Eigenvalues of $ST$ and $TS$ The Next CEO of Stack OverflowEigenvalues and eigenvectors in physics?If $m_1,m_2$ are the minimal polynomials of $ST$ and $TS$ prove $m_2(x)=x^im_1(x)$ where $i=-1,0$ or $1$Finding Eigenvalues and Eigenspaces of Linear OperartorsIf a linear transformation is similar to another, then they have the same eigenvalues.Is the following proof that if $T: V rightarrow V$ as eigenvalue $lambda$, then $aT$ has eigenvalue $alambda$ correctLinear dependency and eigenvaluesProperties of eigenvalues/eigenvectorsEigenvalues-Eigenvectors and Rotation (in $mathbbR^2$)Finding all eigenvalues and eigenvectors by proofingeigenvalues Linear Transformation proof
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Eigenvalues of $ST$ and $TS$
The Next CEO of Stack OverflowEigenvalues and eigenvectors in physics?If $m_1,m_2$ are the minimal polynomials of $ST$ and $TS$ prove $m_2(x)=x^im_1(x)$ where $i=-1,0$ or $1$Finding Eigenvalues and Eigenspaces of Linear OperartorsIf a linear transformation is similar to another, then they have the same eigenvalues.Is the following proof that if $T: V rightarrow V$ as eigenvalue $lambda$, then $aT$ has eigenvalue $alambda$ correctLinear dependency and eigenvaluesProperties of eigenvalues/eigenvectorsEigenvalues-Eigenvectors and Rotation (in $mathbbR^2$)Finding all eigenvalues and eigenvectors by proofingeigenvalues Linear Transformation proof
$begingroup$
Given linear transformations $S,T:V→V$, prove that if $x=0$ is an eigenvalue of $ST$, then it is an eigenvalue of $TS$.
Secondly, prove that if $yneq 0$ is an eigenvalue of $ST$, and that if $v$ is eigenvector of $ST$ for $y$, then $Tv$ is an eigenvector of $TS$ for $y$.
eigenvalues-eigenvectors
edited Mar 27 at 18:53
K.Power
3,630926
asked Mar 27 at 18:20
ga asga as
72
$endgroup$
add a comment |
$begingroup$
Given linear transformations $S,T:V→V$, prove that if $x=0$ is an eigenvalue of $ST$, then it is an eigenvalue of $TS$.
Secondly, prove that if $yneq 0$ is an eigenvalue of $ST$, and that if $v$ is eigenvector of $ST$ for $y$, then $Tv$ is an eigenvector of $TS$ for $y$.
eigenvalues-eigenvectors
edited Mar 27 at 18:53
K.Power
3,630926
asked Mar 27 at 18:20
ga asga as
72
$endgroup$
$begingroup$
What have you tried?
$endgroup$
– K.Power
Mar 27 at 18:25
$begingroup$
In future please use MathJax to format your posts.
$endgroup$
– K.Power
Mar 27 at 18:29
$begingroup$
Anyone can help?
$endgroup$
– ga as
Mar 27 at 19:50
$begingroup$
You're more likely to get an answer if you provide some context, such as showing any ideas you have
$endgroup$
– K.Power
Mar 27 at 19:51
add a comment |
$begingroup$
Given linear transformations $S,T:V→V$, prove that if $x=0$ is an eigenvalue of $ST$, then it is an eigenvalue of $TS$.
Secondly, prove that if $yneq 0$ is an eigenvalue of $ST$, and that if $v$ is eigenvector of $ST$ for $y$, then $Tv$ is an eigenvector of $TS$ for $y$.
eigenvalues-eigenvectors
edited Mar 27 at 18:53
K.Power
3,630926
asked Mar 27 at 18:20
ga asga as
72
$endgroup$
Given linear transformations $S,T:V→V$, prove that if $x=0$ is an eigenvalue of $ST$, then it is an eigenvalue of $TS$.
