Eigenvalues and eigenvectors in combined Linear Transformations
$begingroup$
I am new to linear algebra, and cannot work out the following question, despite the fact that I have been thinking about it for a long while.
Let $V$ be a linear space of n dimensions over R, and let $S,T:V to V$ be linear transformations.
True or False?
- If $v$ is an eigenvector of $S$ and of $T$, then $v$ is also an eigenvector of $S + T$.
- If $λ_1$ is an eigenvalue of $S$ and $λ_2$ is an eigenvalue of $T$, then $λ_1$ + $λ_2$ is a eigenvalue of $S + T$.
I am not just looking for the right answer, but also for the reasoning behind it…
Thank you!
linear-algebra diagonalization
asked Jan 19 at 14:12
daltadalta
1168
$endgroup$
add a comment |
$begingroup$
I am new to linear algebra, and cannot work out the following question, despite the fact that I have been thinking about it for a long while.
Let $V$ be a linear space of n dimensions over R, and let $S,T:V to V$ be linear transformations.
True or False?
- If $v$ is an eigenvector of $S$ and of $T$, then $v$ is also an eigenvector of $S + T$.
- If $λ_1$ is an eigenvalue of $S$ and $λ_2$ is an eigenvalue of $T$, then $λ_1$ + $λ_2$ is a eigenvalue of $S + T$.
I am not just looking for the right answer, but also for the reasoning behind it…
Thank you!
linear-algebra diagonalization
asked Jan 19 at 14:12
daltadalta
1168
$endgroup$
add a comment |
$begingroup$
I am new to linear algebra, and cannot work out the following question, despite the fact that I have been thinking about it for a long while.
Let $V$ be a linear space of n dimensions over R, and let $S,T:V to V$ be linear transformations.
True or False?
- If $v$ is an eigenvector of $S$ and of $T$, then $v$ is also an eigenvector of $S + T$.
- If $λ_1$ is an eigenvalue of $S$ and $λ_2$ is an eigenvalue of $T$, then $λ_1$ + $λ_2$ is a eigenvalue of $S + T$.
I am not just looking for the right answer, but also for the reasoning behind it…
Thank you!
linear-algebra diagonalization
asked Jan 19 at 14:12
daltadalta
1168
$endgroup$
I am new to linear algebra, and cannot work out the following question, despite the fact that I have been thinking about it for a long while.
Let $V$ be a linear space of n dimensions over R, and let $S,T:V to V$ be linear transformations.
True or False?
- If $v$ is an eigenvector of $S$ and of $T$, then $v$ is also an eigenvector of $S + T$.
- If $λ_1$ is an eigenvalue of $S$ and $λ_2$ is an eigenvalue of $T$, then $λ_1$ + $λ_2$ is a eigenvalue of $S + T$.
I am not just looking for the right answer, but also for the reasoning behind it…
Thank you!
linear-algebra diagonalization
linear-algebra diagonalization
asked Jan 19 at 14:12
daltadalta
1168
asked Jan 19 at 14:12
daltadalta
1168
asked Jan 19 at 14:12
daltadalta
1168
asked Jan 19 at 14:12
daltadalta
1168
asked Jan 19 at 14:12
daltadalta
1168
1168
add a comment |
add a comment |
2 Answers
2
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oldest
votes
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As to 1:
If $v$ is an eigenvector of $S$ then $Sv=lambda v $ for some scalar $lambda$.
If $v$ is an eigenvector of $T$ then $Tv=mu v $ for some scalar $mu$.
Then $(S+T)(v)=Sv + Tv = lambda v + mu v = (lambda +mu)v$. So $v$ is indeed an eigenvector of $S+T$ with eigenvalue $lambda+mu$.
As to 2: this reasoning won't work as we will have (in general) different eigenvectors $v$ and $w$, say, and we cannot really say anything about $Tv$ or $Sw$ etc. So there a counterexample will most likely exist. SO in a pure true/false setting go for false.
answered Jan 19 at 14:24
Henno BrandsmaHenno Brandsma
110k347116
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add a comment |
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Hints: for the first, consider the definition of eigenvalues. We have $Sv=lambda_1v$ and $Tv=lambda_2v$ for some $lambda_1, lambda_2$. What can we then say about $(S+T)(v)$? For the second, the same line of reasoning won't work. Try messing around with some common linear transformations and their eigenvalues and you'll quickly find a counter-example.
answered Jan 19 at 14:25
Yuval GatYuval Gat
570213
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As to 1:
If $v$ is an eigenvector of $S$ then $Sv=lambda v $ for some scalar $lambda$.
If $v$ is an eigenvector of $T$ then $Tv=mu v $ for some scalar $mu$.
Then $(S+T)(v)=Sv + Tv = lambda v + mu v = (lambda +mu)v$. So $v$ is indeed an eigenvector of $S+T$ with eigenvalue $lambda+mu$.
