Mixing
Eigenvalues of a matrix with repeating pattern of entries
Clash Royale CLAN TAG#URR8PPP
up vote
10
down vote
favorite
I observed that if
$$A = beginbmatrix a & b \ c & d endbmatrix$$
with non-zero eigenvalues $alpha$ and $beta$, then
$$beginbmatrix A & A\ A & A endbmatrix$$
has eigenvalues $2 alpha$, $2 beta$, and $0$. Also,
$$beginbmatrix A & A & A \ A & A & A \ A & A & A endbmatrix$$
has eigenvalues $3 alpha$, $3 beta$, $0$. Therefore, my conjecture is that for some $r$, $A^[r]$ has eigenvalues $(r+1) alpha$, $(r+1) beta$, $0$.
Is it correct? Is there some theorems related to this? How about their eigenvectors? Can you please send me links that can help me with this kind of problem?
PS. This is my first time asking here. I am an undergrad math student. Please help me. Thank u so much.
linear-algebra matrices eigenvalues-eigenvectors
asked Jul 22 at 11:21
Diggie Cruz
514
 |Â
show 1 more comment
up vote
10
down vote
favorite
I observed that if
$$A = beginbmatrix a & b \ c & d endbmatrix$$
with non-zero eigenvalues $alpha$ and $beta$, then
$$beginbmatrix A & A\ A & A endbmatrix$$
has eigenvalues $2 alpha$, $2 beta$, and $0$. Also,
$$beginbmatrix A & A & A \ A & A & A \ A & A & A endbmatrix$$
has eigenvalues $3 alpha$, $3 beta$, $0$. Therefore, my conjecture is that for some $r$, $A^[r]$ has eigenvalues $(r+1) alpha$, $(r+1) beta$, $0$.
Is it correct? Is there some theorems related to this? How about their eigenvectors? Can you please send me links that can help me with this kind of problem?
PS. This is my first time asking here. I am an undergrad math student. Please help me. Thank u so much.
linear-algebra matrices eigenvalues-eigenvectors
asked Jul 22 at 11:21
Diggie Cruz
514
6
these are Kronecker products.
– Lord Shark the Unknown
Jul 22 at 11:23
I am not exactly sure about your notation: Do you consider a matrix which elements are other matrices or do you want the elements to be the determinant of the original matrix?
– mrtaurho
Jul 22 at 11:24
This looks correct. Some hints: Use the rank to determine the number of zero eigenvalues, and use repeated copies of eigenvectors for the nonzero eigenvectors.
– Michael Burr
Jul 22 at 11:27
Im sorry.. Well, I consider the matrix A as partition matrix of the bigger matrix A*, A**, ...
– Diggie Cruz
Jul 22 at 11:29
2
This is good mathematical thinking, up to the point where you say conclude. The proper word is conjecture. You conclude something when you have a proof, not just a hunch.
– Arnaud Mortier
Jul 22 at 11:30
 |Â
show 1 more comment
up vote
10
down vote
favorite
up vote
10
down vote
favorite
I observed that if
$$A = beginbmatrix a & b \ c & d endbmatrix$$
with non-zero eigenvalues $alpha$ and $beta$, then
$$beginbmatrix A & A\ A & A endbmatrix$$
has eigenvalues $2 alpha$, $2 beta$, and $0$. Also,
$$beginbmatrix A & A & A \ A & A & A \ A & A & A endbmatrix$$
has eigenvalues $3 alpha$, $3 beta$, $0$. Therefore, my conjecture is that for some $r$, $A^[r]$ has eigenvalues $(r+1) alpha$, $(r+1) beta$, $0$.
Is it correct? Is there some theorems related to this? How about their eigenvectors? Can you please send me links that can help me with this kind of problem?
PS. This is my first time asking here. I am an undergrad math student. Please help me. Thank u so much.
linear-algebra matrices eigenvalues-eigenvectors
asked Jul 22 at 11:21
Diggie Cruz
514
I observed that if
$$A = beginbmatrix a & b \ c & d endbmatrix$$
with non-zero eigenvalues $alpha$ and $beta$, then
$$beginbmatrix A & A\ A & A endbmatrix$$
has eigenvalues $2 alpha$, $2 beta$, and $0$. Also,
$$beginbmatrix A & A & A \ A & A & A \ A & A & A endbmatrix$$
has eigenvalues $3 alpha$, $3 beta$, $0$. Therefore, my conjecture is that for some $r$, $A^[r]$ has eigenvalues $(r+1) alpha$, $(r+1) beta$, $0$.
