How can I prove that a linear map $T:Vto V$ such that $T^2 = I$ is equal to the Identity map?
$begingroup$
Suppose $T in mathcal{L}(V)$, $T^2=I$, and $-1$ is not an eigenvalue of $T$. Prove that $T=I$.
I'm not sure where to start, though I know I somehow have to show that $T circ T$ yields the identity matrix, but there isn't much information given.
linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations
edited Dec 3 '18 at 23:46
Batominovski
33.1k33293
asked Dec 3 '18 at 22:34
JaigusJaigus
2259
$endgroup$
add a comment |
$begingroup$
Suppose $T in mathcal{L}(V)$, $T^2=I$, and $-1$ is not an eigenvalue of $T$. Prove that $T=I$.
I'm not sure where to start, though I know I somehow have to show that $T circ T$ yields the identity matrix, but there isn't much information given.
linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations
edited Dec 3 '18 at 23:46
Batominovski
33.1k33293
asked Dec 3 '18 at 22:34
JaigusJaigus
2259
$endgroup$
$begingroup$
I eidited the title of your post to make it better conform to the problem presented in the body. Cheers!
$endgroup$
– Robert Lewis
Dec 3 '18 at 23:07
add a comment |
$begingroup$
Suppose $T in mathcal{L}(V)$, $T^2=I$, and $-1$ is not an eigenvalue of $T$. Prove that $T=I$.
I'm not sure where to start, though I know I somehow have to show that $T circ T$ yields the identity matrix, but there isn't much information given.
linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations
edited Dec 3 '18 at 23:46
Batominovski
33.1k33293
asked Dec 3 '18 at 22:34
JaigusJaigus
2259
$endgroup$
Suppose $T in mathcal{L}(V)$, $T^2=I$, and $-1$ is not an eigenvalue of $T$. Prove that $T=I$.
I'm not sure where to start, though I know I somehow have to show that $T circ T$ yields the identity matrix, but there isn't much information given.
linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations
linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations
edited Dec 3 '18 at 23:46
Batominovski
33.1k33293
asked Dec 3 '18 at 22:34
JaigusJaigus
2259
edited Dec 3 '18 at 23:46
Batominovski
33.1k33293
asked Dec 3 '18 at 22:34
JaigusJaigus
2259
edited Dec 3 '18 at 23:46
Batominovski
33.1k33293
edited Dec 3 '18 at 23:46
Batominovski
33.1k33293
edited Dec 3 '18 at 23:46
Batominovski
33.1k33293
33.1k33293
asked Dec 3 '18 at 22:34
JaigusJaigus
2259
asked Dec 3 '18 at 22:34
JaigusJaigus
2259
asked Dec 3 '18 at 22:34
JaigusJaigus
2259
2259
$begingroup$
I eidited the title of your post to make it better conform to the problem presented in the body. Cheers!
$endgroup$
– Robert Lewis
Dec 3 '18 at 23:07
add a comment |
$begingroup$
I eidited the title of your post to make it better conform to the problem presented in the body. Cheers!
$endgroup$
– Robert Lewis
Dec 3 '18 at 23:07
$begingroup$
I eidited the title of your post to make it better conform to the problem presented in the body. Cheers!
$endgroup$
– Robert Lewis
Dec 3 '18 at 23:07
$begingroup$
I eidited the title of your post to make it better conform to the problem presented in the body. Cheers!
$endgroup$
– Robert Lewis
Dec 3 '18 at 23:07
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: $0 = T^2-I = (T-I)(T+I)$, with $T+I$ invertible (why?).
... or this:
Hint: $0 = T^2-I = (T+I)(T-I)$, with $T+I$ injective (why?).
answered Dec 3 '18 at 22:36
lhflhf
165k10171396
$endgroup$
$begingroup$
I understand the expansion but, I'm not sure how to reason that $T+I$ is invertible. Since it's given that -1 is not an eigenvalue for T, is $T=I$ the only solution for $(T-I)(T+I)$ because of that?
