Why is $(T-lambda_i)$ onto?

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$K_lambda_i$ = $v in V: (T-lambda_i I)^p(v) = 0$ for some positive integer $p$.

$T_W$ is the $T$-invariant subspace.

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.

Now suppose that for some integer $k>1$, the result is established whenever $T$ has fewer than $k$ distinct eigenvalues, and suppose that $T$ has $k$ distinct eigenvalues. Let $m$ be the multiplicity of $lambda_k$, and let $f(t)$ be the characteristic polynomial of $T$. Then $f(t) = (t - lambda_k)^mg(t)$ from some polynomial $g(t)$ not divisible by $(t-lambda_k)$. Let $W = R((T-lambda I)^m)$. Clearly $W$ is $T$-invariant. Observe that $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$. For suppose that $i<k$. Since $(T - lambda I)^m$ maps $K_lambda_i$ into itself and $lambda_k ne lambda_i$, the restriction of $T-lambda_k I$ to $K_lambda_i$ is one-to-one and hence is onto. One consequence of this is that for $i<k$, $K_lambda_i$ is contained in $W$, and hence $lambda_i$ is an eigenvalue of $T_W$ with corresponding generalized eigenspace $K_lambda_i$.

Since $(T - lambda I)^m$ maps $K_lambda_i$ into itself and $lambda_k ne lambda_i$, the restriction of $T-lambda_k I$ to $K_lambda_i$ is one-to-one and hence is onto.

I'm unable to see why $(T-lambda_k)$ being restricted to $K_lambda_i$ and being one-to-one makes $(T-lambda_i)$ onto?

One consequence of this is that for $i<k$, $K_lambda_i$ is contained in $W$, and hence $lambda_i$ is an eigenvalue of $T_W$ with corresponding generalized eigenspace $K_lambda_i$.

I was wondering if someone could also further elaborate why this is so.

linear-algebra

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asked Aug 6 at 3:32

K.M

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$K_lambda_i$ = $v in V: (T-lambda_i I)^p(v) = 0$ for some positive integer $p$.

$T_W$ is the $T$-invariant subspace.

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.

Now suppose that for some integer $k>1$, the result is established whenever $T$ has fewer than $k$ distinct eigenvalues, and suppose that $T$ has $k$ distinct eigenvalues. Let $m$ be the multiplicity of $lambda_k$, and let $f(t)$ be the characteristic polynomial of $T$. Then $f(t) = (t - lambda_k)^mg(t)$ from some polynomial $g(t)$ not divisible by $(t-lambda_k)$. Let $W = R((T-lambda I)^m)$. Clearly $W$ is $T$-invariant. Observe that $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$. For suppose that $i<k$. Since $(T - lambda I)^m$ maps $K_lambda_i$ into itself and $lambda_k ne lambda_i$, the restriction of $T-lambda_k I$ to $K_lambda_i$ is one-to-one and hence is onto. One consequence of this is that for $i<k$, $K_lambda_i$ is contained in $W$, and hence $lambda_i$ is an eigenvalue of $T_W$ with corresponding generalized eigenspace $K_lambda_i$.

Since $(T - lambda I)^m$ maps $K_lambda_i$ into itself and $lambda_k ne lambda_i$, the restriction of $T-lambda_k I$ to $K_lambda_i$ is one-to-one and hence is onto.

I'm unable to see why $(T-lambda_k)$ being restricted to $K_lambda_i$ and being one-to-one makes $(T-lambda_i)$ onto?

One consequence of this is that for $i<k$, $K_lambda_i$ is contained in $W$, and hence $lambda_i$ is an eigenvalue of $T_W$ with corresponding generalized eigenspace $K_lambda_i$.

I was wondering if someone could also further elaborate why this is so.

linear-algebra

share|cite|improve this question

asked Aug 6 at 3:32

K.M

487312

add a comment | 

up vote
0
down vote

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down vote

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$K_lambda_i$ = $v in V: (T-lambda_i I)^p(v) = 0$ for some positive integer $p$.

$T_W$ is the $T$-invariant subspace.

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.

Now suppose that for some integer $k>1$, the result is established whenever $T$ has fewer than $k$ distinct eigenvalues, and suppose that $T$ has $k$ distinct eigenvalues. Let $m$ be the multiplicity of $lambda_k$, and let $f(t)$ be the characteristic polynomial of $T$. Then $f(t) = (t - lambda_k)^mg(t)$ from some polynomial $g(t)$ not divisible by $(t-lambda_k)$. Let $W = R((T-lambda I)^m)$. Clearly $W$ is $T$-invariant. Observe that $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$. For suppose that $i<k$. Since $(T - lambda I)^m$ maps $K_lambda_i$ into itself and $lambda_k ne lambda_i$, the restriction of $T-lambda_k I$ to $K_lambda_i$ is one-to-one and hence is onto. One consequence of this is that for $i<k$, $K_lambda_i$ is contained in $W$, and hence $lambda_i$ is an eigenvalue of $T_W$ with corresponding generalized eigenspace $K_lambda_i$.

Since $(T - lambda I)^m$ maps $K_lambda_i$ into itself and $lambda_k ne lambda_i$, the restriction of $T-lambda_k I$ to $K_lambda_i$ is one-to-one and hence is onto.

I'm unable to see why $(T-lambda_k)$ being restricted to $K_lambda_i$ and being one-to-one makes $(T-lambda_i)$ onto?

One consequence of this is that for $i<k$, $K_lambda_i$ is contained in $W$, and hence $lambda_i$ is an eigenvalue of $T_W$ with corresponding generalized eigenspace $K_lambda_i$.

I was wondering if someone could also further elaborate why this is so.

linear-algebra

share|cite|improve this question

asked Aug 6 at 3:32

K.M

487312

$K_lambda_i$ = $v in V: (T-lambda_i I)^p(v) = 0$ for some positive integer $p$.

