Mixing
$A,Bin mathbb R^{ntimes n}$ share $n$ common linearly-independent eigenvectors $Rightarrow AB=BA$.
$begingroup$
I've been trying to prove the following statement:
Let $A,Bin mathbb R^{ntimes n}$ be square matrices such that they share $n$ common linearly-independent eigenvectors. Then $AB=BA$.
Everything I thought of seemed to be unhelpful, so I have no clue what to do next.
Thank you and have a nice day!.
linear-algebra matrices diagonalization
asked Jan 13 at 18:11
Amit ZachAmit Zach
595
$endgroup$
add a comment |
$begingroup$
I've been trying to prove the following statement:
Let $A,Bin mathbb R^{ntimes n}$ be square matrices such that they share $n$ common linearly-independent eigenvectors. Then $AB=BA$.
Everything I thought of seemed to be unhelpful, so I have no clue what to do next.
Thank you and have a nice day!.
linear-algebra matrices diagonalization
asked Jan 13 at 18:11
Amit ZachAmit Zach
595
$endgroup$
1
$begingroup$
Hint: write the matrices in terms of basis given by those eigenvectors.
$endgroup$
– Wojowu
Jan 13 at 18:18
$begingroup$
@Wojowu I think I got it. Thanks!
$endgroup$
– Amit Zach
Jan 13 at 18:26
add a comment |
$begingroup$
I've been trying to prove the following statement:
Let $A,Bin mathbb R^{ntimes n}$ be square matrices such that they share $n$ common linearly-independent eigenvectors. Then $AB=BA$.
Everything I thought of seemed to be unhelpful, so I have no clue what to do next.
Thank you and have a nice day!.
linear-algebra matrices diagonalization
asked Jan 13 at 18:11
Amit ZachAmit Zach
595
$endgroup$
I've been trying to prove the following statement:
Let $A,Bin mathbb R^{ntimes n}$ be square matrices such that they share $n$ common linearly-independent eigenvectors. Then $AB=BA$.
Everything I thought of seemed to be unhelpful, so I have no clue what to do next.
Thank you and have a nice day!.
linear-algebra matrices diagonalization
linear-algebra matrices diagonalization
asked Jan 13 at 18:11
Amit ZachAmit Zach
595
asked Jan 13 at 18:11
Amit ZachAmit Zach
595
asked Jan 13 at 18:11
Amit ZachAmit Zach
595
asked Jan 13 at 18:11
Amit ZachAmit Zach
595
asked Jan 13 at 18:11
Amit ZachAmit Zach
595
595
1
$begingroup$
Hint: write the matrices in terms of basis given by those eigenvectors.
$endgroup$
– Wojowu
Jan 13 at 18:18
$begingroup$
@Wojowu I think I got it. Thanks!
$endgroup$
– Amit Zach
Jan 13 at 18:26
add a comment |
1
$begingroup$
Hint: write the matrices in terms of basis given by those eigenvectors.
$endgroup$
– Wojowu
Jan 13 at 18:18
$begingroup$
@Wojowu I think I got it. Thanks!
$endgroup$
– Amit Zach
Jan 13 at 18:26
1
1
$begingroup$
Hint: write the matrices in terms of basis given by those eigenvectors.
$endgroup$
– Wojowu
Jan 13 at 18:18
$begingroup$
Hint: write the matrices in terms of basis given by those eigenvectors.
$endgroup$
– Wojowu
Jan 13 at 18:18
$begingroup$
@Wojowu I think I got it. Thanks!
$endgroup$
– Amit Zach
Jan 13 at 18:26
$begingroup$
@Wojowu I think I got it. Thanks!