Is it correct? Is there some theorems related to this? How about their eigenvectors? Can you please send me links that can help me with this kind of problem?
PS. This is my first time asking here. I am an undergrad math student. Please help me. Thank u so much.
linear-algebra matrices eigenvalues-eigenvectors
asked Jul 22 at 11:21
Diggie Cruz
514
edited Jul 22 at 11:37
asked Jul 22 at 11:21
Diggie Cruz
514
asked Jul 22 at 11:21
Diggie Cruz
514
asked Jul 22 at 11:21
Diggie Cruz
514
514
6
these are Kronecker products.
– Lord Shark the Unknown
Jul 22 at 11:23
I am not exactly sure about your notation: Do you consider a matrix which elements are other matrices or do you want the elements to be the determinant of the original matrix?
– mrtaurho
Jul 22 at 11:24
This looks correct. Some hints: Use the rank to determine the number of zero eigenvalues, and use repeated copies of eigenvectors for the nonzero eigenvectors.
– Michael Burr
Jul 22 at 11:27
Im sorry.. Well, I consider the matrix A as partition matrix of the bigger matrix A*, A**, ...
– Diggie Cruz
Jul 22 at 11:29
2
This is good mathematical thinking, up to the point where you say conclude. The proper word is conjecture. You conclude something when you have a proof, not just a hunch.
– Arnaud Mortier
Jul 22 at 11:30
 |Â
show 1 more comment
6
these are Kronecker products.
– Lord Shark the Unknown
Jul 22 at 11:23
I am not exactly sure about your notation: Do you consider a matrix which elements are other matrices or do you want the elements to be the determinant of the original matrix?
– mrtaurho
Jul 22 at 11:24
This looks correct. Some hints: Use the rank to determine the number of zero eigenvalues, and use repeated copies of eigenvectors for the nonzero eigenvectors.
– Michael Burr
Jul 22 at 11:27
Im sorry.. Well, I consider the matrix A as partition matrix of the bigger matrix A*, A**, ...
– Diggie Cruz
Jul 22 at 11:29
2
This is good mathematical thinking, up to the point where you say conclude. The proper word is conjecture. You conclude something when you have a proof, not just a hunch.
– Arnaud Mortier
Jul 22 at 11:30
6
6
these are Kronecker products.
– Lord Shark the Unknown
Jul 22 at 11:23
these are Kronecker products.
– Lord Shark the Unknown
Jul 22 at 11:23
I am not exactly sure about your notation: Do you consider a matrix which elements are other matrices or do you want the elements to be the determinant of the original matrix?
– mrtaurho
Jul 22 at 11:24
I am not exactly sure about your notation: Do you consider a matrix which elements are other matrices or do you want the elements to be the determinant of the original matrix?
– mrtaurho
Jul 22 at 11:24
This looks correct. Some hints: Use the rank to determine the number of zero eigenvalues, and use repeated copies of eigenvectors for the nonzero eigenvectors.
– Michael Burr
Jul 22 at 11:27
This looks correct. Some hints: Use the rank to determine the number of zero eigenvalues, and use repeated copies of eigenvectors for the nonzero eigenvectors.
– Michael Burr
Jul 22 at 11:27
Im sorry.. Well, I consider the matrix A as partition matrix of the bigger matrix A*, A**, ...
– Diggie Cruz
Jul 22 at 11:29
Im sorry.. Well, I consider the matrix A as partition matrix of the bigger matrix A*, A**, ...
– Diggie Cruz
Jul 22 at 11:29
2
2
This is good mathematical thinking, up to the point where you say conclude. The proper word is conjecture. You conclude something when you have a proof, not just a hunch.
