Mixing
If $x$ and $yinBbb{R}^{n}$ are eigenvectors for $lambdaneqmu$, respectively, show $x^{T}cdotp y = 0$
$begingroup$
For $x^{T}cdotp y = 0$, I understand that I can either look at it through matrix multiplication $x^{T}y^{T} = 0$ as you can't do that multiplication. I'm very sure this isn't the right way of looking at it but am unsure how else to think about proving this.
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices transpose
edited Jan 7 at 15:38
Shubham Johri
5,017717
asked Jan 7 at 15:35
L GL G
248
$endgroup$
|
show 4 more comments
$begingroup$
For $x^{T}cdotp y = 0$, I understand that I can either look at it through matrix multiplication $x^{T}y^{T} = 0$ as you can't do that multiplication. I'm very sure this isn't the right way of looking at it but am unsure how else to think about proving this.
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices transpose
edited Jan 7 at 15:38
Shubham Johri
5,017717
asked Jan 7 at 15:35
L GL G
248
$endgroup$
$begingroup$
"Are eigenvectors" makes no sense - you mean they are eigenvectors of $A$. Now what you say you want to prove is false. Look at the problem again, and give a correct statement, without leaving anything out.
$endgroup$
– David C. Ullrich
Jan 7 at 17:59
$begingroup$
@DavidC.Ullrich Im saying that x is an eigenvector for $lambda$ and y is an eigenvector for $mu$
$endgroup$
– L G
Jan 8 at 16:49
$begingroup$
I understood that. Saying $x$ is an eigennvector for $lambda$ makes no sense! Square matrices have eigenvectors and eigenvalues, and there is no matrix here. Once you fix that, by saying you're talking about some matrix $A$, it makes sense but it's false. Wrong. Incorrect. Untrue.
$endgroup$
– David C. Ullrich
Jan 8 at 17:01
$begingroup$
You evidently saw my first comment. It's simply incredible that you can't take the time to look at the exercise again and tell us what the exercise actually says!
$endgroup$
– David C. Ullrich
Jan 8 at 17:02
$begingroup$
@DavidC.Ullrich The exercise word for word states "Q2: Let A be a symmetric n × n matrix over R. (i) Prove that all eigenvalues of A are real. (ii) Let x and y ∈ R n be eigenvectors for ${lambda} {neq} {mu}$, respectively. Show that x T · y = 0.
$endgroup$
– L G
Jan 8 at 17:08
|
show 4 more comments
$begingroup$
For $x^{T}cdotp y = 0$, I understand that I can either look at it through matrix multiplication $x^{T}y^{T} = 0$ as you can't do that multiplication. I'm very sure this isn't the right way of looking at it but am unsure how else to think about proving this.
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices transpose
edited Jan 7 at 15:38
Shubham Johri
5,017717
asked Jan 7 at 15:35
L GL G
248
$endgroup$
For $x^{T}cdotp y = 0$, I understand that I can either look at it through matrix multiplication $x^{T}y^{T} = 0$ as you can't do that multiplication. I'm very sure this isn't the right way of looking at it but am unsure how else to think about proving this.
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices transpose
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices transpose
edited Jan 7 at 15:38
Shubham Johri
5,017717
asked Jan 7 at 15:35
L GL G
248
edited Jan 7 at 15:38
Shubham Johri
5,017717
asked Jan 7 at 15:35
L GL G
248
edited Jan 7 at 15:38
Shubham Johri
5,017717
edited Jan 7 at 15:38
Shubham Johri
5,017717
edited Jan 7 at 15:38
Shubham Johri
5,017717
5,017717
asked Jan 7 at 15:35
L GL G
248
asked Jan 7 at 15:35
L GL G
248
asked Jan 7 at 15:35
L GL G
248
248
$begingroup$
"Are eigenvectors" makes no sense - you mean they are eigenvectors of $A$. Now what you say you want to prove is false. Look at the problem again, and give a correct statement, without leaving anything out.
$endgroup$
– David C. Ullrich
Jan 7 at 17:59
$begingroup$
@DavidC.Ullrich Im saying that x is an eigenvector for $lambda$ and y is an eigenvector for $mu$
$endgroup$
– L G
Jan 8 at 16:49
$begingroup$
I understood that. Saying $x$ is an eigennvector for $lambda$ makes no sense! Square matrices have eigenvectors and eigenvalues, and there is no matrix here. Once you fix that, by saying you're talking about some matrix $A$, it makes sense but it's false. Wrong. Incorrect. Untrue.