$endgroup$
– Amit Zach
Jan 13 at 18:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that though $A, B in Bbb R^{n times n}$, the eigenvalues may be complex and thus the eigenvectors may be in $Bbb C^n$. Bearing this in mind, we proceed:
Let
$vec e_1, vec e_2, ldots, vec e_n tag 1$
be the $n$ common, linearly independent eigenvectors of $A$ and $B$; then we have
$Avec e_i = alpha_i vec e_i, ; 1 le i le n, tag 2$
and
$Bvec e_i = beta_i vec e_i, ; 1 le i le n, tag 3$
for some scalars $alpha_i, beta_i in Bbb C$, $1 le i le n$. Thus
$AB vec e_i = A(beta_i vec e_i) = beta_i A vec e_i = beta_i alpha_i vec e_i = alpha_i beta_i vec e_i = alpha_i Bvec e_i = B alpha_i vec e_i = BAvec e_i. tag 4$
Since the $vec e_i$ are $n$ in number and linearly independent, the form a basis of $Bbb C^n$; thus for any
$vec v in Bbb C^n tag 5$
we may write
$vec v = displaystyle sum_1^n v_i vec e_i, ; v_i in Bbb C, ; 1 le i le n,; tag 6$
thus
$AB vec v = displaystyle sum_1^n v_i AB vec e_i = sum_1^n v_i BA vec e_i = BAvec v; tag 7$
therefore we conclude
$AB = BA. tag 8$
answered Jan 13 at 18:32
Robert LewisRobert Lewis
46.5k23067
$endgroup$
1
$begingroup$
Thank you very much, this is exactly what I did after the hint I was just given. I'm happy to see that my proof was similar to your professional one :)
$endgroup$
– Amit Zach
Jan 13 at 19:29
$begingroup$
Thanks for the kind words. If you really think my proof is "professional", you might consider "accepting it". Cheers! ;)
$endgroup$
– Robert Lewis
Jan 13 at 19:31
add a comment |
Your Answer
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$begingroup$
Note that though $A, B in Bbb R^{n times n}$, the eigenvalues may be complex and thus the eigenvectors may be in $Bbb C^n$. Bearing this in mind, we proceed:
Let
$vec e_1, vec e_2, ldots, vec e_n tag 1$
be the $n$ common, linearly independent eigenvectors of $A$ and $B$; then we have
$Avec e_i = alpha_i vec e_i, ; 1 le i le n, tag 2$
and
$Bvec e_i = beta_i vec e_i, ; 1 le i le n, tag 3$
for some scalars $alpha_i, beta_i in Bbb C$, $1 le i le n$. Thus
$AB vec e_i = A(beta_i vec e_i) = beta_i A vec e_i = beta_i alpha_i vec e_i = alpha_i beta_i vec e_i = alpha_i Bvec e_i = B alpha_i vec e_i = BAvec e_i. tag 4$
Since the $vec e_i$ are $n$ in number and linearly independent, the form a basis of $Bbb C^n$; thus for any
$vec v in Bbb C^n tag 5$
we may write
$vec v = displaystyle sum_1^n v_i vec e_i, ; v_i in Bbb C, ; 1 le i le n,; tag 6$
thus
$AB vec v = displaystyle sum_1^n v_i AB vec e_i = sum_1^n v_i BA vec e_i = BAvec v; tag 7$
therefore we conclude
$AB = BA. tag 8$
answered Jan 13 at 18:32
Robert LewisRobert Lewis
46.5k23067
$endgroup$
1
$begingroup$
Thank you very much, this is exactly what I did after the hint I was just given. I'm happy to see that my proof was similar to your professional one :)
$endgroup$
– Amit Zach
Jan 13 at 19:29
$begingroup$
Thanks for the kind words. If you really think my proof is "professional", you might consider "accepting it". Cheers! ;)
$endgroup$
– Robert Lewis
Jan 13 at 19:31
add a comment |
$begingroup$
Note that though $A, B in Bbb R^{n times n}$, the eigenvalues may be complex and thus the eigenvectors may be in $Bbb C^n$. Bearing this in mind, we proceed:
Let
$vec e_1, vec e_2, ldots, vec e_n tag 1$
be the $n$ common, linearly independent eigenvectors of $A$ and $B$; then we have
$Avec e_i = alpha_i vec e_i, ; 1 le i le n, tag 2$
and
$Bvec e_i = beta_i vec e_i, ; 1 le i le n, tag 3$
for some scalars $alpha_i, beta_i in Bbb C$, $1 le i le n$. Thus
$AB vec e_i = A(beta_i vec e_i) = beta_i A vec e_i = beta_i alpha_i vec e_i = alpha_i beta_i vec e_i = alpha_i Bvec e_i = B alpha_i vec e_i = BAvec e_i. tag 4$
Since the $vec e_i$ are $n$ in number and linearly independent, the form a basis of $Bbb C^n$; thus for any
$vec v in Bbb C^n tag 5$
we may write
$vec v = displaystyle sum_1^n v_i vec e_i, ; v_i in Bbb C, ; 1 le i le n,; tag 6$
thus
$AB vec v = displaystyle sum_1^n v_i AB vec e_i = sum_1^n v_i BA vec e_i = BAvec v; tag 7$
therefore we conclude
$AB = BA. tag 8$
answered Jan 13 at 18:32
Robert LewisRobert Lewis
46.5k23067
$endgroup$
1
$begingroup$
Thank you very much, this is exactly what I did after the hint I was just given. I'm happy to see that my proof was similar to your professional one :)
$endgroup$
– Amit Zach
Jan 13 at 19:29