– Arnaud Mortier
Jul 22 at 11:30
This is good mathematical thinking, up to the point where you say conclude. The proper word is conjecture. You conclude something when you have a proof, not just a hunch.
– Arnaud Mortier
Jul 22 at 11:30
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
8
down vote
Let
$$beginbmatrix mathrm A & mathrm A & dots & mathrm A\ mathrm A & mathrm A & dots & mathrm A\ vdots & vdots & ddots & vdots\ mathrm A & mathrm A & dots & mathrm Aendbmatrix = 1_k 1_k^top otimes mathrm A$$
be a $k times k$ block matrix, where $otimes$ denotes the Kronecker product. Let the spectrum of $rm A$ be $alpha,beta$.
Since the eigenvalues of rank-$1$ matrix $1_k 1_k^top$ are $0$ (with multiplicity $k-1$) and $k$ (with multiplicity $1$), then the eigenvalues of block matrix $1_k 1_k^top otimes mathrm A$ are
- $0$ with multiplicity $2 k - 2$.
- $k alpha$ with multiplicity $1$.
- $k beta$ with multiplicity $1$.
answered Jul 22 at 11:44
Rodrigo de Azevedo
12.6k41751
wow you were really fast, i was about to write the same.
– mathreadler
Jul 22 at 12:27
add a comment |Â
up vote
7
down vote
Yes, your conjecture is correct. Suppose that $A$ is an $ntimes n$ matrix and that you consider the $(nm)times(nm)$ matrix
$$
B=beginbmatrixA&A&dots&A\
A&A&dots&A\
vdots&vdots&ddots&vdots\
A&A&dots&A
endbmatrix
$$
that is a block matrix consisting of $mtimes m$ copies of $A$.
We can determine the eigenvalues of this matrix by inspection. We first note that $operatornamerank(A)=operatornamerank(B)$. This follows from the fact $B$ has lots of copies of the same rows. In fact, if $i$ and $j$ have $iequiv jpmod n$, then the rows $i$ and $j$ are copies of each other. Putting this together, with simple row operations, we can row reduce $B$ to
$$
beginbmatrixA&A&dots&A\
0&0&dots&0\
vdots&vdots&ddots&vdots\
0&0&dots&0
endbmatrix
$$
Then, we observe that the ranks of $A$ and $B$ are the same. Moreover, this row reduction exposes that at least $n(m-1)$ eigenvalues of $B$ are $0$. (Normally, row operations mess up eigenvalues, but this doesn't happen with zero eigenvalues)
Now, let's deal with the eigenvalues from $A$. Suppose that $lambda$ is an eigenvalue from $A$ with eigenvector $v$. Now, consider the vector
$$
w=beginbmatrix
v\v\vdots\v
endbmatrix
$$
consisting of $m$ copies of $v$. Then
$$
Bw=beginbmatrixA&A&dots&A\
A&A&dots&A\
vdots&vdots&ddots&vdots\
A&A&dots&A
endbmatrixbeginbmatrix
v\v\vdots\v
endbmatrix
=beginbmatrix
Av+Av+dots+Av\
Av+Av+dots+Av\
vdotsqquadvdots\
Av+Av+dots+Av
endbmatrix,
$$
where each block row has $m$ copies of $Av$. In other words, this simplifies to
$$
beginbmatrix
mAv\
mAv\
vdots\
mAv
endbmatrix.
$$
Since $v$ is an eigenvector for $A$, we can further simplify to
$$
beginbmatrix
mlambda v\
mlambda v\
vdots\
mlambda v
endbmatrix=mlambda w.
$$
Therefore, $mlambda$ is an eigenvalue for $A$.
Now, you have to convince yourself that we haven't double counted, but I'll leave that for you. If you want to, try to find the eigenvectors for the zeros we found above and observe that the eigenvectors are in distinct spaces. Hint: if $iequiv jpmod n$, use the vector which is $1$ in row $i$, $-1$ in row $j$, and zero elsewhere.
answered Jul 22 at 11:45
Michael Burr
25.6k13262
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
Let
$$beginbmatrix mathrm A & mathrm A & dots & mathrm A\ mathrm A & mathrm A & dots & mathrm A\ vdots & vdots & ddots & vdots\ mathrm A & mathrm A & dots & mathrm Aendbmatrix = 1_k 1_k^top otimes mathrm A$$
be a $k times k$ block matrix, where $otimes$ denotes the Kronecker product. Let the spectrum of $rm A$ be $alpha,beta$.