Secondly, prove that if $yneq 0$ is an eigenvalue of $ST$, and that if $v$ is eigenvector of $ST$ for $y$, then $Tv$ is an eigenvector of $TS$ for $y$.
eigenvalues-eigenvectors
eigenvalues-eigenvectors
edited Mar 27 at 18:53
K.Power
3,630926
asked Mar 27 at 18:20
ga asga as
72
edited Mar 27 at 18:53
K.Power
3,630926
asked Mar 27 at 18:20
ga asga as
72
edited Mar 27 at 18:53
K.Power
3,630926
edited Mar 27 at 18:53
K.Power
3,630926
edited Mar 27 at 18:53
K.Power
3,630926
3,630926
asked Mar 27 at 18:20
ga asga as
72
asked Mar 27 at 18:20
ga asga as
72
asked Mar 27 at 18:20
ga asga as
72
72
$begingroup$
What have you tried?
$endgroup$
– K.Power
Mar 27 at 18:25
$begingroup$
In future please use MathJax to format your posts.
$endgroup$
– K.Power
Mar 27 at 18:29
$begingroup$
Anyone can help?
$endgroup$
– ga as
Mar 27 at 19:50
$begingroup$
You're more likely to get an answer if you provide some context, such as showing any ideas you have
$endgroup$
– K.Power
Mar 27 at 19:51
add a comment |
$begingroup$
What have you tried?
$endgroup$
– K.Power
Mar 27 at 18:25
$begingroup$
In future please use MathJax to format your posts.
$endgroup$
– K.Power
Mar 27 at 18:29
$begingroup$
Anyone can help?
$endgroup$
– ga as
Mar 27 at 19:50
$begingroup$
You're more likely to get an answer if you provide some context, such as showing any ideas you have
$endgroup$
– K.Power
Mar 27 at 19:51
$begingroup$
What have you tried?
$endgroup$
– K.Power
Mar 27 at 18:25
$begingroup$
What have you tried?
$endgroup$
– K.Power
Mar 27 at 18:25
$begingroup$
In future please use MathJax to format your posts.
$endgroup$
– K.Power
Mar 27 at 18:29
$begingroup$
In future please use MathJax to format your posts.
$endgroup$
– K.Power
Mar 27 at 18:29
$begingroup$
Anyone can help?
$endgroup$
– ga as
Mar 27 at 19:50
$begingroup$
Anyone can help?
$endgroup$
– ga as
Mar 27 at 19:50
$begingroup$
You're more likely to get an answer if you provide some context, such as showing any ideas you have
$endgroup$
– K.Power
Mar 27 at 19:51
$begingroup$
You're more likely to get an answer if you provide some context, such as showing any ideas you have
$endgroup$
– K.Power
Mar 27 at 19:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For this first part, there may be a simpler approach, but this is what came to mind. Since $ST$ has zero as an eigenvalue, we have
$$det S det T = det ST = 0.$$
So, either $S$ or $T$ (or both) is singular. If $S$ is singular, take $v_0 neq 0$ so that $Sv_0=0$. Then,
$$TSv_0 = Tvec0 = vec0.$$
Suppose $T$ is singular. If $S$ is also singular, then, by the above, we are done. Suppose $S$ is not singular. Take $v_0 neq 0$ so that $Tv_0 = 0$ and pick $w_0$ so that $Sw_0=v_0$.
For the next part,
$$TS(Tv) = T(STv) = T(yv) = yTv,$$
which implies that $Tv$ is an eigenvector of $TS$ with corresponding eigenvalue $y$.
answered Mar 27 at 19:54
Gary MoonGary Moon
92127
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For this first part, there may be a simpler approach, but this is what came to mind. Since $ST$ has zero as an eigenvalue, we have
$$det S det T = det ST = 0.$$
So, either $S$ or $T$ (or both) is singular. If $S$ is singular, take $v_0 neq 0$ so that $Sv_0=0$. Then,
$$TSv_0 = Tvec0 = vec0.$$
Suppose $T$ is singular. If $S$ is also singular, then, by the above, we are done. Suppose $S$ is not singular. Take $v_0 neq 0$ so that $Tv_0 = 0$ and pick $w_0$ so that $Sw_0=v_0$.
