Show y*x=0 by eigenvectors [closed]
$begingroup$
Let $A$ be a square matrix, $lambda$ and $mu$ be two distinct eigenvalues of $A$.
If $x$ is an eigenvector corresponding to $lambda$, $y^∗$ is a nonzero row vector such that $y^∗ (A − mu I) = 0$, show that $y^∗ x = 0$.
can someone give me some direction to solve this problem?
linear-algebra
edited Nov 29 '18 at 1:59
Siong Thye Goh
101k1466117
asked Nov 29 '18 at 1:52
CColaCCola
346
$endgroup$
closed as off-topic by Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos Nov 29 '18 at 12:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $A$ be a square matrix, $lambda$ and $mu$ be two distinct eigenvalues of $A$.
If $x$ is an eigenvector corresponding to $lambda$, $y^∗$ is a nonzero row vector such that $y^∗ (A − mu I) = 0$, show that $y^∗ x = 0$.
can someone give me some direction to solve this problem?
linear-algebra
edited Nov 29 '18 at 1:59
Siong Thye Goh
101k1466117
asked Nov 29 '18 at 1:52
CColaCCola
346
$endgroup$
closed as off-topic by Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos Nov 29 '18 at 12:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
"Can I simply prove it by letting a square matrix?" ... letting a square matrix WHAT?
$endgroup$
– David G. Stork
Nov 29 '18 at 1:53
add a comment |
$begingroup$
Let $A$ be a square matrix, $lambda$ and $mu$ be two distinct eigenvalues of $A$.
If $x$ is an eigenvector corresponding to $lambda$, $y^∗$ is a nonzero row vector such that $y^∗ (A − mu I) = 0$, show that $y^∗ x = 0$.
can someone give me some direction to solve this problem?
linear-algebra
edited Nov 29 '18 at 1:59
Siong Thye Goh
101k1466117
asked Nov 29 '18 at 1:52
CColaCCola
346
$endgroup$
Let $A$ be a square matrix, $lambda$ and $mu$ be two distinct eigenvalues of $A$.
If $x$ is an eigenvector corresponding to $lambda$, $y^∗$ is a nonzero row vector such that $y^∗ (A − mu I) = 0$, show that $y^∗ x = 0$.
can someone give me some direction to solve this problem?
linear-algebra
linear-algebra
edited Nov 29 '18 at 1:59
Siong Thye Goh
101k1466117
asked Nov 29 '18 at 1:52
CColaCCola
346
edited Nov 29 '18 at 1:59
Siong Thye Goh
101k1466117
asked Nov 29 '18 at 1:52
CColaCCola
346
edited Nov 29 '18 at 1:59
Siong Thye Goh
101k1466117
edited Nov 29 '18 at 1:59
Siong Thye Goh
101k1466117
edited Nov 29 '18 at 1:59
Siong Thye Goh
101k1466117
101k1466117
asked Nov 29 '18 at 1:52
CColaCCola
346
asked Nov 29 '18 at 1:52
CColaCCola
346
asked Nov 29 '18 at 1:52
CColaCCola
346
346
closed as off-topic by Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos Nov 29 '18 at 12:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos Nov 29 '18 at 12:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
"Can I simply prove it by letting a square matrix?" ... letting a square matrix WHAT?
$endgroup$
– David G. Stork
Nov 29 '18 at 1:53
add a comment |
$begingroup$
"Can I simply prove it by letting a square matrix?" ... letting a square matrix WHAT?
$endgroup$
– David G. Stork
Nov 29 '18 at 1:53
$begingroup$
"Can I simply prove it by letting a square matrix?" ... letting a square matrix WHAT?
$endgroup$
– David G. Stork
Nov 29 '18 at 1:53
$begingroup$
"Can I simply prove it by letting a square matrix?" ... letting a square matrix WHAT?
$endgroup$
– David G. Stork
Nov 29 '18 at 1:53
add a comment |
3 Answers
3
active
oldest
votes
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I will assume I is the identity matrix. Since y*(A−μI)=0 then distributing yields y$^*A$=y$^*μI$ , since x is a eigen vector we have y$^*$Ax-y$^*$μIx=0$
so you need to know the fact that if you have a matrix A and a vector v such that Av=0 then A=0 or v=0.