Since the eigenvalues of rank-$1$ matrix $1_k 1_k^top$ are $0$ (with multiplicity $k-1$) and $k$ (with multiplicity $1$), then the eigenvalues of block matrix $1_k 1_k^top otimes mathrm A$ are
- $0$ with multiplicity $2 k - 2$.
- $k alpha$ with multiplicity $1$.
- $k beta$ with multiplicity $1$.
answered Jul 22 at 11:44
Rodrigo de Azevedo
12.6k41751
wow you were really fast, i was about to write the same.
– mathreadler
Jul 22 at 12:27
add a comment |Â
up vote
8
down vote
Let
$$beginbmatrix mathrm A & mathrm A & dots & mathrm A\ mathrm A & mathrm A & dots & mathrm A\ vdots & vdots & ddots & vdots\ mathrm A & mathrm A & dots & mathrm Aendbmatrix = 1_k 1_k^top otimes mathrm A$$
be a $k times k$ block matrix, where $otimes$ denotes the Kronecker product. Let the spectrum of $rm A$ be $alpha,beta$.
Since the eigenvalues of rank-$1$ matrix $1_k 1_k^top$ are $0$ (with multiplicity $k-1$) and $k$ (with multiplicity $1$), then the eigenvalues of block matrix $1_k 1_k^top otimes mathrm A$ are
- $0$ with multiplicity $2 k - 2$.
- $k alpha$ with multiplicity $1$.
- $k beta$ with multiplicity $1$.
answered Jul 22 at 11:44
Rodrigo de Azevedo
12.6k41751
wow you were really fast, i was about to write the same.
– mathreadler
Jul 22 at 12:27
add a comment |Â
up vote
8
down vote
up vote
8
down vote
Let
$$beginbmatrix mathrm A & mathrm A & dots & mathrm A\ mathrm A & mathrm A & dots & mathrm A\ vdots & vdots & ddots & vdots\ mathrm A & mathrm A & dots & mathrm Aendbmatrix = 1_k 1_k^top otimes mathrm A$$
be a $k times k$ block matrix, where $otimes$ denotes the Kronecker product. Let the spectrum of $rm A$ be $alpha,beta$.
Since the eigenvalues of rank-$1$ matrix $1_k 1_k^top$ are $0$ (with multiplicity $k-1$) and $k$ (with multiplicity $1$), then the eigenvalues of block matrix $1_k 1_k^top otimes mathrm A$ are
- $0$ with multiplicity $2 k - 2$.
- $k alpha$ with multiplicity $1$.
- $k beta$ with multiplicity $1$.
answered Jul 22 at 11:44
Rodrigo de Azevedo
12.6k41751
Let
$$beginbmatrix mathrm A & mathrm A & dots & mathrm A\ mathrm A & mathrm A & dots & mathrm A\ vdots & vdots & ddots & vdots\ mathrm A & mathrm A & dots & mathrm Aendbmatrix = 1_k 1_k^top otimes mathrm A$$
be a $k times k$ block matrix, where $otimes$ denotes the Kronecker product. Let the spectrum of $rm A$ be $alpha,beta$.
Since the eigenvalues of rank-$1$ matrix $1_k 1_k^top$ are $0$ (with multiplicity $k-1$) and $k$ (with multiplicity $1$), then the eigenvalues of block matrix $1_k 1_k^top otimes mathrm A$ are
- $0$ with multiplicity $2 k - 2$.
- $k alpha$ with multiplicity $1$.
- $k beta$ with multiplicity $1$.
answered Jul 22 at 11:44
Rodrigo de Azevedo
12.6k41751
answered Jul 22 at 11:44
Rodrigo de Azevedo
12.6k41751
answered Jul 22 at 11:44
Rodrigo de Azevedo
12.6k41751
answered Jul 22 at 11:44
Rodrigo de Azevedo
12.6k41751
12.6k41751
wow you were really fast, i was about to write the same.
