Mixing
An invariant subspace of $mathbb{F}^n$
$begingroup$
Problem: Let $A$ be a square matrix of szie $n$ over a field $mathbb{F}$ and characteristic polynomial $$p_A (t) = a (t -
lambda_1)^{n_1} dots (t - lambda_k)^{n_k}$$ with $a_i in
mathbb{F}, lambda_i in mathbb{F}, lambda_i neq lambda_j, 1 leq
i neq j leq k, a neq 0$ and $n_i in mathbb{N}$ such that
$sum_{i=1}^k n_i = n$. Show that:
$mathcal{R}_{A,lambda_i} = {x in mathbb{F}^n equiv
mathbb{F}^{n times 1} | exists m in mathbb{N}^* : (A -
lambda_i I_n)^m x = 0$ } is a subspace of $mathbb{F}^n$.
$dim (mathcal{R}_{A,lambda_i}) = n_i, forall n = 1, dots, k$.
My attempt: To show that $mathcal{R}_{A,lambda_i}$ is a subspace of $mathbb{F}^n$, let $x,y in mathcal{R}_{A,lambda_i}$. We have
$$(A - lambda_i I_n)^m x + (A - lambda_i I_n)^m y = 0+0 = 0 = (A - lambda_i I_n)^m (x+y).$$
and
$$alpha (A - lambda_i I_n)^m x = alpha0 = 0, alpha in mathbb{F}^n.$$
Is that right? How to show that $dim (mathcal{R}_{A,lambda_i}) = n_i$?
linear-algebra proof-verification
asked Jan 13 at 7:42
MinhMinh
1959
$endgroup$
|
show 10 more comments
$begingroup$
Problem: Let $A$ be a square matrix of szie $n$ over a field $mathbb{F}$ and characteristic polynomial $$p_A (t) = a (t -
lambda_1)^{n_1} dots (t - lambda_k)^{n_k}$$ with $a_i in
mathbb{F}, lambda_i in mathbb{F}, lambda_i neq lambda_j, 1 leq
i neq j leq k, a neq 0$ and $n_i in mathbb{N}$ such that
$sum_{i=1}^k n_i = n$. Show that:
$mathcal{R}_{A,lambda_i} = {x in mathbb{F}^n equiv
mathbb{F}^{n times 1} | exists m in mathbb{N}^* : (A -
lambda_i I_n)^m x = 0$ } is a subspace of $mathbb{F}^n$.
$dim (mathcal{R}_{A,lambda_i}) = n_i, forall n = 1, dots, k$.
My attempt: To show that $mathcal{R}_{A,lambda_i}$ is a subspace of $mathbb{F}^n$, let $x,y in mathcal{R}_{A,lambda_i}$. We have
$$(A - lambda_i I_n)^m x + (A - lambda_i I_n)^m y = 0+0 = 0 = (A - lambda_i I_n)^m (x+y).$$
and
$$alpha (A - lambda_i I_n)^m x = alpha0 = 0, alpha in mathbb{F}^n.$$
Is that right? How to show that $dim (mathcal{R}_{A,lambda_i}) = n_i$?
linear-algebra proof-verification
asked Jan 13 at 7:42
MinhMinh
1959
$endgroup$
1
$begingroup$
That $m$ is different for $x, y$.
$endgroup$
– xbh
Jan 13 at 7:50
$begingroup$
I don't understand. What do you mean?
$endgroup$
– Minh
Jan 13 at 7:51
$begingroup$
For $x$ there is some $m_1$ that $(A-lambda_j I_n)^{m_1} =0$; for $y$ there is another $m_2$ that $(A- lambda_j I_n)^{m_2} = 0$.
$endgroup$
– xbh
Jan 13 at 7:53
$begingroup$
I see, but if $m$ if different for $x,y$, how can we prove that $x+y in mathcal{R}_{A,lambda_i}$
$endgroup$
– Minh
Jan 13 at 7:56
$begingroup$
WOLG, suppose $m_1 geqslant m_2$, then how could you find an $m$ for $x+y$?
$endgroup$
– xbh
Jan 13 at 7:57
|
show 10 more comments
$begingroup$
Problem: Let $A$ be a square matrix of szie $n$ over a field $mathbb{F}$ and characteristic polynomial $$p_A (t) = a (t -
lambda_1)^{n_1} dots (t - lambda_k)^{n_k}$$ with $a_i in
mathbb{F}, lambda_i in mathbb{F}, lambda_i neq lambda_j, 1 leq
i neq j leq k, a neq 0$ and $n_i in mathbb{N}$ such that
$sum_{i=1}^k n_i = n$. Show that:
$mathcal{R}_{A,lambda_i} = {x in mathbb{F}^n equiv
mathbb{F}^{n times 1} | exists m in mathbb{N}^* : (A -
lambda_i I_n)^m x = 0$ } is a subspace of $mathbb{F}^n$.
$dim (mathcal{R}_{A,lambda_i}) = n_i, forall n = 1, dots, k$.