$endgroup$
$begingroup$
ok thanks! i will try to approach in this way
$endgroup$
– CCola
Nov 29 '18 at 4:11
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No problem, if my answer was helpful I would appreciate it if you click on the check mark
$endgroup$
– user612135
Nov 29 '18 at 4:15
add a comment |
$begingroup$
Hint:
Use the property that
$$y^*(Ax)=(y^*A)x.$$
answered Nov 29 '18 at 1:57
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
We may as well start with $y^*(A-mu I) = 0$ and see what happens:
$$
begin{align*}
0 &= y^*(A - mu I)\
&= y^*(A - mu I)x\
&= y^*(Ax - mu x)\
&= ldots
end{align*}
$$
... can you take it from here?
spoiler alert
$$0 = y^*(Ax - mu x) = y^*(lambda x - mu x) = y^*x(lambda - mu)$$ and as $lambda neq mu$ we must have $y^*x = 0$.
answered Nov 29 '18 at 2:00
NealNeal
23.6k23886
$endgroup$
$begingroup$
so we get y*=0, x=0 or (A-μ)=0?
$endgroup$
– CCola
Nov 29 '18 at 5:37
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I will assume I is the identity matrix. Since y*(A−μI)=0 then distributing yields y$^*A$=y$^*μI$ , since x is a eigen vector we have y$^*$Ax-y$^*$μIx=0$
so you need to know the fact that if you have a matrix A and a vector v such that Av=0 then A=0 or v=0.
$endgroup$
$begingroup$
ok thanks! i will try to approach in this way
$endgroup$
– CCola
Nov 29 '18 at 4:11
$begingroup$
No problem, if my answer was helpful I would appreciate it if you click on the check mark
$endgroup$
– user612135
Nov 29 '18 at 4:15
add a comment |
$begingroup$
I will assume I is the identity matrix. Since y*(A−μI)=0 then distributing yields y$^*A$=y$^*μI$ , since x is a eigen vector we have y$^*$Ax-y$^*$μIx=0$
so you need to know the fact that if you have a matrix A and a vector v such that Av=0 then A=0 or v=0.
$endgroup$
$begingroup$
ok thanks! i will try to approach in this way
$endgroup$
– CCola
Nov 29 '18 at 4:11
$begingroup$
No problem, if my answer was helpful I would appreciate it if you click on the check mark
$endgroup$
– user612135
Nov 29 '18 at 4:15
add a comment |
$begingroup$
I will assume I is the identity matrix. Since y*(A−μI)=0 then distributing yields y$^*A$=y$^*μI$ , since x is a eigen vector we have y$^*$Ax-y$^*$μIx=0$
so you need to know the fact that if you have a matrix A and a vector v such that Av=0 then A=0 or v=0.
$endgroup$
I will assume I is the identity matrix. Since y*(A−μI)=0 then distributing yields y$^*A$=y$^*μI$ , since x is a eigen vector we have y$^*$Ax-y$^*$μIx=0$
so you need to know the fact that if you have a matrix A and a vector v such that Av=0 then A=0 or v=0.
answered Nov 29 '18 at 2:10
user612135
$begingroup$
ok thanks! i will try to approach in this way
$endgroup$
– CCola
Nov 29 '18 at 4:11
$begingroup$
No problem, if my answer was helpful I would appreciate it if you click on the check mark
$endgroup$
– user612135
Nov 29 '18 at 4:15
add a comment |
$begingroup$
ok thanks! i will try to approach in this way
$endgroup$
– CCola
Nov 29 '18 at 4:11
$begingroup$
No problem, if my answer was helpful I would appreciate it if you click on the check mark
$endgroup$
– user612135
Nov 29 '18 at 4:15
$begingroup$
ok thanks! i will try to approach in this way
$endgroup$
– CCola
Nov 29 '18 at 4:11
$begingroup$
ok thanks! i will try to approach in this way
$endgroup$
– CCola
Nov 29 '18 at 4:11
$begingroup$
No problem, if my answer was helpful I would appreciate it if you click on the check mark
$endgroup$
– user612135
Nov 29 '18 at 4:15
$begingroup$
No problem, if my answer was helpful I would appreciate it if you click on the check mark
$endgroup$
– user612135
Nov 29 '18 at 4:15
add a comment |
$begingroup$
Hint:
Use the property that
$$y^*(Ax)=(y^*A)x.$$
answered Nov 29 '18 at 1:57
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
Hint:
Use the property that
$$y^*(Ax)=(y^*A)x.$$
answered Nov 29 '18 at 1:57
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
Hint:
Use the property that
$$y^*(Ax)=(y^*A)x.$$
answered Nov 29 '18 at 1:57
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
Hint:
Use the property that
$$y^*(Ax)=(y^*A)x.$$
answered Nov 29 '18 at 1:57
Siong Thye GohSiong Thye Goh
101k1466117
answered Nov 29 '18 at 1:57
Siong Thye GohSiong Thye Goh
101k1466117
answered Nov 29 '18 at 1:57
Siong Thye GohSiong Thye Goh
101k1466117
answered Nov 29 '18 at 1:57
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
add a comment |
add a comment |
$begingroup$
We may as well start with $y^*(A-mu I) = 0$ and see what happens:
$$
begin{align*}
0 &= y^*(A - mu I)\
&= y^*(A - mu I)x\
&= y^*(Ax - mu x)\
&= ldots
end{align*}
$$
... can you take it from here?