– mathreadler
Jul 22 at 12:27
add a comment |Â
wow you were really fast, i was about to write the same.
– mathreadler
Jul 22 at 12:27
wow you were really fast, i was about to write the same.
– mathreadler
Jul 22 at 12:27
wow you were really fast, i was about to write the same.
– mathreadler
Jul 22 at 12:27
add a comment |Â
up vote
7
down vote
Yes, your conjecture is correct. Suppose that $A$ is an $ntimes n$ matrix and that you consider the $(nm)times(nm)$ matrix
$$
B=beginbmatrixA&A&dots&A\
A&A&dots&A\
vdots&vdots&ddots&vdots\
A&A&dots&A
endbmatrix
$$
that is a block matrix consisting of $mtimes m$ copies of $A$.
We can determine the eigenvalues of this matrix by inspection. We first note that $operatornamerank(A)=operatornamerank(B)$. This follows from the fact $B$ has lots of copies of the same rows. In fact, if $i$ and $j$ have $iequiv jpmod n$, then the rows $i$ and $j$ are copies of each other. Putting this together, with simple row operations, we can row reduce $B$ to
$$
beginbmatrixA&A&dots&A\
0&0&dots&0\
vdots&vdots&ddots&vdots\
0&0&dots&0
endbmatrix
$$
Then, we observe that the ranks of $A$ and $B$ are the same. Moreover, this row reduction exposes that at least $n(m-1)$ eigenvalues of $B$ are $0$. (Normally, row operations mess up eigenvalues, but this doesn't happen with zero eigenvalues)
Now, let's deal with the eigenvalues from $A$. Suppose that $lambda$ is an eigenvalue from $A$ with eigenvector $v$. Now, consider the vector
$$
w=beginbmatrix
v\v\vdots\v
endbmatrix
$$
consisting of $m$ copies of $v$. Then
$$
Bw=beginbmatrixA&A&dots&A\
A&A&dots&A\
vdots&vdots&ddots&vdots\
A&A&dots&A
endbmatrixbeginbmatrix
v\v\vdots\v
endbmatrix
=beginbmatrix
Av+Av+dots+Av\
Av+Av+dots+Av\
vdotsqquadvdots\
Av+Av+dots+Av
endbmatrix,
$$
where each block row has $m$ copies of $Av$. In other words, this simplifies to
$$
beginbmatrix
mAv\
mAv\
vdots\
mAv
endbmatrix.
$$
Since $v$ is an eigenvector for $A$, we can further simplify to
$$
beginbmatrix
mlambda v\
mlambda v\
vdots\
mlambda v
endbmatrix=mlambda w.
$$
Therefore, $mlambda$ is an eigenvalue for $A$.
Now, you have to convince yourself that we haven't double counted, but I'll leave that for you. If you want to, try to find the eigenvectors for the zeros we found above and observe that the eigenvectors are in distinct spaces. Hint: if $iequiv jpmod n$, use the vector which is $1$ in row $i$, $-1$ in row $j$, and zero elsewhere.
answered Jul 22 at 11:45
Michael Burr
25.6k13262
add a comment |Â
up vote
7
down vote
Yes, your conjecture is correct. Suppose that $A$ is an $ntimes n$ matrix and that you consider the $(nm)times(nm)$ matrix
$$
B=beginbmatrixA&A&dots&A\
A&A&dots&A\
vdots&vdots&ddots&vdots\
A&A&dots&A
endbmatrix
$$
that is a block matrix consisting of $mtimes m$ copies of $A$.
