Mixing
Suppose $v,w, v + w$ are all eigenvectors of the linear operator $phi:V to V$. Prove that $v, w, v + w$ all...
$begingroup$
The Problem:
Just as it is in the title: Suppose $v,w, v + w$ are all eigenvectors of the linear operator $phi:V to V$. Prove that $v, w, v + w$ all have the same eigenvalue.
My Approach:
Let $phi(v) = alpha v$, $phi(w) = beta w$, and $phi(v + w) = gamma (v + w)$. We then have that
$$ phi(v) + phi(w) = gamma v + gamma w implies alpha v + beta w = gamma v + gamma w implies (alpha - gamma)v = (gamma - beta)w. $$
This means that $v$ and $w$ are scalar multiplies of one another; say, $w = lambda v$...
I feel like this is supposed to tell me something. My thought is that, since $v$ and $w$ are linearly dependent, they must occupy the same eigenspace; but I can't seem to prove that...
linear-algebra eigenvalues-eigenvectors
asked Jan 22 at 14:22
thisisourconcerndudethisisourconcerndude
1,1321122
$endgroup$
add a comment |
$begingroup$
The Problem:
Just as it is in the title: Suppose $v,w, v + w$ are all eigenvectors of the linear operator $phi:V to V$. Prove that $v, w, v + w$ all have the same eigenvalue.
My Approach:
Let $phi(v) = alpha v$, $phi(w) = beta w$, and $phi(v + w) = gamma (v + w)$. We then have that
$$ phi(v) + phi(w) = gamma v + gamma w implies alpha v + beta w = gamma v + gamma w implies (alpha - gamma)v = (gamma - beta)w. $$
This means that $v$ and $w$ are scalar multiplies of one another; say, $w = lambda v$...
I feel like this is supposed to tell me something. My thought is that, since $v$ and $w$ are linearly dependent, they must occupy the same eigenspace; but I can't seem to prove that...
linear-algebra eigenvalues-eigenvectors
asked Jan 22 at 14:22
thisisourconcerndudethisisourconcerndude
1,1321122
$endgroup$
add a comment |
$begingroup$
The Problem:
Just as it is in the title: Suppose $v,w, v + w$ are all eigenvectors of the linear operator $phi:V to V$. Prove that $v, w, v + w$ all have the same eigenvalue.
My Approach:
Let $phi(v) = alpha v$, $phi(w) = beta w$, and $phi(v + w) = gamma (v + w)$. We then have that
$$ phi(v) + phi(w) = gamma v + gamma w implies alpha v + beta w = gamma v + gamma w implies (alpha - gamma)v = (gamma - beta)w. $$
This means that $v$ and $w$ are scalar multiplies of one another; say, $w = lambda v$...
I feel like this is supposed to tell me something. My thought is that, since $v$ and $w$ are linearly dependent, they must occupy the same eigenspace; but I can't seem to prove that...
linear-algebra eigenvalues-eigenvectors
asked Jan 22 at 14:22
thisisourconcerndudethisisourconcerndude
1,1321122
$endgroup$
The Problem:
Just as it is in the title: Suppose $v,w, v + w$ are all eigenvectors of the linear operator $phi:V to V$. Prove that $v, w, v + w$ all have the same eigenvalue.
My Approach:
Let $phi(v) = alpha v$, $phi(w) = beta w$, and $phi(v + w) = gamma (v + w)$. We then have that
$$ phi(v) + phi(w) = gamma v + gamma w implies alpha v + beta w = gamma v + gamma w implies (alpha - gamma)v = (gamma - beta)w. $$
This means that $v$ and $w$ are scalar multiplies of one another; say, $w = lambda v$...
I feel like this is supposed to tell me something. My thought is that, since $v$ and $w$ are linearly dependent, they must occupy the same eigenspace; but I can't seem to prove that...