My attempt: To show that $mathcal{R}_{A,lambda_i}$ is a subspace of $mathbb{F}^n$, let $x,y in mathcal{R}_{A,lambda_i}$. We have
$$(A - lambda_i I_n)^m x + (A - lambda_i I_n)^m y = 0+0 = 0 = (A - lambda_i I_n)^m (x+y).$$
and
$$alpha (A - lambda_i I_n)^m x = alpha0 = 0, alpha in mathbb{F}^n.$$
Is that right? How to show that $dim (mathcal{R}_{A,lambda_i}) = n_i$?
linear-algebra proof-verification
asked Jan 13 at 7:42
MinhMinh
1959
$endgroup$
Problem: Let $A$ be a square matrix of szie $n$ over a field $mathbb{F}$ and characteristic polynomial $$p_A (t) = a (t -
lambda_1)^{n_1} dots (t - lambda_k)^{n_k}$$ with $a_i in
mathbb{F}, lambda_i in mathbb{F}, lambda_i neq lambda_j, 1 leq
i neq j leq k, a neq 0$ and $n_i in mathbb{N}$ such that
$sum_{i=1}^k n_i = n$. Show that:
$mathcal{R}_{A,lambda_i} = {x in mathbb{F}^n equiv
mathbb{F}^{n times 1} | exists m in mathbb{N}^* : (A -
lambda_i I_n)^m x = 0$ } is a subspace of $mathbb{F}^n$.
$dim (mathcal{R}_{A,lambda_i}) = n_i, forall n = 1, dots, k$.
My attempt: To show that $mathcal{R}_{A,lambda_i}$ is a subspace of $mathbb{F}^n$, let $x,y in mathcal{R}_{A,lambda_i}$. We have
$$(A - lambda_i I_n)^m x + (A - lambda_i I_n)^m y = 0+0 = 0 = (A - lambda_i I_n)^m (x+y).$$
and
$$alpha (A - lambda_i I_n)^m x = alpha0 = 0, alpha in mathbb{F}^n.$$
Is that right? How to show that $dim (mathcal{R}_{A,lambda_i}) = n_i$?
linear-algebra proof-verification
linear-algebra proof-verification
asked Jan 13 at 7:42
MinhMinh
1959
asked Jan 13 at 7:42
MinhMinh
1959
edited Jan 13 at 8:57
Minh
asked Jan 13 at 7:42
MinhMinh
1959
asked Jan 13 at 7:42
MinhMinh
1959
asked Jan 13 at 7:42
MinhMinh
1959
1959
1
$begingroup$
That $m$ is different for $x, y$.
$endgroup$
– xbh
Jan 13 at 7:50
$begingroup$
I don't understand. What do you mean?
$endgroup$
– Minh
Jan 13 at 7:51
$begingroup$
For $x$ there is some $m_1$ that $(A-lambda_j I_n)^{m_1} =0$; for $y$ there is another $m_2$ that $(A- lambda_j I_n)^{m_2} = 0$.
$endgroup$
– xbh
Jan 13 at 7:53
$begingroup$
I see, but if $m$ if different for $x,y$, how can we prove that $x+y in mathcal{R}_{A,lambda_i}$
$endgroup$
– Minh
Jan 13 at 7:56
$begingroup$
WOLG, suppose $m_1 geqslant m_2$, then how could you find an $m$ for $x+y$?
$endgroup$
– xbh
Jan 13 at 7:57
|
show 10 more comments
1
$begingroup$
That $m$ is different for $x, y$.
$endgroup$
– xbh
Jan 13 at 7:50
$begingroup$
I don't understand. What do you mean?
$endgroup$
– Minh
Jan 13 at 7:51
$begingroup$
For $x$ there is some $m_1$ that $(A-lambda_j I_n)^{m_1} =0$; for $y$ there is another $m_2$ that $(A- lambda_j I_n)^{m_2} = 0$.
$endgroup$
– xbh
Jan 13 at 7:53
$begingroup$
I see, but if $m$ if different for $x,y$, how can we prove that $x+y in mathcal{R}_{A,lambda_i}$
$endgroup$
– Minh
Jan 13 at 7:56
$begingroup$
WOLG, suppose $m_1 geqslant m_2$, then how could you find an $m$ for $x+y$?
$endgroup$
– xbh
Jan 13 at 7:57
1
1
$begingroup$
That $m$ is different for $x, y$.
$endgroup$
– xbh
Jan 13 at 7:50
$begingroup$
That $m$ is different for $x, y$.
$endgroup$
– xbh
Jan 13 at 7:50
$begingroup$
I don't understand. What do you mean?
$endgroup$
– Minh
Jan 13 at 7:51
$begingroup$
I don't understand. What do you mean?
$endgroup$
– Minh
Jan 13 at 7:51
$begingroup$
For $x$ there is some $m_1$ that $(A-lambda_j I_n)^{m_1} =0$; for $y$ there is another $m_2$ that $(A- lambda_j I_n)^{m_2} = 0$.
$endgroup$
– xbh
Jan 13 at 7:53
$begingroup$
For $x$ there is some $m_1$ that $(A-lambda_j I_n)^{m_1} =0$; for $y$ there is another $m_2$ that $(A- lambda_j I_n)^{m_2} = 0$.