We can determine the eigenvalues of this matrix by inspection. We first note that $operatornamerank(A)=operatornamerank(B)$. This follows from the fact $B$ has lots of copies of the same rows. In fact, if $i$ and $j$ have $iequiv jpmod n$, then the rows $i$ and $j$ are copies of each other. Putting this together, with simple row operations, we can row reduce $B$ to
$$
beginbmatrixA&A&dots&A\
0&0&dots&0\
vdots&vdots&ddots&vdots\
0&0&dots&0
endbmatrix
$$
Then, we observe that the ranks of $A$ and $B$ are the same. Moreover, this row reduction exposes that at least $n(m-1)$ eigenvalues of $B$ are $0$. (Normally, row operations mess up eigenvalues, but this doesn't happen with zero eigenvalues)
Now, let's deal with the eigenvalues from $A$. Suppose that $lambda$ is an eigenvalue from $A$ with eigenvector $v$. Now, consider the vector
$$
w=beginbmatrix
v\v\vdots\v
endbmatrix
$$
consisting of $m$ copies of $v$. Then
$$
Bw=beginbmatrixA&A&dots&A\
A&A&dots&A\
vdots&vdots&ddots&vdots\
A&A&dots&A
endbmatrixbeginbmatrix
v\v\vdots\v
endbmatrix
=beginbmatrix
Av+Av+dots+Av\
Av+Av+dots+Av\
vdotsqquadvdots\
Av+Av+dots+Av
endbmatrix,
$$
where each block row has $m$ copies of $Av$. In other words, this simplifies to
$$
beginbmatrix
mAv\
mAv\
vdots\
mAv
endbmatrix.
$$
Since $v$ is an eigenvector for $A$, we can further simplify to
$$
beginbmatrix
mlambda v\
mlambda v\
vdots\
mlambda v
endbmatrix=mlambda w.
$$
Therefore, $mlambda$ is an eigenvalue for $A$.
Now, you have to convince yourself that we haven't double counted, but I'll leave that for you. If you want to, try to find the eigenvectors for the zeros we found above and observe that the eigenvectors are in distinct spaces. Hint: if $iequiv jpmod n$, use the vector which is $1$ in row $i$, $-1$ in row $j$, and zero elsewhere.
answered Jul 22 at 11:45
Michael Burr
25.6k13262
add a comment |Â
up vote
7
down vote
up vote
7
down vote
Yes, your conjecture is correct. Suppose that $A$ is an $ntimes n$ matrix and that you consider the $(nm)times(nm)$ matrix
$$
B=beginbmatrixA&A&dots&A\
A&A&dots&A\
vdots&vdots&ddots&vdots\
A&A&dots&A
endbmatrix
$$
that is a block matrix consisting of $mtimes m$ copies of $A$.
We can determine the eigenvalues of this matrix by inspection. We first note that $operatornamerank(A)=operatornamerank(B)$. This follows from the fact $B$ has lots of copies of the same rows. In fact, if $i$ and $j$ have $iequiv jpmod n$, then the rows $i$ and $j$ are copies of each other. Putting this together, with simple row operations, we can row reduce $B$ to
$$
beginbmatrixA&A&dots&A\
0&0&dots&0\
vdots&vdots&ddots&vdots\
0&0&dots&0
endbmatrix
$$
Then, we observe that the ranks of $A$ and $B$ are the same. Moreover, this row reduction exposes that at least $n(m-1)$ eigenvalues of $B$ are $0$. (Normally, row operations mess up eigenvalues, but this doesn't happen with zero eigenvalues)
Now, let's deal with the eigenvalues from $A$. Suppose that $lambda$ is an eigenvalue from $A$ with eigenvector $v$. Now, consider the vector
$$
w=beginbmatrix
v\v\vdots\v
endbmatrix
$$
consisting of $m$ copies of $v$. Then
$$
Bw=beginbmatrixA&A&dots&A\
A&A&dots&A\
vdots&vdots&ddots&vdots\
A&A&dots&A
endbmatrixbeginbmatrix
v\v\vdots\v
endbmatrix
=beginbmatrix
Av+Av+dots+Av\
Av+Av+dots+Av\
vdotsqquadvdots\
Av+Av+dots+Av
endbmatrix,
$$
where each block row has $m$ copies of $Av$. In other words, this simplifies to
$$
beginbmatrix
mAv\
mAv\
vdots\
mAv
endbmatrix.
$$
Since $v$ is an eigenvector for $A$, we can further simplify to
$$
beginbmatrix
mlambda v\
mlambda v\
vdots\
mlambda v
endbmatrix=mlambda w.
$$
Therefore, $mlambda$ is an eigenvalue for $A$.