$endgroup$
– Jaigus
Dec 3 '18 at 23:04
$begingroup$
@Jaigus not quite. Because $-1$ is not an eigenvalue, the only solution for $(T+I)x = 0$ is $x = 0$. That is, the matrix $T+I$ has a trivial kernel, which is to say that $T+I$ is invertible.
$endgroup$
– Omnomnomnom
Dec 4 '18 at 1:05
add a comment |
$begingroup$
Note that
$(T + I)(T - I) = T^2 - I; tag 1$
since
$T^2 = I, tag 2$
$T^2 - I = 0; tag 3$
now suppose
$T ne I; tag 4$
then
$exists ; x in v, ; Tx ne Ix = x; tag 5$
therefore,
$w = (T - I) x = Tx - x ne 0; tag 6$
thus, by (1) and (3),
$(T + I)w = (T + I)(T - I)x = (T^2 - I)x = 0(x) = 0; tag 7$
but
$(T + I)w = 0 Longrightarrow Tw + w = 0 Longrightarrow Tw = -w, tag 8$
which asserts that $w ne 0$ is an eigenvector of $T$ corresponding to eigenvalue $-1$, in contradiction to the hypothesis placed upon $T$. We see then that (5) cannot bind, whence
$forall x in v, ; Tx = Ix = x, tag 9$
as was to be proved.
answered Dec 3 '18 at 23:01
Robert LewisRobert Lewis
47k23067
$endgroup$
add a comment |
$begingroup$
Assume that the base field is of characteristic not equal to $2$. First, using the hypothesis $T^2=I$ and the fact that $x-1$ and $x+1$ are coprime polynomials, show that $$V=text{im}(T+I)oplus ker(T+I),.$$ Then, prove that $text{im}(T-I)=ker(T+I)$. Since $-1$ is not an eigenvalue of $T$, this means $ker(T+I)=0$. Therefore, $text{im}(T-I)=0$, proving that $T=I$.
Observe that, for any $vin V$, $$v=frac{1}{2}(T+I)(v)-frac{1}{2}(T-I)(v),,tag{*}$$ with $frac{1}{2}(T+I)(v)in text{im}(T+I)$ and $-frac12(T-I)(v)in text{im}(T-I)subseteq ker(T+I)$ (since $(T-I)(T+I)=T^2-I=0$). This shows that $$V=text{im}(T+I)+ker(T+I),.tag{#}$$ If $vin text{im}(T+I)cap ker(T+I)$, then $v=(T+I)(u)$ for some $uin V$, so $$(T+I)^2(u)=(T+I)(v)=0,.$$ Since $T^2-I=0$, we get $$v=(T+I)(u)=frac{1}{2}big((T+I)^2-(T^2-I)big)(u)=0,.$$ Thus, the sum (#) is direct.
From (*), we can also see that $V=text{im}(T-I)+text{im}(T+I)$ and $text{im}(T-I)subseteq ker(T+I)$. This implies $text{im}(T-I)=ker(T+I)$.
answered Dec 3 '18 at 23:37
BatominovskiBatominovski
33.1k33293
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $0 = T^2-I = (T-I)(T+I)$, with $T+I$ invertible (why?).
... or this:
Hint: $0 = T^2-I = (T+I)(T-I)$, with $T+I$ injective (why?).
answered Dec 3 '18 at 22:36
lhflhf
165k10171396
$endgroup$
$begingroup$
I understand the expansion but, I'm not sure how to reason that $T+I$ is invertible. Since it's given that -1 is not an eigenvalue for T, is $T=I$ the only solution for $(T-I)(T+I)$ because of that?
$endgroup$
– Jaigus
Dec 3 '18 at 23:04
$begingroup$
@Jaigus not quite. Because $-1$ is not an eigenvalue, the only solution for $(T+I)x = 0$ is $x = 0$. That is, the matrix $T+I$ has a trivial kernel, which is to say that $T+I$ is invertible.