$endgroup$
– xbh
Jan 13 at 7:53
$begingroup$
I see, but if $m$ if different for $x,y$, how can we prove that $x+y in mathcal{R}_{A,lambda_i}$
$endgroup$
– Minh
Jan 13 at 7:56
$begingroup$
I see, but if $m$ if different for $x,y$, how can we prove that $x+y in mathcal{R}_{A,lambda_i}$
$endgroup$
– Minh
Jan 13 at 7:56
$begingroup$
WOLG, suppose $m_1 geqslant m_2$, then how could you find an $m$ for $x+y$?
$endgroup$
– xbh
Jan 13 at 7:57
$begingroup$
WOLG, suppose $m_1 geqslant m_2$, then how could you find an $m$ for $x+y$?
$endgroup$
– xbh
Jan 13 at 7:57
|
show 10 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I'm here to give a demo. This may not be the quickest one.
For a matrix $newcommandbmboldsymbol
DeclareMathOperatorKer{Ker}
newcommandF{{mathbb F}}
newcommandN {{mathbb N}}
bm A in mathrm M_n (F)$ [i.e. $n times n$ with entries in $F$], let $Ker bm A = {bm x in F^n colon bm {Ax} = bm 0}$ From now on, let $bm N_j = bm A - lambda_j bm I_n$.
First we figure out what exactly $W_j = R_{bm A,lambda_j} $ is. For any matrix $bm P in mathrm M_n (F)$, consider $Ker(bm P^s)$ and $Ker(bm P^{s+1})$ where $s in N$. If $bm y in Ker(bm P^s)$, i.e. $bm P^s bm y = bm 0$, then clearly $bm P^{s+1}bm y = bm P(bm P^s bm y) = bm 0$. In other words, $Ker(bm P^s) subseteq Ker(bm P^{s+1})$. So we have a chain:
$$
varnothing subseteq Ker bm P subseteq Ker (bm P^2) subseteq dots subseteq Ker(bm P^s) subseteq cdots
$$
Easy to see that each $Ker (bm P^s)$ is a subspace of $F^n$. Thus $dim(Ker (bm P ^s)) leqslant n$ for all $s$. Therefore there exists some smallest $min N$ s.t. $dim(Ker(bm P^m)) = dim(Ker (bm P^{m+p}))$ for all $p in N^*$.
Additionally, we see that
if $dim(Ker (bm P^s))= dim(Ker (bm P^{s+1}))$ for some $sin N^*$, then $dim(Ker (bm P^s)) = dim (Ker (bm P^{s+p}))$ for all $p in N^*$.
Indeed, under the assumption, we have $Ker (bm P^s) = Ker (bm P^{s+1})$. Then for each $ellin N^*$,if $bm yin Ker(bm P^{s+ell})$, i.e. $bm P^{s+ell} bm y = bm 0$, then $bm P^{s+1} (bm P^{ell-1} bm y)= bm 0$, hence $bm P^{ell-1} bm y in Ker(bm P^{s+1})$. Use the assumption $Ker(bm P^s) = Ker(bm P^{s+1})$, we have $bm P^{ell-1}bm y in Ker(bm P^s)$ or $bm P^{s+ell -1} bm y =bm 0$, thus $bm y in Ker(bm P^{s+ell -1})$. Conclusively
$$
Ker (bm P^{s+ell-1}) = Ker (bm P^{s+ell}), quad ell in N^*,
$$
now let $ell = 1, 2, dots, p-1$ and the claim is proved. To be clearer, we have the following chain:
$$
varnothing subset Ker(bm P) subset dots subset Ker(bm P^{m-1}) subset Ker(bm P^m) = Ker (bm P^{m+1}) = cdots subseteq F^n,
$$
where $subset$ means $subsetneq$.
Now focus on each $bm N_j$. Let the corresponding integer $m$ be $m_j$. Consider $W_j$ and $Ker(bm N_j^{m_j})$. By the definition of $W_j$, clearly $Ker(bm N_j^{m_j}) subseteq W_j$. Conversely if $bm x in W_j$, then there is some $pin N^*$ s.t. $bm N_j^p bm x = bm 0$ by definition, or $bm x in Ker (bm N_j^p)$. According to the chain above, $Ker (bm N_j^p) subseteq Ker(bm N_j^{m_j})$. Therefore we actually have
$$
W_j = Ker(bm N_j^{m_j}), j = 1,2, dots, k.
$$
Now we prove that
$$
W = sum_1^k Ker(bm N_j ^{m_j}) = sum_1^k W_j tag {1}
$$
is also a direct sum.
Equivalently we need to show
If $bm 0 = sum_1^k bm w_j$ where $w_j in W_j$ for each $j$, then $bm w_j =bm 0$ for all $j$.
Let $f_j (t) = (t - lambda_j)^{m_j} in F[t]$ and $g_j (t) = prod_{ell neq j} f_ell (t)$ for each $j$. Clearly $mathrm{gcd}(g_j, f_j) = 1$. Thus there exists $u_j, v_j in F[t]$ such that
$$
u_j f_j + v_j g_j = 1.
$$
Hence
$$
u_j (bm A) f_j(bm A) + v_j (bm A) g_j(bm A) = bm I_n,
$$
and for each $bm y in F^n$,
$$
bm y = u_j (bm A) f_j(bm A)bm y + v_j (bm A) g_j(bm A) bm y. tag{2}
$$
If $bm y in Ker(f_j (bm A)) cap Ker (g_j(bm A))$, i.e. $f_j(bm A) bm y =g_j(bm A)bm y = bm 0$, then $(2)$ becomes $bm y = bm 0$. Hence $Ker (f_j(bm A)) cap Ker(g_j (bm A)) = {bm 0}$. Now suppose $bm 0 = sum_1^k bm w_j$ where $bm w_j in W_j$ for each $j$, then by the definition of $bm w_j$, $g_ell (bm A) bm w_j = bm 0$ if $ell neq j$. Thus $bm w_ell in Ker(g_ell (bm A)) cap Ker(f_ell (bm A)) = 0$, then $bm w_ell = bm 0$ for each $ell$. Therefore $(1)$ is proved.
