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Let $V$ be a finite dimensional real vector space. If $S,T∈L(V,V)$ prove that $ST$ and $TS$ have the same eigenvalues whenever $T$ is an isomorphism
Isomorphisms Between a Finite-Dimensional Vector Space and its DualLet f,g be bilinear forms on a finite dimensional vector space. Show that there exist unique linear operatorsHow does the following definition of isomorphism between vector space imply “structure preserving bijection”?Isomorphism is an equivalence relation on finite dimensional vector spaces over $F$.Finite-dimensional vector space isomorphism problemInfinite-dimensional Vector space that has a finite-dimensional subspaceWhere did I use the assumption of finite-dimensional vector space? Isomorphism in a quotient spaceDoes every finite dimensional real vector space have a canonical basis?Isomorphism between functions set and the field of real number.Show that the vector space of polynomials R[x] is isomorphic to a proper subspace of itself
$begingroup$
I know the result is
true even when T is not an isomorphism but how would I show it if it was one, however.
linear-algebra vector-space-isomorphism
edited Mar 15 at 12:02
YuiTo Cheng
2,1092837
asked Mar 15 at 6:47
Kevin WangKevin Wang
241
$endgroup$
add a comment |
$begingroup$
I know the result is
true even when T is not an isomorphism but how would I show it if it was one, however.
linear-algebra vector-space-isomorphism
edited Mar 15 at 12:02
YuiTo Cheng
2,1092837
asked Mar 15 at 6:47
Kevin WangKevin Wang
241
$endgroup$
1
$begingroup$
They have the same characteristic polynomial, they have the same eigenvalues. Perhaps you should try that.
$endgroup$
– Joppy
Mar 15 at 6:50
add a comment |
$begingroup$
I know the result is
true even when T is not an isomorphism but how would I show it if it was one, however.
linear-algebra vector-space-isomorphism
edited Mar 15 at 12:02
YuiTo Cheng
2,1092837
asked Mar 15 at 6:47
Kevin WangKevin Wang
241
$endgroup$
I know the result is
true even when T is not an isomorphism but how would I show it if it was one, however.
linear-algebra vector-space-isomorphism
linear-algebra vector-space-isomorphism
edited Mar 15 at 12:02
YuiTo Cheng
2,1092837
asked Mar 15 at 6:47
Kevin WangKevin Wang
241
edited Mar 15 at 12:02
YuiTo Cheng
2,1092837
asked Mar 15 at 6:47
Kevin WangKevin Wang
241
edited Mar 15 at 12:02
YuiTo Cheng
2,1092837
edited Mar 15 at 12:02
YuiTo Cheng
2,1092837
edited Mar 15 at 12:02
YuiTo Cheng
2,1092837
2,1092837
asked Mar 15 at 6:47
Kevin WangKevin Wang
241
asked Mar 15 at 6:47
Kevin WangKevin Wang
241
asked Mar 15 at 6:47
Kevin WangKevin Wang
241
241
1
$begingroup$
They have the same characteristic polynomial, they have the same eigenvalues. Perhaps you should try that.
$endgroup$
– Joppy
Mar 15 at 6:50
add a comment |
1
$begingroup$
They have the same characteristic polynomial, they have the same eigenvalues. Perhaps you should try that.
$endgroup$
– Joppy
Mar 15 at 6:50
1
1
$begingroup$
They have the same characteristic polynomial, they have the same eigenvalues. Perhaps you should try that.
$endgroup$
– Joppy
Mar 15 at 6:50
$begingroup$
They have the same characteristic polynomial, they have the same eigenvalues. Perhaps you should try that.
$endgroup$
– Joppy
Mar 15 at 6:50
add a comment |
1 Answer
1
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oldest
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$begingroup$
We have
$$lambda I-TS=T(lambda I-ST)T^-1.$$
Can you take it from here ?
answered Mar 15 at 6:59
FredFred
48.6k11849
$endgroup$
add a comment |
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$begingroup$
We have
$$lambda I-TS=T(lambda I-ST)T^-1.$$
Can you take it from here ?
answered Mar 15 at 6:59
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
We have
$$lambda I-TS=T(lambda I-ST)T^-1.$$
Can you take it from here ?
answered Mar 15 at 6:59
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
We have
$$lambda I-TS=T(lambda I-ST)T^-1.$$
Can you take it from here ?
answered Mar 15 at 6:59
FredFred
48.6k11849
$endgroup$
We have
$$lambda I-TS=T(lambda I-ST)T^-1.$$
Can you take it from here ?
answered Mar 15 at 6:59
FredFred
48.6k11849
answered Mar 15 at 6:59
FredFred
48.6k11849
answered Mar 15 at 6:59
FredFred
48.6k11849
answered Mar 15 at 6:59
FredFred
48.6k11849
48.6k11849
add a comment |
add a comment |
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$begingroup$
They have the same characteristic polynomial, they have the same eigenvalues. Perhaps you should try that.
$endgroup$
– Joppy
Mar 15 at 6:50