$endgroup$
– David C. Ullrich
Jan 8 at 17:01
$begingroup$
You evidently saw my first comment. It's simply incredible that you can't take the time to look at the exercise again and tell us what the exercise actually says!
$endgroup$
– David C. Ullrich
Jan 8 at 17:02
$begingroup$
@DavidC.Ullrich The exercise word for word states "Q2: Let A be a symmetric n × n matrix over R. (i) Prove that all eigenvalues of A are real. (ii) Let x and y ∈ R n be eigenvectors for ${lambda} {neq} {mu}$, respectively. Show that x T · y = 0.
$endgroup$
– L G
Jan 8 at 17:08
|
show 4 more comments
$begingroup$
"Are eigenvectors" makes no sense - you mean they are eigenvectors of $A$. Now what you say you want to prove is false. Look at the problem again, and give a correct statement, without leaving anything out.
$endgroup$
– David C. Ullrich
Jan 7 at 17:59
$begingroup$
@DavidC.Ullrich Im saying that x is an eigenvector for $lambda$ and y is an eigenvector for $mu$
$endgroup$
– L G
Jan 8 at 16:49
$begingroup$
I understood that. Saying $x$ is an eigennvector for $lambda$ makes no sense! Square matrices have eigenvectors and eigenvalues, and there is no matrix here. Once you fix that, by saying you're talking about some matrix $A$, it makes sense but it's false. Wrong. Incorrect. Untrue.
$endgroup$
– David C. Ullrich
Jan 8 at 17:01
$begingroup$
You evidently saw my first comment. It's simply incredible that you can't take the time to look at the exercise again and tell us what the exercise actually says!
$endgroup$
– David C. Ullrich
Jan 8 at 17:02
$begingroup$
@DavidC.Ullrich The exercise word for word states "Q2: Let A be a symmetric n × n matrix over R. (i) Prove that all eigenvalues of A are real. (ii) Let x and y ∈ R n be eigenvectors for ${lambda} {neq} {mu}$, respectively. Show that x T · y = 0.
$endgroup$
– L G
Jan 8 at 17:08
$begingroup$
"Are eigenvectors" makes no sense - you mean they are eigenvectors of $A$. Now what you say you want to prove is false. Look at the problem again, and give a correct statement, without leaving anything out.
$endgroup$
– David C. Ullrich
Jan 7 at 17:59
$begingroup$
"Are eigenvectors" makes no sense - you mean they are eigenvectors of $A$. Now what you say you want to prove is false. Look at the problem again, and give a correct statement, without leaving anything out.
$endgroup$
– David C. Ullrich
Jan 7 at 17:59
$begingroup$
@DavidC.Ullrich Im saying that x is an eigenvector for $lambda$ and y is an eigenvector for $mu$
$endgroup$
– L G
Jan 8 at 16:49
$begingroup$
@DavidC.Ullrich Im saying that x is an eigenvector for $lambda$ and y is an eigenvector for $mu$
$endgroup$
– L G
Jan 8 at 16:49
$begingroup$
I understood that. Saying $x$ is an eigennvector for $lambda$ makes no sense! Square matrices have eigenvectors and eigenvalues, and there is no matrix here. Once you fix that, by saying you're talking about some matrix $A$, it makes sense but it's false. Wrong. Incorrect. Untrue.
$endgroup$
– David C. Ullrich
Jan 8 at 17:01
$begingroup$
I understood that. Saying $x$ is an eigennvector for $lambda$ makes no sense! Square matrices have eigenvectors and eigenvalues, and there is no matrix here. Once you fix that, by saying you're talking about some matrix $A$, it makes sense but it's false. Wrong. Incorrect. Untrue.
$endgroup$
– David C. Ullrich
Jan 8 at 17:01
$begingroup$
You evidently saw my first comment. It's simply incredible that you can't take the time to look at the exercise again and tell us what the exercise actually says!
$endgroup$
– David C. Ullrich
Jan 8 at 17:02
$begingroup$
You evidently saw my first comment. It's simply incredible that you can't take the time to look at the exercise again and tell us what the exercise actually says!
$endgroup$
– David C. Ullrich
Jan 8 at 17:02
$begingroup$
@DavidC.Ullrich The exercise word for word states "Q2: Let A be a symmetric n × n matrix over R. (i) Prove that all eigenvalues of A are real. (ii) Let x and y ∈ R n be eigenvectors for ${lambda} {neq} {mu}$, respectively. Show that x T · y = 0.