For the next part,
$$TS(Tv) = T(STv) = T(yv) = yTv,$$
which implies that $Tv$ is an eigenvector of $TS$ with corresponding eigenvalue $y$.
answered Mar 27 at 19:54
Gary MoonGary Moon
92127
$endgroup$
add a comment |
$begingroup$
For this first part, there may be a simpler approach, but this is what came to mind. Since $ST$ has zero as an eigenvalue, we have
$$det S det T = det ST = 0.$$
So, either $S$ or $T$ (or both) is singular. If $S$ is singular, take $v_0 neq 0$ so that $Sv_0=0$. Then,
$$TSv_0 = Tvec0 = vec0.$$
Suppose $T$ is singular. If $S$ is also singular, then, by the above, we are done. Suppose $S$ is not singular. Take $v_0 neq 0$ so that $Tv_0 = 0$ and pick $w_0$ so that $Sw_0=v_0$.
For the next part,
$$TS(Tv) = T(STv) = T(yv) = yTv,$$
which implies that $Tv$ is an eigenvector of $TS$ with corresponding eigenvalue $y$.
answered Mar 27 at 19:54
Gary MoonGary Moon
92127
$endgroup$
add a comment |
$begingroup$
For this first part, there may be a simpler approach, but this is what came to mind. Since $ST$ has zero as an eigenvalue, we have
$$det S det T = det ST = 0.$$
So, either $S$ or $T$ (or both) is singular. If $S$ is singular, take $v_0 neq 0$ so that $Sv_0=0$. Then,
$$TSv_0 = Tvec0 = vec0.$$
Suppose $T$ is singular. If $S$ is also singular, then, by the above, we are done. Suppose $S$ is not singular. Take $v_0 neq 0$ so that $Tv_0 = 0$ and pick $w_0$ so that $Sw_0=v_0$.
For the next part,
$$TS(Tv) = T(STv) = T(yv) = yTv,$$
which implies that $Tv$ is an eigenvector of $TS$ with corresponding eigenvalue $y$.
answered Mar 27 at 19:54
Gary MoonGary Moon
92127
$endgroup$
For this first part, there may be a simpler approach, but this is what came to mind. Since $ST$ has zero as an eigenvalue, we have
$$det S det T = det ST = 0.$$
So, either $S$ or $T$ (or both) is singular. If $S$ is singular, take $v_0 neq 0$ so that $Sv_0=0$. Then,
$$TSv_0 = Tvec0 = vec0.$$
Suppose $T$ is singular. If $S$ is also singular, then, by the above, we are done. Suppose $S$ is not singular. Take $v_0 neq 0$ so that $Tv_0 = 0$ and pick $w_0$ so that $Sw_0=v_0$.
For the next part,
$$TS(Tv) = T(STv) = T(yv) = yTv,$$
which implies that $Tv$ is an eigenvector of $TS$ with corresponding eigenvalue $y$.
answered Mar 27 at 19:54
Gary MoonGary Moon
92127
edited Mar 27 at 20:05
answered Mar 27 at 19:54
Gary MoonGary Moon
92127
answered Mar 27 at 19:54
Gary MoonGary Moon
92127
answered Mar 27 at 19:54
Gary MoonGary Moon
92127
92127
add a comment |
add a comment |
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$begingroup$
What have you tried?
$endgroup$
– K.Power
Mar 27 at 18:25
$begingroup$
In future please use MathJax to format your posts.
$endgroup$
– K.Power
Mar 27 at 18:29
$begingroup$
Anyone can help?
$endgroup$
– ga as
Mar 27 at 19:50
$begingroup$
You're more likely to get an answer if you provide some context, such as showing any ideas you have
$endgroup$
– K.Power
Mar 27 at 19:51