BTW)
$endgroup$
– John
Feb 14 '14 at 20:50






4




4




$begingroup$
@John The answer was obtained by observing that the third-order Newton difference of the sequence $0,1,4,9,16,ldots$ is zero.
$endgroup$
– user1551
Feb 14 '14 at 20:54




$begingroup$
@John The answer was obtained by observing that the third-order Newton difference of the sequence $0,1,4,9,16,ldots$ is zero.
$endgroup$
– user1551
Feb 14 '14 at 20:54












$begingroup$
A very elegant solution.
$endgroup$
– John
Feb 14 '14 at 21:02




$begingroup$
A very elegant solution.
$endgroup$
– John
Feb 14 '14 at 21:02




3




3




$begingroup$
I see no way that someone could top this. Most excellent.
$endgroup$
– Vladhagen
Feb 14 '14 at 21:18




$begingroup$
I see no way that someone could top this. Most excellent.
$endgroup$
– Vladhagen
Feb 14 '14 at 21:18


















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