$endgroup$
– Omnomnomnom
Dec 4 '18 at 1:05
add a comment |
$begingroup$
Hint: $0 = T^2-I = (T-I)(T+I)$, with $T+I$ invertible (why?).
... or this:
Hint: $0 = T^2-I = (T+I)(T-I)$, with $T+I$ injective (why?).
answered Dec 3 '18 at 22:36
lhflhf
165k10171396
$endgroup$
$begingroup$
I understand the expansion but, I'm not sure how to reason that $T+I$ is invertible. Since it's given that -1 is not an eigenvalue for T, is $T=I$ the only solution for $(T-I)(T+I)$ because of that?
$endgroup$
– Jaigus
Dec 3 '18 at 23:04
$begingroup$
@Jaigus not quite. Because $-1$ is not an eigenvalue, the only solution for $(T+I)x = 0$ is $x = 0$. That is, the matrix $T+I$ has a trivial kernel, which is to say that $T+I$ is invertible.
$endgroup$
– Omnomnomnom
Dec 4 '18 at 1:05
add a comment |
$begingroup$
Hint: $0 = T^2-I = (T-I)(T+I)$, with $T+I$ invertible (why?).
... or this:
Hint: $0 = T^2-I = (T+I)(T-I)$, with $T+I$ injective (why?).
answered Dec 3 '18 at 22:36
lhflhf
165k10171396
$endgroup$
Hint: $0 = T^2-I = (T-I)(T+I)$, with $T+I$ invertible (why?).
... or this:
Hint: $0 = T^2-I = (T+I)(T-I)$, with $T+I$ injective (why?).
answered Dec 3 '18 at 22:36
lhflhf
165k10171396
edited Dec 4 '18 at 10:11
answered Dec 3 '18 at 22:36
lhflhf
165k10171396
answered Dec 3 '18 at 22:36
lhflhf
165k10171396
answered Dec 3 '18 at 22:36
lhflhf
165k10171396
165k10171396
$begingroup$
I understand the expansion but, I'm not sure how to reason that $T+I$ is invertible. Since it's given that -1 is not an eigenvalue for T, is $T=I$ the only solution for $(T-I)(T+I)$ because of that?
$endgroup$
– Jaigus
Dec 3 '18 at 23:04
$begingroup$
@Jaigus not quite. Because $-1$ is not an eigenvalue, the only solution for $(T+I)x = 0$ is $x = 0$. That is, the matrix $T+I$ has a trivial kernel, which is to say that $T+I$ is invertible.
$endgroup$
– Omnomnomnom
Dec 4 '18 at 1:05
add a comment |
$begingroup$
I understand the expansion but, I'm not sure how to reason that $T+I$ is invertible. Since it's given that -1 is not an eigenvalue for T, is $T=I$ the only solution for $(T-I)(T+I)$ because of that?
$endgroup$
– Jaigus
Dec 3 '18 at 23:04
$begingroup$
@Jaigus not quite. Because $-1$ is not an eigenvalue, the only solution for $(T+I)x = 0$ is $x = 0$. That is, the matrix $T+I$ has a trivial kernel, which is to say that $T+I$ is invertible.
$endgroup$
– Omnomnomnom
Dec 4 '18 at 1:05
$begingroup$
I understand the expansion but, I'm not sure how to reason that $T+I$ is invertible. Since it's given that -1 is not an eigenvalue for T, is $T=I$ the only solution for $(T-I)(T+I)$ because of that?
$endgroup$
– Jaigus
Dec 3 '18 at 23:04
$begingroup$
I understand the expansion but, I'm not sure how to reason that $T+I$ is invertible. Since it's given that -1 is not an eigenvalue for T, is $T=I$ the only solution for $(T-I)(T+I)$ because of that?
$endgroup$
– Jaigus
Dec 3 '18 at 23:04
$begingroup$
@Jaigus not quite. Because $-1$ is not an eigenvalue, the only solution for $(T+I)x = 0$ is $x = 0$. That is, the matrix $T+I$ has a trivial kernel, which is to say that $T+I$ is invertible.