$T_W$ is the $T$-invariant subspace.

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.

Now suppose that for some integer $k>1$, the result is established whenever $T$ has fewer than $k$ distinct eigenvalues, and suppose that $T$ has $k$ distinct eigenvalues. Let $m$ be the multiplicity of $lambda_k$, and let $f(t)$ be the characteristic polynomial of $T$. Then $f(t) = (t - lambda_k)^mg(t)$ from some polynomial $g(t)$ not divisible by $(t-lambda_k)$. Let $W = R((T-lambda I)^m)$. Clearly $W$ is $T$-invariant. Observe that $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$. For suppose that $i<k$. Since $(T - lambda I)^m$ maps $K_lambda_i$ into itself and $lambda_k ne lambda_i$, the restriction of $T-lambda_k I$ to $K_lambda_i$ is one-to-one and hence is onto. One consequence of this is that for $i<k$, $K_lambda_i$ is contained in $W$, and hence $lambda_i$ is an eigenvalue of $T_W$ with corresponding generalized eigenspace $K_lambda_i$.

Since $(T - lambda I)^m$ maps $K_lambda_i$ into itself and $lambda_k ne lambda_i$, the restriction of $T-lambda_k I$ to $K_lambda_i$ is one-to-one and hence is onto.

I'm unable to see why $(T-lambda_k)$ being restricted to $K_lambda_i$ and being one-to-one makes $(T-lambda_i)$ onto?

One consequence of this is that for $i<k$, $K_lambda_i$ is contained in $W$, and hence $lambda_i$ is an eigenvalue of $T_W$ with corresponding generalized eigenspace $K_lambda_i$.

I was wondering if someone could also further elaborate why this is so.

linear-algebra

share|cite|improve this question

asked Aug 6 at 3:32

K.M

487312

share|cite|improve this question

share|cite|improve this question

asked Aug 6 at 3:32

K.M

487312

asked Aug 6 at 3:32

K.M

487312

asked Aug 6 at 3:32

K.M

487312

487312

add a comment | 

add a comment | 

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You write

I'm unable to see why $(T−lambda_i)$ being restricted to $K_lambda_i$ and being one-to-one makes $(T−lambda_i)$ onto?

This is because the vector space is finite-dimensional, and a one-to-one
linear map from a finite-dimensional vector space to itself is onto.
(Rank-nullity formula).

share|cite|improve this answer

answered Aug 6 at 3:38

Lord Shark the Unknown

86k951112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  from the rank-nullity formula, I know that dim V = dim null(T) + dim range(T), but how do I apply that to the linear map being onto?
  – K.M
  Aug 6 at 3:45
  

  
  
  
  

add a comment | 

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You write

I'm unable to see why $(T−lambda_i)$ being restricted to $K_lambda_i$ and being one-to-one makes $(T−lambda_i)$ onto?

This is because the vector space is finite-dimensional, and a one-to-one
linear map from a finite-dimensional vector space to itself is onto.
(Rank-nullity formula).

share|cite|improve this answer

answered Aug 6 at 3:38

Lord Shark the Unknown

86k951112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  from the rank-nullity formula, I know that dim V = dim null(T) + dim range(T), but how do I apply that to the linear map being onto?
  – K.M
  Aug 6 at 3:45
  

  
  
  
  

add a comment | 

up vote
1
down vote

You write

I'm unable to see why $(T−lambda_i)$ being restricted to $K_lambda_i$ and being one-to-one makes $(T−lambda_i)$ onto?

This is because the vector space is finite-dimensional, and a one-to-one
linear map from a finite-dimensional vector space to itself is onto.
(Rank-nullity formula).

share|cite|improve this answer

answered Aug 6 at 3:38

Lord Shark the Unknown

86k951112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  from the rank-nullity formula, I know that dim V = dim null(T) + dim range(T), but how do I apply that to the linear map being onto?
  – K.M
  Aug 6 at 3:45
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

You write

I'm unable to see why $(T−lambda_i)$ being restricted to $K_lambda_i$ and being one-to-one makes $(T−lambda_i)$ onto?

This is because the vector space is finite-dimensional, and a one-to-one
linear map from a finite-dimensional vector space to itself is onto.
(Rank-nullity formula).

share|cite|improve this answer

answered Aug 6 at 3:38

Lord Shark the Unknown

86k951112

You write

I'm unable to see why $(T−lambda_i)$ being restricted to $K_lambda_i$ and being one-to-one makes $(T−lambda_i)$ onto?

This is because the vector space is finite-dimensional, and a one-to-one
linear map from a finite-dimensional vector space to itself is onto.
(Rank-nullity formula).

share|cite|improve this answer

answered Aug 6 at 3:38

Lord Shark the Unknown

86k951112

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 3:38

Lord Shark the Unknown

86k951112

answered Aug 6 at 3:38

Lord Shark the Unknown

86k951112

answered Aug 6 at 3:38

Lord Shark the Unknown

86k951112

86k951112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  from the rank-nullity formula, I know that dim V = dim null(T) + dim range(T), but how do I apply that to the linear map being onto?
  – K.M
  Aug 6 at 3:45
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  from the rank-nullity formula, I know that dim V = dim null(T) + dim range(T), but how do I apply that to the linear map being onto?
  – K.M
  Aug 6 at 3:45
  

  
  
  
  

from the rank-nullity formula, I know that dim V = dim null(T) + dim range(T), but how do I apply that to the linear map being onto?
– K.M
Aug 6 at 3:45

from the rank-nullity formula, I know that dim V = dim null(T) + dim range(T), but how do I apply that to the linear map being onto?
– K.M
Aug 6 at 3:45

add a comment | 

 

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