$endgroup$
– L G
Jan 8 at 17:08
$begingroup$
@DavidC.Ullrich The exercise word for word states "Q2: Let A be a symmetric n × n matrix over R. (i) Prove that all eigenvalues of A are real. (ii) Let x and y ∈ R n be eigenvectors for ${lambda} {neq} {mu}$, respectively. Show that x T · y = 0.
$endgroup$
– L G
Jan 8 at 17:08
|
show 4 more comments
1 Answer
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active
oldest
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$begingroup$
The statement in the question is unclear, speaking of eigenvalues and eigenvectors without mentioning a matrix. For some reason the OP is unwilling to fix the question.
A reader not already familiar with the correct version of the result in question might get the impression we want to prove this:
False Fact If $A$ is a square matrix, $Ax=lambda x$, $Ay=mu y$ and $lambdane mu$ then $x^Ty=0$.
Of course that's false: Consider $A=begin{bmatrix}1&1\0&2end{bmatrix}$, $x=(1,0)^T$, $y=(1,1)^T$.
The correct version is this:
True Fact If $A$ is a symmetric square matrix, $Ax=lambda x$, $Ay=mu y$ and $lambdane mu$ then $x^Ty=0$.
This is easy. First, the whole point to symmetric matricies:
Lemma. If $A$ is symmetric then $x^TAy=y^TAx$.
Proof: Since a $1times 1$ matrix is its own transpose and $(AB)^T=B^TA^T$ we see that $$x^TAy=(x^TAy)^T=y^TA^Tx=y^TAx.$$
Having established that the True Fact is easy:
$$0=x^TAy-y^TAx=mu x^Ty-lambda y^Tx=(mu-lambda)x^Ty,$$which implies that $x^Ty=0$, since $mu-lambdane0$.
answered Jan 9 at 13:47
David C. UllrichDavid C. Ullrich
60.1k43994
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
The statement in the question is unclear, speaking of eigenvalues and eigenvectors without mentioning a matrix. For some reason the OP is unwilling to fix the question.
A reader not already familiar with the correct version of the result in question might get the impression we want to prove this:
False Fact If $A$ is a square matrix, $Ax=lambda x$, $Ay=mu y$ and $lambdane mu$ then $x^Ty=0$.
Of course that's false: Consider $A=begin{bmatrix}1&1\0&2end{bmatrix}$, $x=(1,0)^T$, $y=(1,1)^T$.
The correct version is this:
True Fact If $A$ is a symmetric square matrix, $Ax=lambda x$, $Ay=mu y$ and $lambdane mu$ then $x^Ty=0$.
This is easy. First, the whole point to symmetric matricies:
Lemma. If $A$ is symmetric then $x^TAy=y^TAx$.
Proof: Since a $1times 1$ matrix is its own transpose and $(AB)^T=B^TA^T$ we see that $$x^TAy=(x^TAy)^T=y^TA^Tx=y^TAx.$$
Having established that the True Fact is easy:
$$0=x^TAy-y^TAx=mu x^Ty-lambda y^Tx=(mu-lambda)x^Ty,$$which implies that $x^Ty=0$, since $mu-lambdane0$.
answered Jan 9 at 13:47
David C. UllrichDavid C. Ullrich
60.1k43994
$endgroup$
add a comment |
$begingroup$
The statement in the question is unclear, speaking of eigenvalues and eigenvectors without mentioning a matrix. For some reason the OP is unwilling to fix the question.
A reader not already familiar with the correct version of the result in question might get the impression we want to prove this:
False Fact If $A$ is a square matrix, $Ax=lambda x$, $Ay=mu y$ and $lambdane mu$ then $x^Ty=0$.
Of course that's false: Consider $A=begin{bmatrix}1&1\0&2end{bmatrix}$, $x=(1,0)^T$, $y=(1,1)^T$.
The correct version is this:
True Fact If $A$ is a symmetric square matrix, $Ax=lambda x$, $Ay=mu y$ and $lambdane mu$ then $x^Ty=0$.
This is easy. First, the whole point to symmetric matricies:
Lemma. If $A$ is symmetric then $x^TAy=y^TAx$.
Proof: Since a $1times 1$ matrix is its own transpose and $(AB)^T=B^TA^T$ we see that $$x^TAy=(x^TAy)^T=y^TA^Tx=y^TAx.$$
Having established that the True Fact is easy:
$$0=x^TAy-y^TAx=mu x^Ty-lambda y^Tx=(mu-lambda)x^Ty,$$which implies that $x^Ty=0$, since $mu-lambdane0$.
answered Jan 9 at 13:47
David C. UllrichDavid C. Ullrich
60.1k43994
$endgroup$
add a comment |
$begingroup$
The statement in the question is unclear, speaking of eigenvalues and eigenvectors without mentioning a matrix. For some reason the OP is unwilling to fix the question.