$endgroup$
– Omnomnomnom
Dec 4 '18 at 1:05
$begingroup$
@Jaigus not quite. Because $-1$ is not an eigenvalue, the only solution for $(T+I)x = 0$ is $x = 0$. That is, the matrix $T+I$ has a trivial kernel, which is to say that $T+I$ is invertible.
$endgroup$
– Omnomnomnom
Dec 4 '18 at 1:05
add a comment |
$begingroup$
Note that
$(T + I)(T - I) = T^2 - I; tag 1$
since
$T^2 = I, tag 2$
$T^2 - I = 0; tag 3$
now suppose
$T ne I; tag 4$
then
$exists ; x in v, ; Tx ne Ix = x; tag 5$
therefore,
$w = (T - I) x = Tx - x ne 0; tag 6$
thus, by (1) and (3),
$(T + I)w = (T + I)(T - I)x = (T^2 - I)x = 0(x) = 0; tag 7$
but
$(T + I)w = 0 Longrightarrow Tw + w = 0 Longrightarrow Tw = -w, tag 8$
which asserts that $w ne 0$ is an eigenvector of $T$ corresponding to eigenvalue $-1$, in contradiction to the hypothesis placed upon $T$. We see then that (5) cannot bind, whence
$forall x in v, ; Tx = Ix = x, tag 9$
as was to be proved.
answered Dec 3 '18 at 23:01
Robert LewisRobert Lewis
47k23067
$endgroup$
add a comment |
$begingroup$
Note that
$(T + I)(T - I) = T^2 - I; tag 1$
since
$T^2 = I, tag 2$
$T^2 - I = 0; tag 3$
now suppose
$T ne I; tag 4$
then
$exists ; x in v, ; Tx ne Ix = x; tag 5$
therefore,
$w = (T - I) x = Tx - x ne 0; tag 6$
thus, by (1) and (3),
$(T + I)w = (T + I)(T - I)x = (T^2 - I)x = 0(x) = 0; tag 7$
but
$(T + I)w = 0 Longrightarrow Tw + w = 0 Longrightarrow Tw = -w, tag 8$
which asserts that $w ne 0$ is an eigenvector of $T$ corresponding to eigenvalue $-1$, in contradiction to the hypothesis placed upon $T$. We see then that (5) cannot bind, whence
$forall x in v, ; Tx = Ix = x, tag 9$
as was to be proved.
answered Dec 3 '18 at 23:01
Robert LewisRobert Lewis
47k23067
$endgroup$
add a comment |
$begingroup$
Note that
$(T + I)(T - I) = T^2 - I; tag 1$
since
$T^2 = I, tag 2$
$T^2 - I = 0; tag 3$
now suppose
$T ne I; tag 4$
then
$exists ; x in v, ; Tx ne Ix = x; tag 5$
therefore,
$w = (T - I) x = Tx - x ne 0; tag 6$
thus, by (1) and (3),
$(T + I)w = (T + I)(T - I)x = (T^2 - I)x = 0(x) = 0; tag 7$
but
$(T + I)w = 0 Longrightarrow Tw + w = 0 Longrightarrow Tw = -w, tag 8$
which asserts that $w ne 0$ is an eigenvector of $T$ corresponding to eigenvalue $-1$, in contradiction to the hypothesis placed upon $T$. We see then that (5) cannot bind, whence
$forall x in v, ; Tx = Ix = x, tag 9$
as was to be proved.
answered Dec 3 '18 at 23:01
Robert LewisRobert Lewis
47k23067
$endgroup$
Note that
$(T + I)(T - I) = T^2 - I; tag 1$
since
$T^2 = I, tag 2$
$T^2 - I = 0; tag 3$
now suppose
$T ne I; tag 4$
then
$exists ; x in v, ; Tx ne Ix = x; tag 5$
therefore,
$w = (T - I) x = Tx - x ne 0; tag 6$
thus, by (1) and (3),
$(T + I)w = (T + I)(T - I)x = (T^2 - I)x = 0(x) = 0; tag 7$
but
$(T + I)w = 0 Longrightarrow Tw + w = 0 Longrightarrow Tw = -w, tag 8$
which asserts that $w ne 0$ is an eigenvector of $T$ corresponding to eigenvalue $-1$, in contradiction to the hypothesis placed upon $T$. We see then that (5) cannot bind, whence
$forall x in v, ; Tx = Ix = x, tag 9$
as was to be proved.