Finally we get the goal: by the definition of $W_j$, we have $Ker(bm N_j ^{n_j}) subseteq W_j$ for each $j$. Now compute the dimension by using the dimension formula of direct sums, we have
$$
n = dim (F^n) geqslant dim W = dim left(bigoplus _1^k W_jright) = sum_1^k dim left( W_jright) geqslant sum_1^k dim (Ker(bm N_j ^{n_j})) = sum_1^k n_j = n,
$$
so actually the $geqslant$'s are all $=$. Hence each $dim(Ker (bm N_j^{n_j})) = dim (W_j) = n_j$, QED.
answered Jan 14 at 6:03
xbhxbh
6,1351522
$endgroup$
add a comment |
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1 Answer
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$begingroup$
I'm here to give a demo. This may not be the quickest one.
For a matrix $newcommandbmboldsymbol
DeclareMathOperatorKer{Ker}
newcommandF{{mathbb F}}
newcommandN {{mathbb N}}
bm A in mathrm M_n (F)$ [i.e. $n times n$ with entries in $F$], let $Ker bm A = {bm x in F^n colon bm {Ax} = bm 0}$ From now on, let $bm N_j = bm A - lambda_j bm I_n$.
First we figure out what exactly $W_j = R_{bm A,lambda_j} $ is. For any matrix $bm P in mathrm M_n (F)$, consider $Ker(bm P^s)$ and $Ker(bm P^{s+1})$ where $s in N$. If $bm y in Ker(bm P^s)$, i.e. $bm P^s bm y = bm 0$, then clearly $bm P^{s+1}bm y = bm P(bm P^s bm y) = bm 0$. In other words, $Ker(bm P^s) subseteq Ker(bm P^{s+1})$. So we have a chain:
$$
varnothing subseteq Ker bm P subseteq Ker (bm P^2) subseteq dots subseteq Ker(bm P^s) subseteq cdots
$$
Easy to see that each $Ker (bm P^s)$ is a subspace of $F^n$. Thus $dim(Ker (bm P ^s)) leqslant n$ for all $s$. Therefore there exists some smallest $min N$ s.t. $dim(Ker(bm P^m)) = dim(Ker (bm P^{m+p}))$ for all $p in N^*$.
Additionally, we see that
if $dim(Ker (bm P^s))= dim(Ker (bm P^{s+1}))$ for some $sin N^*$, then $dim(Ker (bm P^s)) = dim (Ker (bm P^{s+p}))$ for all $p in N^*$.
Indeed, under the assumption, we have $Ker (bm P^s) = Ker (bm P^{s+1})$. Then for each $ellin N^*$,if $bm yin Ker(bm P^{s+ell})$, i.e. $bm P^{s+ell} bm y = bm 0$, then $bm P^{s+1} (bm P^{ell-1} bm y)= bm 0$, hence $bm P^{ell-1} bm y in Ker(bm P^{s+1})$. Use the assumption $Ker(bm P^s) = Ker(bm P^{s+1})$, we have $bm P^{ell-1}bm y in Ker(bm P^s)$ or $bm P^{s+ell -1} bm y =bm 0$, thus $bm y in Ker(bm P^{s+ell -1})$. Conclusively
$$
Ker (bm P^{s+ell-1}) = Ker (bm P^{s+ell}), quad ell in N^*,
$$
now let $ell = 1, 2, dots, p-1$ and the claim is proved. To be clearer, we have the following chain:
$$
varnothing subset Ker(bm P) subset dots subset Ker(bm P^{m-1}) subset Ker(bm P^m) = Ker (bm P^{m+1}) = cdots subseteq F^n,
$$
where $subset$ means $subsetneq$.
Now focus on each $bm N_j$. Let the corresponding integer $m$ be $m_j$. Consider $W_j$ and $Ker(bm N_j^{m_j})$. By the definition of $W_j$, clearly $Ker(bm N_j^{m_j}) subseteq W_j$. Conversely if $bm x in W_j$, then there is some $pin N^*$ s.t. $bm N_j^p bm x = bm 0$ by definition, or $bm x in Ker (bm N_j^p)$. According to the chain above, $Ker (bm N_j^p) subseteq Ker(bm N_j^{m_j})$. Therefore we actually have
$$
W_j = Ker(bm N_j^{m_j}), j = 1,2, dots, k.
$$
Now we prove that
$$
W = sum_1^k Ker(bm N_j ^{m_j}) = sum_1^k W_j tag {1}
$$
is also a direct sum.
Equivalently we need to show
If $bm 0 = sum_1^k bm w_j$ where $w_j in W_j$ for each $j$, then $bm w_j =bm 0$ for all $j$.