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Jan 22 at 14:22
thisisourconcerndudethisisourconcerndude
1,1321122
asked Jan 22 at 14:22
thisisourconcerndudethisisourconcerndude
1,1321122
asked Jan 22 at 14:22
thisisourconcerndudethisisourconcerndude
1,1321122
asked Jan 22 at 14:22
thisisourconcerndudethisisourconcerndude
1,1321122
asked Jan 22 at 14:22
thisisourconcerndudethisisourconcerndude
1,1321122
1,1321122
add a comment |
add a comment |
4 Answers
4
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$begingroup$
From $(alpha-gamma)v=(gamma-beta)w$ you cannot necessarily conclude that $v$ and $w$ are scalar multiples of one another. There are two possibilities:
- If $alpha-gamma=0$, then the LHS is the zero vector. Since $w neq 0$ (it is an eigenvector), it must also be that $gamma-beta=0$. So $alpha=beta=gamma$.
- If $alpha-gamma neq 0$, then the LHS is not zero, so the RHS is not zero either (in particlar $gamma-beta neq 0$). So $v=frac{gamma-beta}{alpha-gamma} w = cw$ where $c=frac{gamma-beta}{alpha-gamma} neq 0$. Now re-write the equation $phi(v)=alpha v$ using $v=cw$, and similarly rewrite $phi(v+w)=gamma(v+w)$. This will show you $alpha=beta=gamma$.
answered Jan 22 at 14:29
kccukccu
10.6k11229
$endgroup$
add a comment |
$begingroup$
Apply $phi$ once more on $(alpha-gamma)v = (gamma-beta)w$ to obtain
$$(alpha-gamma)alpha v = (gamma-beta)beta w$$
On the other hand, multiplying the first identity by $alpha$ yields $$(alpha-gamma)alpha v = (gamma-beta)alpha w$$
so $$(gamma-beta)alpha w = (gamma-beta)beta w implies (gamma-beta)(alpha-beta)w = 0$$
Since $w$ is an eigenvector, we have $w ne 0$ so $(gamma-beta)(alpha-beta)=0$.
Hence $alpha = beta$ or $beta = gamma$. From either of those it easily follows $alpha=beta=gamma$.
answered Jan 22 at 14:32
mechanodroidmechanodroid
28.5k62548
$endgroup$
add a comment |
$begingroup$
If $v,w$ are not linearly independent, the result is trivial $w=cv,phi(w)=phi(cv)=cphi(v)=calpha v=alpha w$,...
$(alpha-gamma)v+(beta-gamma)w=$ since $v,w$ are linearly independent, $alpha=beta=gamma$.
answered Jan 22 at 14:25
Tsemo AristideTsemo Aristide
59.2k11445
$endgroup$
$begingroup$
You don't know that $v$ and $w$ are linearly independent.
$endgroup$
– kccu
Jan 22 at 14:25
add a comment |
$begingroup$
Suppose $phi(v) = lambda v$, $phi(w) = mu w$ and that $phi(v + w) = kappa(v + w)$, where $lambda$, $mu$, and $kappa$ are all scalars. Then we ahve
$$ lambda v + mu w = T(v + w) = kappa(v + w).$$
We therefore deduce that
$$ (kappa - lambda) v + (kappa - mu) w = 0.$$
If $v$ and $w$ are linearly independant, then $kappa=lambda=mu$. If not, then $v$ and $w$ are scalar multiples; in this case they are in the same eigenspace, and the result folllows.
answered Jan 22 at 14:34
ncmathsadistncmathsadist
42.9k260103
$endgroup$
add a comment |
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4 Answers
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4 Answers
4
active
oldest
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$begingroup$
From $(alpha-gamma)v=(gamma-beta)w$ you cannot necessarily conclude that $v$ and $w$ are scalar multiples of one another. There are two possibilities:
- If $alpha-gamma=0$, then the LHS is the zero vector. Since $w neq 0$ (it is an eigenvector), it must also be that $gamma-beta=0$. So $alpha=beta=gamma$.
- If $alpha-gamma neq 0$, then the LHS is not zero, so the RHS is not zero either (in particlar $gamma-beta neq 0$). So $v=frac{gamma-beta}{alpha-gamma} w = cw$ where $c=frac{gamma-beta}{alpha-gamma} neq 0$. Now re-write the equation $phi(v)=alpha v$ using $v=cw$, and similarly rewrite $phi(v+w)=gamma(v+w)$. This will show you $alpha=beta=gamma$.
answered Jan 22 at 14:29
kccukccu
10.6k11229
$endgroup$
add a comment |
$begingroup$
From $(alpha-gamma)v=(gamma-beta)w$ you cannot necessarily conclude that $v$ and $w$ are scalar multiples of one another. There are two possibilities:
- If $alpha-gamma=0$, then the LHS is the zero vector. Since $w neq 0$ (it is an eigenvector), it must also be that $gamma-beta=0$. So $alpha=beta=gamma$.