answered Dec 3 '18 at 23:01
Robert LewisRobert Lewis
47k23067
edited Dec 3 '18 at 23:16
answered Dec 3 '18 at 23:01
Robert LewisRobert Lewis
47k23067
answered Dec 3 '18 at 23:01
Robert LewisRobert Lewis
47k23067
answered Dec 3 '18 at 23:01
Robert LewisRobert Lewis
47k23067
47k23067
add a comment |
add a comment |
$begingroup$
Assume that the base field is of characteristic not equal to $2$. First, using the hypothesis $T^2=I$ and the fact that $x-1$ and $x+1$ are coprime polynomials, show that $$V=text{im}(T+I)oplus ker(T+I),.$$ Then, prove that $text{im}(T-I)=ker(T+I)$. Since $-1$ is not an eigenvalue of $T$, this means $ker(T+I)=0$. Therefore, $text{im}(T-I)=0$, proving that $T=I$.
Observe that, for any $vin V$, $$v=frac{1}{2}(T+I)(v)-frac{1}{2}(T-I)(v),,tag{*}$$ with $frac{1}{2}(T+I)(v)in text{im}(T+I)$ and $-frac12(T-I)(v)in text{im}(T-I)subseteq ker(T+I)$ (since $(T-I)(T+I)=T^2-I=0$). This shows that $$V=text{im}(T+I)+ker(T+I),.tag{#}$$ If $vin text{im}(T+I)cap ker(T+I)$, then $v=(T+I)(u)$ for some $uin V$, so $$(T+I)^2(u)=(T+I)(v)=0,.$$ Since $T^2-I=0$, we get $$v=(T+I)(u)=frac{1}{2}big((T+I)^2-(T^2-I)big)(u)=0,.$$ Thus, the sum (#) is direct.
From (*), we can also see that $V=text{im}(T-I)+text{im}(T+I)$ and $text{im}(T-I)subseteq ker(T+I)$. This implies $text{im}(T-I)=ker(T+I)$.
answered Dec 3 '18 at 23:37
BatominovskiBatominovski
33.1k33293
$endgroup$
add a comment |
$begingroup$
Assume that the base field is of characteristic not equal to $2$. First, using the hypothesis $T^2=I$ and the fact that $x-1$ and $x+1$ are coprime polynomials, show that $$V=text{im}(T+I)oplus ker(T+I),.$$ Then, prove that $text{im}(T-I)=ker(T+I)$. Since $-1$ is not an eigenvalue of $T$, this means $ker(T+I)=0$. Therefore, $text{im}(T-I)=0$, proving that $T=I$.
Observe that, for any $vin V$, $$v=frac{1}{2}(T+I)(v)-frac{1}{2}(T-I)(v),,tag{*}$$ with $frac{1}{2}(T+I)(v)in text{im}(T+I)$ and $-frac12(T-I)(v)in text{im}(T-I)subseteq ker(T+I)$ (since $(T-I)(T+I)=T^2-I=0$). This shows that $$V=text{im}(T+I)+ker(T+I),.tag{#}$$ If $vin text{im}(T+I)cap ker(T+I)$, then $v=(T+I)(u)$ for some $uin V$, so $$(T+I)^2(u)=(T+I)(v)=0,.$$ Since $T^2-I=0$, we get $$v=(T+I)(u)=frac{1}{2}big((T+I)^2-(T^2-I)big)(u)=0,.$$ Thus, the sum (#) is direct.