Let $f_j (t) = (t - lambda_j)^{m_j} in F[t]$ and $g_j (t) = prod_{ell neq j} f_ell (t)$ for each $j$. Clearly $mathrm{gcd}(g_j, f_j) = 1$. Thus there exists $u_j, v_j in F[t]$ such that
$$
u_j f_j + v_j g_j = 1.
$$
Hence
$$
u_j (bm A) f_j(bm A) + v_j (bm A) g_j(bm A) = bm I_n,
$$
and for each $bm y in F^n$,
$$
bm y = u_j (bm A) f_j(bm A)bm y + v_j (bm A) g_j(bm A) bm y. tag{2}
$$
If $bm y in Ker(f_j (bm A)) cap Ker (g_j(bm A))$, i.e. $f_j(bm A) bm y =g_j(bm A)bm y = bm 0$, then $(2)$ becomes $bm y = bm 0$. Hence $Ker (f_j(bm A)) cap Ker(g_j (bm A)) = {bm 0}$. Now suppose $bm 0 = sum_1^k bm w_j$ where $bm w_j in W_j$ for each $j$, then by the definition of $bm w_j$, $g_ell (bm A) bm w_j = bm 0$ if $ell neq j$. Thus $bm w_ell in Ker(g_ell (bm A)) cap Ker(f_ell (bm A)) = 0$, then $bm w_ell = bm 0$ for each $ell$. Therefore $(1)$ is proved.
Finally we get the goal: by the definition of $W_j$, we have $Ker(bm N_j ^{n_j}) subseteq W_j$ for each $j$. Now compute the dimension by using the dimension formula of direct sums, we have
$$
n = dim (F^n) geqslant dim W = dim left(bigoplus _1^k W_jright) = sum_1^k dim left( W_jright) geqslant sum_1^k dim (Ker(bm N_j ^{n_j})) = sum_1^k n_j = n,
$$
so actually the $geqslant$'s are all $=$. Hence each $dim(Ker (bm N_j^{n_j})) = dim (W_j) = n_j$, QED.
answered Jan 14 at 6:03
xbhxbh
6,1351522
$endgroup$
add a comment |
$begingroup$
I'm here to give a demo. This may not be the quickest one.
For a matrix $newcommandbmboldsymbol
DeclareMathOperatorKer{Ker}
newcommandF{{mathbb F}}
newcommandN {{mathbb N}}
bm A in mathrm M_n (F)$ [i.e. $n times n$ with entries in $F$], let $Ker bm A = {bm x in F^n colon bm {Ax} = bm 0}$ From now on, let $bm N_j = bm A - lambda_j bm I_n$.
First we figure out what exactly $W_j = R_{bm A,lambda_j} $ is. For any matrix $bm P in mathrm M_n (F)$, consider $Ker(bm P^s)$ and $Ker(bm P^{s+1})$ where $s in N$. If $bm y in Ker(bm P^s)$, i.e. $bm P^s bm y = bm 0$, then clearly $bm P^{s+1}bm y = bm P(bm P^s bm y) = bm 0$. In other words, $Ker(bm P^s) subseteq Ker(bm P^{s+1})$. So we have a chain:
$$
varnothing subseteq Ker bm P subseteq Ker (bm P^2) subseteq dots subseteq Ker(bm P^s) subseteq cdots
$$
Easy to see that each $Ker (bm P^s)$ is a subspace of $F^n$. Thus $dim(Ker (bm P ^s)) leqslant n$ for all $s$. Therefore there exists some smallest $min N$ s.t. $dim(Ker(bm P^m)) = dim(Ker (bm P^{m+p}))$ for all $p in N^*$.
Additionally, we see that
if $dim(Ker (bm P^s))= dim(Ker (bm P^{s+1}))$ for some $sin N^*$, then $dim(Ker (bm P^s)) = dim (Ker (bm P^{s+p}))$ for all $p in N^*$.
Indeed, under the assumption, we have $Ker (bm P^s) = Ker (bm P^{s+1})$. Then for each $ellin N^*$,if $bm yin Ker(bm P^{s+ell})$, i.e. $bm P^{s+ell} bm y = bm 0$, then $bm P^{s+1} (bm P^{ell-1} bm y)= bm 0$, hence $bm P^{ell-1} bm y in Ker(bm P^{s+1})$. Use the assumption $Ker(bm P^s) = Ker(bm P^{s+1})$, we have $bm P^{ell-1}bm y in Ker(bm P^s)$ or $bm P^{s+ell -1} bm y =bm 0$, thus $bm y in Ker(bm P^{s+ell -1})$. Conclusively
$$
Ker (bm P^{s+ell-1}) = Ker (bm P^{s+ell}), quad ell in N^*,
$$
now let $ell = 1, 2, dots, p-1$ and the claim is proved. To be clearer, we have the following chain:
$$
varnothing subset Ker(bm P) subset dots subset Ker(bm P^{m-1}) subset Ker(bm P^m) = Ker (bm P^{m+1}) = cdots subseteq F^n,
$$
where $subset$ means $subsetneq$.