- If $alpha-gamma neq 0$, then the LHS is not zero, so the RHS is not zero either (in particlar $gamma-beta neq 0$). So $v=frac{gamma-beta}{alpha-gamma} w = cw$ where $c=frac{gamma-beta}{alpha-gamma} neq 0$. Now re-write the equation $phi(v)=alpha v$ using $v=cw$, and similarly rewrite $phi(v+w)=gamma(v+w)$. This will show you $alpha=beta=gamma$.
answered Jan 22 at 14:29
kccukccu
10.6k11229
$endgroup$
add a comment |
$begingroup$
From $(alpha-gamma)v=(gamma-beta)w$ you cannot necessarily conclude that $v$ and $w$ are scalar multiples of one another. There are two possibilities:
- If $alpha-gamma=0$, then the LHS is the zero vector. Since $w neq 0$ (it is an eigenvector), it must also be that $gamma-beta=0$. So $alpha=beta=gamma$.
- If $alpha-gamma neq 0$, then the LHS is not zero, so the RHS is not zero either (in particlar $gamma-beta neq 0$). So $v=frac{gamma-beta}{alpha-gamma} w = cw$ where $c=frac{gamma-beta}{alpha-gamma} neq 0$. Now re-write the equation $phi(v)=alpha v$ using $v=cw$, and similarly rewrite $phi(v+w)=gamma(v+w)$. This will show you $alpha=beta=gamma$.
answered Jan 22 at 14:29
kccukccu
10.6k11229
$endgroup$
From $(alpha-gamma)v=(gamma-beta)w$ you cannot necessarily conclude that $v$ and $w$ are scalar multiples of one another. There are two possibilities:
- If $alpha-gamma=0$, then the LHS is the zero vector. Since $w neq 0$ (it is an eigenvector), it must also be that $gamma-beta=0$. So $alpha=beta=gamma$.
- If $alpha-gamma neq 0$, then the LHS is not zero, so the RHS is not zero either (in particlar $gamma-beta neq 0$). So $v=frac{gamma-beta}{alpha-gamma} w = cw$ where $c=frac{gamma-beta}{alpha-gamma} neq 0$. Now re-write the equation $phi(v)=alpha v$ using $v=cw$, and similarly rewrite $phi(v+w)=gamma(v+w)$. This will show you $alpha=beta=gamma$.
answered Jan 22 at 14:29
kccukccu
10.6k11229
answered Jan 22 at 14:29
kccukccu
10.6k11229
answered Jan 22 at 14:29
kccukccu
10.6k11229
answered Jan 22 at 14:29
kccukccu
10.6k11229
10.6k11229
add a comment |
add a comment |
$begingroup$
Apply $phi$ once more on $(alpha-gamma)v = (gamma-beta)w$ to obtain
$$(alpha-gamma)alpha v = (gamma-beta)beta w$$
On the other hand, multiplying the first identity by $alpha$ yields $$(alpha-gamma)alpha v = (gamma-beta)alpha w$$
so $$(gamma-beta)alpha w = (gamma-beta)beta w implies (gamma-beta)(alpha-beta)w = 0$$
Since $w$ is an eigenvector, we have $w ne 0$ so $(gamma-beta)(alpha-beta)=0$.
Hence $alpha = beta$ or $beta = gamma$. From either of those it easily follows $alpha=beta=gamma$.
answered Jan 22 at 14:32
mechanodroidmechanodroid
28.5k62548
$endgroup$
add a comment |
$begingroup$
Apply $phi$ once more on $(alpha-gamma)v = (gamma-beta)w$ to obtain
$$(alpha-gamma)alpha v = (gamma-beta)beta w$$
On the other hand, multiplying the first identity by $alpha$ yields $$(alpha-gamma)alpha v = (gamma-beta)alpha w$$
so $$(gamma-beta)alpha w = (gamma-beta)beta w implies (gamma-beta)(alpha-beta)w = 0$$
Since $w$ is an eigenvector, we have $w ne 0$ so $(gamma-beta)(alpha-beta)=0$.