A reader not already familiar with the correct version of the result in question might get the impression we want to prove this:
False Fact If $A$ is a square matrix, $Ax=lambda x$, $Ay=mu y$ and $lambdane mu$ then $x^Ty=0$.
Of course that's false: Consider $A=begin{bmatrix}1&1\0&2end{bmatrix}$, $x=(1,0)^T$, $y=(1,1)^T$.
The correct version is this:
True Fact If $A$ is a symmetric square matrix, $Ax=lambda x$, $Ay=mu y$ and $lambdane mu$ then $x^Ty=0$.
This is easy. First, the whole point to symmetric matricies:
Lemma. If $A$ is symmetric then $x^TAy=y^TAx$.
Proof: Since a $1times 1$ matrix is its own transpose and $(AB)^T=B^TA^T$ we see that $$x^TAy=(x^TAy)^T=y^TA^Tx=y^TAx.$$
Having established that the True Fact is easy:
$$0=x^TAy-y^TAx=mu x^Ty-lambda y^Tx=(mu-lambda)x^Ty,$$which implies that $x^Ty=0$, since $mu-lambdane0$.
answered Jan 9 at 13:47
David C. UllrichDavid C. Ullrich
60.1k43994
$endgroup$
The statement in the question is unclear, speaking of eigenvalues and eigenvectors without mentioning a matrix. For some reason the OP is unwilling to fix the question.
A reader not already familiar with the correct version of the result in question might get the impression we want to prove this:
False Fact If $A$ is a square matrix, $Ax=lambda x$, $Ay=mu y$ and $lambdane mu$ then $x^Ty=0$.
Of course that's false: Consider $A=begin{bmatrix}1&1\0&2end{bmatrix}$, $x=(1,0)^T$, $y=(1,1)^T$.
The correct version is this:
True Fact If $A$ is a symmetric square matrix, $Ax=lambda x$, $Ay=mu y$ and $lambdane mu$ then $x^Ty=0$.
This is easy. First, the whole point to symmetric matricies:
Lemma. If $A$ is symmetric then $x^TAy=y^TAx$.
Proof: Since a $1times 1$ matrix is its own transpose and $(AB)^T=B^TA^T$ we see that $$x^TAy=(x^TAy)^T=y^TA^Tx=y^TAx.$$
Having established that the True Fact is easy:
$$0=x^TAy-y^TAx=mu x^Ty-lambda y^Tx=(mu-lambda)x^Ty,$$which implies that $x^Ty=0$, since $mu-lambdane0$.
answered Jan 9 at 13:47
David C. UllrichDavid C. Ullrich
60.1k43994
answered Jan 9 at 13:47
David C. UllrichDavid C. Ullrich
60.1k43994
answered Jan 9 at 13:47
David C. UllrichDavid C. Ullrich
60.1k43994
answered Jan 9 at 13:47
David C. UllrichDavid C. Ullrich
60.1k43994
60.1k43994
add a comment |
add a comment |
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$begingroup$
"Are eigenvectors" makes no sense - you mean they are eigenvectors of $A$. Now what you say you want to prove is false. Look at the problem again, and give a correct statement, without leaving anything out.
$endgroup$
– David C. Ullrich
Jan 7 at 17:59
$begingroup$
@DavidC.Ullrich Im saying that x is an eigenvector for $lambda$ and y is an eigenvector for $mu$
$endgroup$
– L G
Jan 8 at 16:49
$begingroup$
I understood that. Saying $x$ is an eigennvector for $lambda$ makes no sense! Square matrices have eigenvectors and eigenvalues, and there is no matrix here. Once you fix that, by saying you're talking about some matrix $A$, it makes sense but it's false. Wrong. Incorrect. Untrue.
$endgroup$
– David C. Ullrich
Jan 8 at 17:01
$begingroup$
You evidently saw my first comment. It's simply incredible that you can't take the time to look at the exercise again and tell us what the exercise actually says!
$endgroup$
– David C. Ullrich
Jan 8 at 17:02
$begingroup$
@DavidC.Ullrich The exercise word for word states "Q2: Let A be a symmetric n × n matrix over R. (i) Prove that all eigenvalues of A are real. (ii) Let x and y ∈ R n be eigenvectors for ${lambda} {neq} {mu}$, respectively. Show that x T · y = 0.
$endgroup$
– L G
Jan 8 at 17:08