Now, you have to convince yourself that we haven't double counted, but I'll leave that for you. If you want to, try to find the eigenvectors for the zeros we found above and observe that the eigenvectors are in distinct spaces. Hint: if $iequiv jpmod n$, use the vector which is $1$ in row $i$, $-1$ in row $j$, and zero elsewhere.
answered Jul 22 at 11:45
Michael Burr
25.6k13262
Yes, your conjecture is correct. Suppose that $A$ is an $ntimes n$ matrix and that you consider the $(nm)times(nm)$ matrix
$$
B=beginbmatrixA&A&dots&A\
A&A&dots&A\
vdots&vdots&ddots&vdots\
A&A&dots&A
endbmatrix
$$
that is a block matrix consisting of $mtimes m$ copies of $A$.
We can determine the eigenvalues of this matrix by inspection. We first note that $operatornamerank(A)=operatornamerank(B)$. This follows from the fact $B$ has lots of copies of the same rows. In fact, if $i$ and $j$ have $iequiv jpmod n$, then the rows $i$ and $j$ are copies of each other. Putting this together, with simple row operations, we can row reduce $B$ to
$$
beginbmatrixA&A&dots&A\
0&0&dots&0\
vdots&vdots&ddots&vdots\
0&0&dots&0
endbmatrix
$$
Then, we observe that the ranks of $A$ and $B$ are the same. Moreover, this row reduction exposes that at least $n(m-1)$ eigenvalues of $B$ are $0$. (Normally, row operations mess up eigenvalues, but this doesn't happen with zero eigenvalues)
Now, let's deal with the eigenvalues from $A$. Suppose that $lambda$ is an eigenvalue from $A$ with eigenvector $v$. Now, consider the vector
$$
w=beginbmatrix
v\v\vdots\v
endbmatrix
$$
consisting of $m$ copies of $v$. Then
$$
Bw=beginbmatrixA&A&dots&A\
A&A&dots&A\
vdots&vdots&ddots&vdots\
A&A&dots&A
endbmatrixbeginbmatrix
v\v\vdots\v
endbmatrix
=beginbmatrix
Av+Av+dots+Av\
Av+Av+dots+Av\
vdotsqquadvdots\
Av+Av+dots+Av
endbmatrix,
$$
where each block row has $m$ copies of $Av$. In other words, this simplifies to
$$
beginbmatrix
mAv\
mAv\
vdots\
mAv
endbmatrix.
$$
Since $v$ is an eigenvector for $A$, we can further simplify to
$$
beginbmatrix
mlambda v\
mlambda v\
vdots\
mlambda v
endbmatrix=mlambda w.
$$
Therefore, $mlambda$ is an eigenvalue for $A$.
Now, you have to convince yourself that we haven't double counted, but I'll leave that for you. If you want to, try to find the eigenvectors for the zeros we found above and observe that the eigenvectors are in distinct spaces. Hint: if $iequiv jpmod n$, use the vector which is $1$ in row $i$, $-1$ in row $j$, and zero elsewhere.
answered Jul 22 at 11:45
Michael Burr
25.6k13262
answered Jul 22 at 11:45
Michael Burr
25.6k13262
answered Jul 22 at 11:45
Michael Burr
25.6k13262
answered Jul 22 at 11:45
Michael Burr
25.6k13262
25.6k13262
add a comment |Â
add a comment |Â
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6
these are Kronecker products.
– Lord Shark the Unknown
Jul 22 at 11:23
I am not exactly sure about your notation: Do you consider a matrix which elements are other matrices or do you want the elements to be the determinant of the original matrix?
– mrtaurho
Jul 22 at 11:24
This looks correct. Some hints: Use the rank to determine the number of zero eigenvalues, and use repeated copies of eigenvectors for the nonzero eigenvectors.
– Michael Burr
Jul 22 at 11:27
Im sorry.. Well, I consider the matrix A as partition matrix of the bigger matrix A*, A**, ...
– Diggie Cruz
Jul 22 at 11:29
2
This is good mathematical thinking, up to the point where you say conclude. The proper word is conjecture. You conclude something when you have a proof, not just a hunch.
– Arnaud Mortier
Jul 22 at 11:30