$begingroup$
Thanks for the kind words. If you really think my proof is "professional", you might consider "accepting it". Cheers! ;)
$endgroup$
– Robert Lewis
Jan 13 at 19:31
add a comment |
$begingroup$
Note that though $A, B in Bbb R^{n times n}$, the eigenvalues may be complex and thus the eigenvectors may be in $Bbb C^n$. Bearing this in mind, we proceed:
Let
$vec e_1, vec e_2, ldots, vec e_n tag 1$
be the $n$ common, linearly independent eigenvectors of $A$ and $B$; then we have
$Avec e_i = alpha_i vec e_i, ; 1 le i le n, tag 2$
and
$Bvec e_i = beta_i vec e_i, ; 1 le i le n, tag 3$
for some scalars $alpha_i, beta_i in Bbb C$, $1 le i le n$. Thus
$AB vec e_i = A(beta_i vec e_i) = beta_i A vec e_i = beta_i alpha_i vec e_i = alpha_i beta_i vec e_i = alpha_i Bvec e_i = B alpha_i vec e_i = BAvec e_i. tag 4$
Since the $vec e_i$ are $n$ in number and linearly independent, the form a basis of $Bbb C^n$; thus for any
$vec v in Bbb C^n tag 5$
we may write
$vec v = displaystyle sum_1^n v_i vec e_i, ; v_i in Bbb C, ; 1 le i le n,; tag 6$
thus
$AB vec v = displaystyle sum_1^n v_i AB vec e_i = sum_1^n v_i BA vec e_i = BAvec v; tag 7$
therefore we conclude
$AB = BA. tag 8$
answered Jan 13 at 18:32
Robert LewisRobert Lewis
46.5k23067
$endgroup$
Note that though $A, B in Bbb R^{n times n}$, the eigenvalues may be complex and thus the eigenvectors may be in $Bbb C^n$. Bearing this in mind, we proceed:
Let
$vec e_1, vec e_2, ldots, vec e_n tag 1$
be the $n$ common, linearly independent eigenvectors of $A$ and $B$; then we have
$Avec e_i = alpha_i vec e_i, ; 1 le i le n, tag 2$
and
$Bvec e_i = beta_i vec e_i, ; 1 le i le n, tag 3$
for some scalars $alpha_i, beta_i in Bbb C$, $1 le i le n$. Thus
$AB vec e_i = A(beta_i vec e_i) = beta_i A vec e_i = beta_i alpha_i vec e_i = alpha_i beta_i vec e_i = alpha_i Bvec e_i = B alpha_i vec e_i = BAvec e_i. tag 4$
Since the $vec e_i$ are $n$ in number and linearly independent, the form a basis of $Bbb C^n$; thus for any
$vec v in Bbb C^n tag 5$
we may write
$vec v = displaystyle sum_1^n v_i vec e_i, ; v_i in Bbb C, ; 1 le i le n,; tag 6$
thus
$AB vec v = displaystyle sum_1^n v_i AB vec e_i = sum_1^n v_i BA vec e_i = BAvec v; tag 7$
therefore we conclude
$AB = BA. tag 8$
answered Jan 13 at 18:32
Robert LewisRobert Lewis
46.5k23067
edited Jan 13 at 18:50
answered Jan 13 at 18:32
Robert LewisRobert Lewis
46.5k23067
answered Jan 13 at 18:32
Robert LewisRobert Lewis
46.5k23067
answered Jan 13 at 18:32
Robert LewisRobert Lewis
46.5k23067
46.5k23067
1
$begingroup$
Thank you very much, this is exactly what I did after the hint I was just given. I'm happy to see that my proof was similar to your professional one :)
$endgroup$
– Amit Zach
Jan 13 at 19:29
$begingroup$
Thanks for the kind words. If you really think my proof is "professional", you might consider "accepting it". Cheers! ;)
$endgroup$
– Robert Lewis
Jan 13 at 19:31
add a comment |
1
$begingroup$
Thank you very much, this is exactly what I did after the hint I was just given. I'm happy to see that my proof was similar to your professional one :)
$endgroup$
– Amit Zach
Jan 13 at 19:29
$begingroup$
Thanks for the kind words. If you really think my proof is "professional", you might consider "accepting it". Cheers! ;)
$endgroup$
– Robert Lewis
Jan 13 at 19:31
1
1
$begingroup$
Thank you very much, this is exactly what I did after the hint I was just given. I'm happy to see that my proof was similar to your professional one :)
$endgroup$
– Amit Zach
Jan 13 at 19:29
$begingroup$
Thank you very much, this is exactly what I did after the hint I was just given. I'm happy to see that my proof was similar to your professional one :)
$endgroup$
– Amit Zach
Jan 13 at 19:29
$begingroup$
Thanks for the kind words. If you really think my proof is "professional", you might consider "accepting it". Cheers! ;)
$endgroup$
– Robert Lewis
Jan 13 at 19:31
$begingroup$
Thanks for the kind words. If you really think my proof is "professional", you might consider "accepting it". Cheers! ;)
$endgroup$
– Robert Lewis
Jan 13 at 19:31
add a comment |
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$begingroup$
Hint: write the matrices in terms of basis given by those eigenvectors.
$endgroup$
– Wojowu
Jan 13 at 18:18
$begingroup$
@Wojowu I think I got it. Thanks!
$endgroup$
– Amit Zach
Jan 13 at 18:26