Hence $alpha = beta$ or $beta = gamma$. From either of those it easily follows $alpha=beta=gamma$.
answered Jan 22 at 14:32
mechanodroidmechanodroid
28.5k62548
$endgroup$
add a comment |
$begingroup$
Apply $phi$ once more on $(alpha-gamma)v = (gamma-beta)w$ to obtain
$$(alpha-gamma)alpha v = (gamma-beta)beta w$$
On the other hand, multiplying the first identity by $alpha$ yields $$(alpha-gamma)alpha v = (gamma-beta)alpha w$$
so $$(gamma-beta)alpha w = (gamma-beta)beta w implies (gamma-beta)(alpha-beta)w = 0$$
Since $w$ is an eigenvector, we have $w ne 0$ so $(gamma-beta)(alpha-beta)=0$.
Hence $alpha = beta$ or $beta = gamma$. From either of those it easily follows $alpha=beta=gamma$.
answered Jan 22 at 14:32
mechanodroidmechanodroid
28.5k62548
$endgroup$
Apply $phi$ once more on $(alpha-gamma)v = (gamma-beta)w$ to obtain
$$(alpha-gamma)alpha v = (gamma-beta)beta w$$
On the other hand, multiplying the first identity by $alpha$ yields $$(alpha-gamma)alpha v = (gamma-beta)alpha w$$
so $$(gamma-beta)alpha w = (gamma-beta)beta w implies (gamma-beta)(alpha-beta)w = 0$$
Since $w$ is an eigenvector, we have $w ne 0$ so $(gamma-beta)(alpha-beta)=0$.
Hence $alpha = beta$ or $beta = gamma$. From either of those it easily follows $alpha=beta=gamma$.
answered Jan 22 at 14:32
mechanodroidmechanodroid
28.5k62548
answered Jan 22 at 14:32
mechanodroidmechanodroid
28.5k62548
answered Jan 22 at 14:32
mechanodroidmechanodroid
28.5k62548
answered Jan 22 at 14:32
mechanodroidmechanodroid
28.5k62548
28.5k62548
add a comment |
add a comment |
$begingroup$
If $v,w$ are not linearly independent, the result is trivial $w=cv,phi(w)=phi(cv)=cphi(v)=calpha v=alpha w$,...
$(alpha-gamma)v+(beta-gamma)w=$ since $v,w$ are linearly independent, $alpha=beta=gamma$.
answered Jan 22 at 14:25
Tsemo AristideTsemo Aristide
59.2k11445
$endgroup$
$begingroup$
You don't know that $v$ and $w$ are linearly independent.
$endgroup$
– kccu
Jan 22 at 14:25
add a comment |
$begingroup$
If $v,w$ are not linearly independent, the result is trivial $w=cv,phi(w)=phi(cv)=cphi(v)=calpha v=alpha w$,...
$(alpha-gamma)v+(beta-gamma)w=$ since $v,w$ are linearly independent, $alpha=beta=gamma$.
answered Jan 22 at 14:25
Tsemo AristideTsemo Aristide
59.2k11445
$endgroup$
$begingroup$
You don't know that $v$ and $w$ are linearly independent.
$endgroup$
– kccu
Jan 22 at 14:25
add a comment |
$begingroup$
If $v,w$ are not linearly independent, the result is trivial $w=cv,phi(w)=phi(cv)=cphi(v)=calpha v=alpha w$,...
$(alpha-gamma)v+(beta-gamma)w=$ since $v,w$ are linearly independent, $alpha=beta=gamma$.
answered Jan 22 at 14:25
Tsemo AristideTsemo Aristide
59.2k11445
$endgroup$
If $v,w$ are not linearly independent, the result is trivial $w=cv,phi(w)=phi(cv)=cphi(v)=calpha v=alpha w$,...