From (*), we can also see that $V=text{im}(T-I)+text{im}(T+I)$ and $text{im}(T-I)subseteq ker(T+I)$. This implies $text{im}(T-I)=ker(T+I)$.
answered Dec 3 '18 at 23:37
BatominovskiBatominovski
33.1k33293
$endgroup$
add a comment |
$begingroup$
Assume that the base field is of characteristic not equal to $2$. First, using the hypothesis $T^2=I$ and the fact that $x-1$ and $x+1$ are coprime polynomials, show that $$V=text{im}(T+I)oplus ker(T+I),.$$ Then, prove that $text{im}(T-I)=ker(T+I)$. Since $-1$ is not an eigenvalue of $T$, this means $ker(T+I)=0$. Therefore, $text{im}(T-I)=0$, proving that $T=I$.
Observe that, for any $vin V$, $$v=frac{1}{2}(T+I)(v)-frac{1}{2}(T-I)(v),,tag{*}$$ with $frac{1}{2}(T+I)(v)in text{im}(T+I)$ and $-frac12(T-I)(v)in text{im}(T-I)subseteq ker(T+I)$ (since $(T-I)(T+I)=T^2-I=0$). This shows that $$V=text{im}(T+I)+ker(T+I),.tag{#}$$ If $vin text{im}(T+I)cap ker(T+I)$, then $v=(T+I)(u)$ for some $uin V$, so $$(T+I)^2(u)=(T+I)(v)=0,.$$ Since $T^2-I=0$, we get $$v=(T+I)(u)=frac{1}{2}big((T+I)^2-(T^2-I)big)(u)=0,.$$ Thus, the sum (#) is direct.
From (*), we can also see that $V=text{im}(T-I)+text{im}(T+I)$ and $text{im}(T-I)subseteq ker(T+I)$. This implies $text{im}(T-I)=ker(T+I)$.
answered Dec 3 '18 at 23:37
BatominovskiBatominovski
33.1k33293
$endgroup$
Assume that the base field is of characteristic not equal to $2$. First, using the hypothesis $T^2=I$ and the fact that $x-1$ and $x+1$ are coprime polynomials, show that $$V=text{im}(T+I)oplus ker(T+I),.$$ Then, prove that $text{im}(T-I)=ker(T+I)$. Since $-1$ is not an eigenvalue of $T$, this means $ker(T+I)=0$. Therefore, $text{im}(T-I)=0$, proving that $T=I$.
Observe that, for any $vin V$, $$v=frac{1}{2}(T+I)(v)-frac{1}{2}(T-I)(v),,tag{*}$$ with $frac{1}{2}(T+I)(v)in text{im}(T+I)$ and $-frac12(T-I)(v)in text{im}(T-I)subseteq ker(T+I)$ (since $(T-I)(T+I)=T^2-I=0$). This shows that $$V=text{im}(T+I)+ker(T+I),.tag{#}$$ If $vin text{im}(T+I)cap ker(T+I)$, then $v=(T+I)(u)$ for some $uin V$, so $$(T+I)^2(u)=(T+I)(v)=0,.$$ Since $T^2-I=0$, we get $$v=(T+I)(u)=frac{1}{2}big((T+I)^2-(T^2-I)big)(u)=0,.$$ Thus, the sum (#) is direct.
From (*), we can also see that $V=text{im}(T-I)+text{im}(T+I)$ and $text{im}(T-I)subseteq ker(T+I)$. This implies $text{im}(T-I)=ker(T+I)$.
answered Dec 3 '18 at 23:37
BatominovskiBatominovski
33.1k33293
edited Dec 4 '18 at 0:03
answered Dec 3 '18 at 23:37
BatominovskiBatominovski
33.1k33293
answered Dec 3 '18 at 23:37
BatominovskiBatominovski
33.1k33293
answered Dec 3 '18 at 23:37
BatominovskiBatominovski
33.1k33293
33.1k33293
add a comment |
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$begingroup$
I eidited the title of your post to make it better conform to the problem presented in the body. Cheers!
$endgroup$
– Robert Lewis
Dec 3 '18 at 23:07