Mixing
Why does $V=K_lambda_i$ for $k=1$?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Theorem 7.3. Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits, and lett $lambda_1,lambda_2,...,lambda_k$ be the distinct eigenvalues of $T$. Then, for every $xin V$, there exist vectors $v_i in K_lambda_i$, $1le i le k$, such that
$x = v_1 + v_2 + cdots +v_k$.
Proof. The proof is by mathematical induction on the number $k$ of distinct eigenvalues of $T$. First suppose that $k = 1$, and let $m$ be the multiplicity of $lambda_1$. Then $(lambda_1 - t)^m$is the characteristic polynomial of $T$, and hence $(lambda_1 I - T)^m = T_0$ by the Cayley-Hamilton theorem(p.317). Thus $V=K_lambda_i$, and the result follows.
where $K_lambda_i$ = $v in V: (T-lambda_1 I)^p(v) = 0$ for some positive integer $p$.
I wanted to know why it follows that $V=K_lambda_i$?
linear-algebra
asked 13 hours ago
K.M
461312
add a comment |Â
up vote
0
down vote
favorite
Theorem 7.3. Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits, and lett $lambda_1,lambda_2,...,lambda_k$ be the distinct eigenvalues of $T$. Then, for every $xin V$, there exist vectors $v_i in K_lambda_i$, $1le i le k$, such that
$x = v_1 + v_2 + cdots +v_k$.
Proof. The proof is by mathematical induction on the number $k$ of distinct eigenvalues of $T$. First suppose that $k = 1$, and let $m$ be the multiplicity of $lambda_1$. Then $(lambda_1 - t)^m$is the characteristic polynomial of $T$, and hence $(lambda_1 I - T)^m = T_0$ by the Cayley-Hamilton theorem(p.317). Thus $V=K_lambda_i$, and the result follows.
where $K_lambda_i$ = $v in V: (T-lambda_1 I)^p(v) = 0$ for some positive integer $p$.
I wanted to know why it follows that $V=K_lambda_i$?
linear-algebra
asked 13 hours ago
K.M
461312
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Theorem 7.3. Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits, and lett $lambda_1,lambda_2,...,lambda_k$ be the distinct eigenvalues of $T$. Then, for every $xin V$, there exist vectors $v_i in K_lambda_i$, $1le i le k$, such that
$x = v_1 + v_2 + cdots +v_k$.
Proof. The proof is by mathematical induction on the number $k$ of distinct eigenvalues of $T$. First suppose that $k = 1$, and let $m$ be the multiplicity of $lambda_1$. Then $(lambda_1 - t)^m$is the characteristic polynomial of $T$, and hence $(lambda_1 I - T)^m = T_0$ by the Cayley-Hamilton theorem(p.317). Thus $V=K_lambda_i$, and the result follows.
where $K_lambda_i$ = $v in V: (T-lambda_1 I)^p(v) = 0$ for some positive integer $p$.
I wanted to know why it follows that $V=K_lambda_i$?
linear-algebra
asked 13 hours ago
K.M
461312
Theorem 7.3. Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits, and lett $lambda_1,lambda_2,...,lambda_k$ be the distinct eigenvalues of $T$. Then, for every $xin V$, there exist vectors $v_i in K_lambda_i$, $1le i le k$, such that
$x = v_1 + v_2 + cdots +v_k$.
Proof. The proof is by mathematical induction on the number $k$ of distinct eigenvalues of $T$. First suppose that $k = 1$, and let $m$ be the multiplicity of $lambda_1$. Then $(lambda_1 - t)^m$is the characteristic polynomial of $T$, and hence $(lambda_1 I - T)^m = T_0$ by the Cayley-Hamilton theorem(p.317). Thus $V=K_lambda_i$, and the result follows.
where $K_lambda_i$ = $v in V: (T-lambda_1 I)^p(v) = 0$ for some positive integer $p$.
I wanted to know why it follows that $V=K_lambda_i$?
linear-algebra
asked 13 hours ago
K.M
461312
asked 13 hours ago
K.M
461312
asked 13 hours ago
K.M
461312
asked 13 hours ago
K.M
461312
461312
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Cayley-Hamilton tells you that the characteristic polynomial of your linear operator annihilates it. Hence, what you must write is $(lambda_1I-T)^m=0_operatornameEnd(V)$. Now, consider any $vin V$. We have $$0=0_operatornameEnd(V)(v)=(lambda_1I-T)^m(v)$$
Hence, by the definition you gave of $K_lambda_1$, we may conclude that $vin K_lambda_1$.
answered 13 hours ago
Suzet
2,003324
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Cayley-Hamilton tells you that the characteristic polynomial of your linear operator annihilates it. Hence, what you must write is $(lambda_1I-T)^m=0_operatornameEnd(V)$. Now, consider any $vin V$. We have $$0=0_operatornameEnd(V)(v)=(lambda_1I-T)^m(v)$$
Hence, by the definition you gave of $K_lambda_1$, we may conclude that $vin K_lambda_1$.
answered 13 hours ago
Suzet
2,003324
add a comment |Â
up vote
1
down vote
Cayley-Hamilton tells you that the characteristic polynomial of your linear operator annihilates it. Hence, what you must write is $(lambda_1I-T)^m=0_operatornameEnd(V)$. Now, consider any $vin V$. We have $$0=0_operatornameEnd(V)(v)=(lambda_1I-T)^m(v)$$
Hence, by the definition you gave of $K_lambda_1$, we may conclude that $vin K_lambda_1$.
answered 13 hours ago
Suzet
2,003324
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Cayley-Hamilton tells you that the characteristic polynomial of your linear operator annihilates it. Hence, what you must write is $(lambda_1I-T)^m=0_operatornameEnd(V)$. Now, consider any $vin V$. We have $$0=0_operatornameEnd(V)(v)=(lambda_1I-T)^m(v)$$
Hence, by the definition you gave of $K_lambda_1$, we may conclude that $vin K_lambda_1$.
answered 13 hours ago
Suzet
2,003324
Cayley-Hamilton tells you that the characteristic polynomial of your linear operator annihilates it. Hence, what you must write is $(lambda_1I-T)^m=0_operatornameEnd(V)$. Now, consider any $vin V$. We have $$0=0_operatornameEnd(V)(v)=(lambda_1I-T)^m(v)$$
Hence, by the definition you gave of $K_lambda_1$, we may conclude that $vin K_lambda_1$.
answered 13 hours ago
Suzet
2,003324
answered 13 hours ago
Suzet
2,003324
answered 13 hours ago
Suzet
2,003324
answered 13 hours ago
Suzet
2,003324
2,003324
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873108%2fwhy-does-v-k-lambda-i-for-k-1%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password