$(alpha-gamma)v+(beta-gamma)w=$ since $v,w$ are linearly independent, $alpha=beta=gamma$.
answered Jan 22 at 14:25
Tsemo AristideTsemo Aristide
59.2k11445
edited Jan 22 at 14:26
answered Jan 22 at 14:25
Tsemo AristideTsemo Aristide
59.2k11445
answered Jan 22 at 14:25
Tsemo AristideTsemo Aristide
59.2k11445
answered Jan 22 at 14:25
Tsemo AristideTsemo Aristide
59.2k11445
59.2k11445
$begingroup$
You don't know that $v$ and $w$ are linearly independent.
$endgroup$
– kccu
Jan 22 at 14:25
add a comment |
$begingroup$
You don't know that $v$ and $w$ are linearly independent.
$endgroup$
– kccu
Jan 22 at 14:25
$begingroup$
You don't know that $v$ and $w$ are linearly independent.
$endgroup$
– kccu
Jan 22 at 14:25
$begingroup$
You don't know that $v$ and $w$ are linearly independent.
$endgroup$
– kccu
Jan 22 at 14:25
add a comment |
$begingroup$
Suppose $phi(v) = lambda v$, $phi(w) = mu w$ and that $phi(v + w) = kappa(v + w)$, where $lambda$, $mu$, and $kappa$ are all scalars. Then we ahve
$$ lambda v + mu w = T(v + w) = kappa(v + w).$$
We therefore deduce that
$$ (kappa - lambda) v + (kappa - mu) w = 0.$$
If $v$ and $w$ are linearly independant, then $kappa=lambda=mu$. If not, then $v$ and $w$ are scalar multiples; in this case they are in the same eigenspace, and the result folllows.
answered Jan 22 at 14:34
ncmathsadistncmathsadist
42.9k260103
$endgroup$
add a comment |
$begingroup$
Suppose $phi(v) = lambda v$, $phi(w) = mu w$ and that $phi(v + w) = kappa(v + w)$, where $lambda$, $mu$, and $kappa$ are all scalars. Then we ahve
$$ lambda v + mu w = T(v + w) = kappa(v + w).$$
We therefore deduce that
$$ (kappa - lambda) v + (kappa - mu) w = 0.$$
If $v$ and $w$ are linearly independant, then $kappa=lambda=mu$. If not, then $v$ and $w$ are scalar multiples; in this case they are in the same eigenspace, and the result folllows.
answered Jan 22 at 14:34
ncmathsadistncmathsadist
42.9k260103
$endgroup$
add a comment |
$begingroup$
Suppose $phi(v) = lambda v$, $phi(w) = mu w$ and that $phi(v + w) = kappa(v + w)$, where $lambda$, $mu$, and $kappa$ are all scalars. Then we ahve
$$ lambda v + mu w = T(v + w) = kappa(v + w).$$
We therefore deduce that
$$ (kappa - lambda) v + (kappa - mu) w = 0.$$
If $v$ and $w$ are linearly independant, then $kappa=lambda=mu$. If not, then $v$ and $w$ are scalar multiples; in this case they are in the same eigenspace, and the result folllows.
answered Jan 22 at 14:34
ncmathsadistncmathsadist
42.9k260103
$endgroup$
Suppose $phi(v) = lambda v$, $phi(w) = mu w$ and that $phi(v + w) = kappa(v + w)$, where $lambda$, $mu$, and $kappa$ are all scalars. Then we ahve
$$ lambda v + mu w = T(v + w) = kappa(v + w).$$
We therefore deduce that
$$ (kappa - lambda) v + (kappa - mu) w = 0.$$
If $v$ and $w$ are linearly independant, then $kappa=lambda=mu$. If not, then $v$ and $w$ are scalar multiples; in this case they are in the same eigenspace, and the result folllows.
answered Jan 22 at 14:34
ncmathsadistncmathsadist
42.9k260103
answered Jan 22 at 14:34
ncmathsadistncmathsadist
42.9k260103
answered Jan 22 at 14:34
ncmathsadistncmathsadist
42.9k260103
answered Jan 22 at 14:34
ncmathsadistncmathsadist
42.9k260103
42.9k260103
add a comment |
add a comment |
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