As to 2: this reasoning won't work as we will have (in general) different eigenvectors $v$ and $w$, say, and we cannot really say anything about $Tv$ or $Sw$ etc. So there a counterexample will most likely exist. SO in a pure true/false setting go for false.
answered Jan 19 at 14:24
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
As to 1:
If $v$ is an eigenvector of $S$ then $Sv=lambda v $ for some scalar $lambda$.
If $v$ is an eigenvector of $T$ then $Tv=mu v $ for some scalar $mu$.
Then $(S+T)(v)=Sv + Tv = lambda v + mu v = (lambda +mu)v$. So $v$ is indeed an eigenvector of $S+T$ with eigenvalue $lambda+mu$.
As to 2: this reasoning won't work as we will have (in general) different eigenvectors $v$ and $w$, say, and we cannot really say anything about $Tv$ or $Sw$ etc. So there a counterexample will most likely exist. SO in a pure true/false setting go for false.
answered Jan 19 at 14:24
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
As to 1:
If $v$ is an eigenvector of $S$ then $Sv=lambda v $ for some scalar $lambda$.
If $v$ is an eigenvector of $T$ then $Tv=mu v $ for some scalar $mu$.
Then $(S+T)(v)=Sv + Tv = lambda v + mu v = (lambda +mu)v$. So $v$ is indeed an eigenvector of $S+T$ with eigenvalue $lambda+mu$.
As to 2: this reasoning won't work as we will have (in general) different eigenvectors $v$ and $w$, say, and we cannot really say anything about $Tv$ or $Sw$ etc. So there a counterexample will most likely exist. SO in a pure true/false setting go for false.
answered Jan 19 at 14:24
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
As to 1:
If $v$ is an eigenvector of $S$ then $Sv=lambda v $ for some scalar $lambda$.
If $v$ is an eigenvector of $T$ then $Tv=mu v $ for some scalar $mu$.
Then $(S+T)(v)=Sv + Tv = lambda v + mu v = (lambda +mu)v$. So $v$ is indeed an eigenvector of $S+T$ with eigenvalue $lambda+mu$.
As to 2: this reasoning won't work as we will have (in general) different eigenvectors $v$ and $w$, say, and we cannot really say anything about $Tv$ or $Sw$ etc. So there a counterexample will most likely exist. SO in a pure true/false setting go for false.
answered Jan 19 at 14:24
Henno BrandsmaHenno Brandsma
110k347116
answered Jan 19 at 14:24
Henno BrandsmaHenno Brandsma
110k347116
answered Jan 19 at 14:24
Henno BrandsmaHenno Brandsma
110k347116
answered Jan 19 at 14:24
Henno BrandsmaHenno Brandsma
110k347116
110k347116
add a comment |
add a comment |
$begingroup$
Hints: for the first, consider the definition of eigenvalues. We have $Sv=lambda_1v$ and $Tv=lambda_2v$ for some $lambda_1, lambda_2$. What can we then say about $(S+T)(v)$? For the second, the same line of reasoning won't work. Try messing around with some common linear transformations and their eigenvalues and you'll quickly find a counter-example.
answered Jan 19 at 14:25
Yuval GatYuval Gat
570213
$endgroup$
add a comment |
$begingroup$
Hints: for the first, consider the definition of eigenvalues. We have $Sv=lambda_1v$ and $Tv=lambda_2v$ for some $lambda_1, lambda_2$. What can we then say about $(S+T)(v)$? For the second, the same line of reasoning won't work. Try messing around with some common linear transformations and their eigenvalues and you'll quickly find a counter-example.
answered Jan 19 at 14:25
Yuval GatYuval Gat
570213
$endgroup$
add a comment |
$begingroup$
Hints: for the first, consider the definition of eigenvalues. We have $Sv=lambda_1v$ and $Tv=lambda_2v$ for some $lambda_1, lambda_2$. What can we then say about $(S+T)(v)$? For the second, the same line of reasoning won't work. Try messing around with some common linear transformations and their eigenvalues and you'll quickly find a counter-example.
answered Jan 19 at 14:25
Yuval GatYuval Gat
570213
$endgroup$
Hints: for the first, consider the definition of eigenvalues. We have $Sv=lambda_1v$ and $Tv=lambda_2v$ for some $lambda_1, lambda_2$. What can we then say about $(S+T)(v)$? For the second, the same line of reasoning won't work. Try messing around with some common linear transformations and their eigenvalues and you'll quickly find a counter-example.
answered Jan 19 at 14:25
Yuval GatYuval Gat
570213
answered Jan 19 at 14:25
Yuval GatYuval Gat
570213
answered Jan 19 at 14:25
Yuval GatYuval Gat
570213
answered Jan 19 at 14:25
Yuval GatYuval Gat
570213
570213
add a comment |
add a comment |
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