Now focus on each $bm N_j$. Let the corresponding integer $m$ be $m_j$. Consider $W_j$ and $Ker(bm N_j^{m_j})$. By the definition of $W_j$, clearly $Ker(bm N_j^{m_j}) subseteq W_j$. Conversely if $bm x in W_j$, then there is some $pin N^*$ s.t. $bm N_j^p bm x = bm 0$ by definition, or $bm x in Ker (bm N_j^p)$. According to the chain above, $Ker (bm N_j^p) subseteq Ker(bm N_j^{m_j})$. Therefore we actually have
$$
W_j = Ker(bm N_j^{m_j}), j = 1,2, dots, k.
$$
Now we prove that
$$
W = sum_1^k Ker(bm N_j ^{m_j}) = sum_1^k W_j tag {1}
$$
is also a direct sum.
Equivalently we need to show
If $bm 0 = sum_1^k bm w_j$ where $w_j in W_j$ for each $j$, then $bm w_j =bm 0$ for all $j$.
Let $f_j (t) = (t - lambda_j)^{m_j} in F[t]$ and $g_j (t) = prod_{ell neq j} f_ell (t)$ for each $j$. Clearly $mathrm{gcd}(g_j, f_j) = 1$. Thus there exists $u_j, v_j in F[t]$ such that
$$
u_j f_j + v_j g_j = 1.
$$
Hence
$$
u_j (bm A) f_j(bm A) + v_j (bm A) g_j(bm A) = bm I_n,
$$
and for each $bm y in F^n$,
$$
bm y = u_j (bm A) f_j(bm A)bm y + v_j (bm A) g_j(bm A) bm y. tag{2}
$$
If $bm y in Ker(f_j (bm A)) cap Ker (g_j(bm A))$, i.e. $f_j(bm A) bm y =g_j(bm A)bm y = bm 0$, then $(2)$ becomes $bm y = bm 0$. Hence $Ker (f_j(bm A)) cap Ker(g_j (bm A)) = {bm 0}$. Now suppose $bm 0 = sum_1^k bm w_j$ where $bm w_j in W_j$ for each $j$, then by the definition of $bm w_j$, $g_ell (bm A) bm w_j = bm 0$ if $ell neq j$. Thus $bm w_ell in Ker(g_ell (bm A)) cap Ker(f_ell (bm A)) = 0$, then $bm w_ell = bm 0$ for each $ell$. Therefore $(1)$ is proved.
Finally we get the goal: by the definition of $W_j$, we have $Ker(bm N_j ^{n_j}) subseteq W_j$ for each $j$. Now compute the dimension by using the dimension formula of direct sums, we have
$$
n = dim (F^n) geqslant dim W = dim left(bigoplus _1^k W_jright) = sum_1^k dim left( W_jright) geqslant sum_1^k dim (Ker(bm N_j ^{n_j})) = sum_1^k n_j = n,
$$
so actually the $geqslant$'s are all $=$. Hence each $dim(Ker (bm N_j^{n_j})) = dim (W_j) = n_j$, QED.
answered Jan 14 at 6:03
xbhxbh
6,1351522
$endgroup$
add a comment |
$begingroup$
I'm here to give a demo. This may not be the quickest one.
For a matrix $newcommandbmboldsymbol
DeclareMathOperatorKer{Ker}
newcommandF{{mathbb F}}
newcommandN {{mathbb N}}
bm A in mathrm M_n (F)$ [i.e. $n times n$ with entries in $F$], let $Ker bm A = {bm x in F^n colon bm {Ax} = bm 0}$ From now on, let $bm N_j = bm A - lambda_j bm I_n$.
First we figure out what exactly $W_j = R_{bm A,lambda_j} $ is. For any matrix $bm P in mathrm M_n (F)$, consider $Ker(bm P^s)$ and $Ker(bm P^{s+1})$ where $s in N$. If $bm y in Ker(bm P^s)$, i.e. $bm P^s bm y = bm 0$, then clearly $bm P^{s+1}bm y = bm P(bm P^s bm y) = bm 0$. In other words, $Ker(bm P^s) subseteq Ker(bm P^{s+1})$. So we have a chain:
$$
varnothing subseteq Ker bm P subseteq Ker (bm P^2) subseteq dots subseteq Ker(bm P^s) subseteq cdots
$$
Easy to see that each $Ker (bm P^s)$ is a subspace of $F^n$. Thus $dim(Ker (bm P ^s)) leqslant n$ for all $s$. Therefore there exists some smallest $min N$ s.t. $dim(Ker(bm P^m)) = dim(Ker (bm P^{m+p}))$ for all $p in N^*$.
Additionally, we see that
if $dim(Ker (bm P^s))= dim(Ker (bm P^{s+1}))$ for some $sin N^*$, then $dim(Ker (bm P^s)) = dim (Ker (bm P^{s+p}))$ for all $p in N^*$.