spoiler alert
$$0 = y^*(Ax - mu x) = y^*(lambda x - mu x) = y^*x(lambda - mu)$$ and as $lambda neq mu$ we must have $y^*x = 0$.
answered Nov 29 '18 at 2:00
NealNeal
23.6k23886
$endgroup$
$begingroup$
so we get y*=0, x=0 or (A-μ)=0?
$endgroup$
– CCola
Nov 29 '18 at 5:37
add a comment |
$begingroup$
We may as well start with $y^*(A-mu I) = 0$ and see what happens:
$$
begin{align*}
0 &= y^*(A - mu I)\
&= y^*(A - mu I)x\
&= y^*(Ax - mu x)\
&= ldots
end{align*}
$$
... can you take it from here?
spoiler alert
$$0 = y^*(Ax - mu x) = y^*(lambda x - mu x) = y^*x(lambda - mu)$$ and as $lambda neq mu$ we must have $y^*x = 0$.
answered Nov 29 '18 at 2:00
NealNeal
23.6k23886
$endgroup$
$begingroup$
so we get y*=0, x=0 or (A-μ)=0?
$endgroup$
– CCola
Nov 29 '18 at 5:37
add a comment |
$begingroup$
We may as well start with $y^*(A-mu I) = 0$ and see what happens:
$$
begin{align*}
0 &= y^*(A - mu I)\
&= y^*(A - mu I)x\
&= y^*(Ax - mu x)\
&= ldots
end{align*}
$$
... can you take it from here?
spoiler alert
$$0 = y^*(Ax - mu x) = y^*(lambda x - mu x) = y^*x(lambda - mu)$$ and as $lambda neq mu$ we must have $y^*x = 0$.
answered Nov 29 '18 at 2:00
NealNeal
23.6k23886
$endgroup$
We may as well start with $y^*(A-mu I) = 0$ and see what happens:
$$
begin{align*}
0 &= y^*(A - mu I)\
&= y^*(A - mu I)x\
&= y^*(Ax - mu x)\
&= ldots
end{align*}
$$
... can you take it from here?
spoiler alert
$$0 = y^*(Ax - mu x) = y^*(lambda x - mu x) = y^*x(lambda - mu)$$ and as $lambda neq mu$ we must have $y^*x = 0$.
answered Nov 29 '18 at 2:00
NealNeal
23.6k23886
edited Nov 29 '18 at 13:16
answered Nov 29 '18 at 2:00
NealNeal
23.6k23886
answered Nov 29 '18 at 2:00
NealNeal
23.6k23886
answered Nov 29 '18 at 2:00
NealNeal
23.6k23886
23.6k23886
$begingroup$
so we get y*=0, x=0 or (A-μ)=0?
$endgroup$
– CCola
Nov 29 '18 at 5:37
add a comment |
$begingroup$
so we get y*=0, x=0 or (A-μ)=0?
$endgroup$
– CCola
Nov 29 '18 at 5:37
$begingroup$
so we get y*=0, x=0 or (A-μ)=0?
$endgroup$
– CCola
Nov 29 '18 at 5:37
$begingroup$
so we get y*=0, x=0 or (A-μ)=0?
$endgroup$
– CCola
Nov 29 '18 at 5:37
add a comment |
$begingroup$
"Can I simply prove it by letting a square matrix?" ... letting a square matrix WHAT?
$endgroup$
– David G. Stork
Nov 29 '18 at 1:53