Indeed, under the assumption, we have $Ker (bm P^s) = Ker (bm P^{s+1})$. Then for each $ellin N^*$,if $bm yin Ker(bm P^{s+ell})$, i.e. $bm P^{s+ell} bm y = bm 0$, then $bm P^{s+1} (bm P^{ell-1} bm y)= bm 0$, hence $bm P^{ell-1} bm y in Ker(bm P^{s+1})$. Use the assumption $Ker(bm P^s) = Ker(bm P^{s+1})$, we have $bm P^{ell-1}bm y in Ker(bm P^s)$ or $bm P^{s+ell -1} bm y =bm 0$, thus $bm y in Ker(bm P^{s+ell -1})$. Conclusively
$$
Ker (bm P^{s+ell-1}) = Ker (bm P^{s+ell}), quad ell in N^*,
$$
now let $ell = 1, 2, dots, p-1$ and the claim is proved. To be clearer, we have the following chain:
$$
varnothing subset Ker(bm P) subset dots subset Ker(bm P^{m-1}) subset Ker(bm P^m) = Ker (bm P^{m+1}) = cdots subseteq F^n,
$$
where $subset$ means $subsetneq$.
Now focus on each $bm N_j$. Let the corresponding integer $m$ be $m_j$. Consider $W_j$ and $Ker(bm N_j^{m_j})$. By the definition of $W_j$, clearly $Ker(bm N_j^{m_j}) subseteq W_j$. Conversely if $bm x in W_j$, then there is some $pin N^*$ s.t. $bm N_j^p bm x = bm 0$ by definition, or $bm x in Ker (bm N_j^p)$. According to the chain above, $Ker (bm N_j^p) subseteq Ker(bm N_j^{m_j})$. Therefore we actually have
$$
W_j = Ker(bm N_j^{m_j}), j = 1,2, dots, k.
$$
Now we prove that
$$
W = sum_1^k Ker(bm N_j ^{m_j}) = sum_1^k W_j tag {1}
$$
is also a direct sum.
Equivalently we need to show
If $bm 0 = sum_1^k bm w_j$ where $w_j in W_j$ for each $j$, then $bm w_j =bm 0$ for all $j$.
Let $f_j (t) = (t - lambda_j)^{m_j} in F[t]$ and $g_j (t) = prod_{ell neq j} f_ell (t)$ for each $j$. Clearly $mathrm{gcd}(g_j, f_j) = 1$. Thus there exists $u_j, v_j in F[t]$ such that
$$
u_j f_j + v_j g_j = 1.
$$
Hence
$$
u_j (bm A) f_j(bm A) + v_j (bm A) g_j(bm A) = bm I_n,
$$
and for each $bm y in F^n$,
$$
bm y = u_j (bm A) f_j(bm A)bm y + v_j (bm A) g_j(bm A) bm y. tag{2}
$$
If $bm y in Ker(f_j (bm A)) cap Ker (g_j(bm A))$, i.e. $f_j(bm A) bm y =g_j(bm A)bm y = bm 0$, then $(2)$ becomes $bm y = bm 0$. Hence $Ker (f_j(bm A)) cap Ker(g_j (bm A)) = {bm 0}$. Now suppose $bm 0 = sum_1^k bm w_j$ where $bm w_j in W_j$ for each $j$, then by the definition of $bm w_j$, $g_ell (bm A) bm w_j = bm 0$ if $ell neq j$. Thus $bm w_ell in Ker(g_ell (bm A)) cap Ker(f_ell (bm A)) = 0$, then $bm w_ell = bm 0$ for each $ell$. Therefore $(1)$ is proved.
Finally we get the goal: by the definition of $W_j$, we have $Ker(bm N_j ^{n_j}) subseteq W_j$ for each $j$. Now compute the dimension by using the dimension formula of direct sums, we have
$$
n = dim (F^n) geqslant dim W = dim left(bigoplus _1^k W_jright) = sum_1^k dim left( W_jright) geqslant sum_1^k dim (Ker(bm N_j ^{n_j})) = sum_1^k n_j = n,
$$
so actually the $geqslant$'s are all $=$. Hence each $dim(Ker (bm N_j^{n_j})) = dim (W_j) = n_j$, QED.
answered Jan 14 at 6:03
xbhxbh
6,1351522
$endgroup$
I'm here to give a demo. This may not be the quickest one.
For a matrix $newcommandbmboldsymbol
DeclareMathOperatorKer{Ker}
newcommandF{{mathbb F}}
newcommandN {{mathbb N}}
bm A in mathrm M_n (F)$ [i.e. $n times n$ with entries in $F$], let $Ker bm A = {bm x in F^n colon bm {Ax} = bm 0}$ From now on, let $bm N_j = bm A - lambda_j bm I_n$.
First we figure out what exactly $W_j = R_{bm A,lambda_j} $ is. For any matrix $bm P in mathrm M_n (F)$, consider $Ker(bm P^s)$ and $Ker(bm P^{s+1})$ where $s in N$. If $bm y in Ker(bm P^s)$, i.e. $bm P^s bm y = bm 0$, then clearly $bm P^{s+1}bm y = bm P(bm P^s bm y) = bm 0$. In other words, $Ker(bm P^s) subseteq Ker(bm P^{s+1})$. So we have a chain:
$$
varnothing subseteq Ker bm P subseteq Ker (bm P^2) subseteq dots subseteq Ker(bm P^s) subseteq cdots
$$
Easy to see that each $Ker (bm P^s)$ is a subspace of $F^n$. Thus $dim(Ker (bm P ^s)) leqslant n$ for all $s$. Therefore there exists some smallest $min N$ s.t. $dim(Ker(bm P^m)) = dim(Ker (bm P^{m+p}))$ for all $p in N^*$.
Additionally, we see that
if $dim(Ker (bm P^s))= dim(Ker (bm P^{s+1}))$ for some $sin N^*$, then $dim(Ker (bm P^s)) = dim (Ker (bm P^{s+p}))$ for all $p in N^*$.
Indeed, under the assumption, we have $Ker (bm P^s) = Ker (bm P^{s+1})$. Then for each $ellin N^*$,if $bm yin Ker(bm P^{s+ell})$, i.e. $bm P^{s+ell} bm y = bm 0$, then $bm P^{s+1} (bm P^{ell-1} bm y)= bm 0$, hence $bm P^{ell-1} bm y in Ker(bm P^{s+1})$. Use the assumption $Ker(bm P^s) = Ker(bm P^{s+1})$, we have $bm P^{ell-1}bm y in Ker(bm P^s)$ or $bm P^{s+ell -1} bm y =bm 0$, thus $bm y in Ker(bm P^{s+ell -1})$. Conclusively
$$
Ker (bm P^{s+ell-1}) = Ker (bm P^{s+ell}), quad ell in N^*,
$$
now let $ell = 1, 2, dots, p-1$ and the claim is proved. To be clearer, we have the following chain:
$$
varnothing subset Ker(bm P) subset dots subset Ker(bm P^{m-1}) subset Ker(bm P^m) = Ker (bm P^{m+1}) = cdots subseteq F^n,
$$
where $subset$ means $subsetneq$.
Now focus on each $bm N_j$. Let the corresponding integer $m$ be $m_j$. Consider $W_j$ and $Ker(bm N_j^{m_j})$. By the definition of $W_j$, clearly $Ker(bm N_j^{m_j}) subseteq W_j$. Conversely if $bm x in W_j$, then there is some $pin N^*$ s.t. $bm N_j^p bm x = bm 0$ by definition, or $bm x in Ker (bm N_j^p)$. According to the chain above, $Ker (bm N_j^p) subseteq Ker(bm N_j^{m_j})$. Therefore we actually have
$$
W_j = Ker(bm N_j^{m_j}), j = 1,2, dots, k.
$$
Now we prove that
$$
W = sum_1^k Ker(bm N_j ^{m_j}) = sum_1^k W_j tag {1}
$$
is also a direct sum.
Equivalently we need to show
If $bm 0 = sum_1^k bm w_j$ where $w_j in W_j$ for each $j$, then $bm w_j =bm 0$ for all $j$.
Let $f_j (t) = (t - lambda_j)^{m_j} in F[t]$ and $g_j (t) = prod_{ell neq j} f_ell (t)$ for each $j$. Clearly $mathrm{gcd}(g_j, f_j) = 1$. Thus there exists $u_j, v_j in F[t]$ such that
$$
u_j f_j + v_j g_j = 1.
$$
Hence
$$
u_j (bm A) f_j(bm A) + v_j (bm A) g_j(bm A) = bm I_n,
$$
and for each $bm y in F^n$,
$$
bm y = u_j (bm A) f_j(bm A)bm y + v_j (bm A) g_j(bm A) bm y. tag{2}
$$
If $bm y in Ker(f_j (bm A)) cap Ker (g_j(bm A))$, i.e. $f_j(bm A) bm y =g_j(bm A)bm y = bm 0$, then $(2)$ becomes $bm y = bm 0$. Hence $Ker (f_j(bm A)) cap Ker(g_j (bm A)) = {bm 0}$. Now suppose $bm 0 = sum_1^k bm w_j$ where $bm w_j in W_j$ for each $j$, then by the definition of $bm w_j$, $g_ell (bm A) bm w_j = bm 0$ if $ell neq j$. Thus $bm w_ell in Ker(g_ell (bm A)) cap Ker(f_ell (bm A)) = 0$, then $bm w_ell = bm 0$ for each $ell$. Therefore $(1)$ is proved.
Finally we get the goal: by the definition of $W_j$, we have $Ker(bm N_j ^{n_j}) subseteq W_j$ for each $j$. Now compute the dimension by using the dimension formula of direct sums, we have
$$
n = dim (F^n) geqslant dim W = dim left(bigoplus _1^k W_jright) = sum_1^k dim left( W_jright) geqslant sum_1^k dim (Ker(bm N_j ^{n_j})) = sum_1^k n_j = n,
$$
so actually the $geqslant$'s are all $=$. Hence each $dim(Ker (bm N_j^{n_j})) = dim (W_j) = n_j$, QED.
answered Jan 14 at 6:03
xbhxbh
6,1351522
answered Jan 14 at 6:03
xbhxbh
6,1351522
answered Jan 14 at 6:03
xbhxbh
6,1351522
answered Jan 14 at 6:03
xbhxbh
6,1351522
6,1351522
add a comment |
add a comment |
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1
$begingroup$
That $m$ is different for $x, y$.
$endgroup$
– xbh
Jan 13 at 7:50
$begingroup$
I don't understand. What do you mean?
$endgroup$
– Minh
Jan 13 at 7:51
$begingroup$
For $x$ there is some $m_1$ that $(A-lambda_j I_n)^{m_1} =0$; for $y$ there is another $m_2$ that $(A- lambda_j I_n)^{m_2} = 0$.
$endgroup$
– xbh
Jan 13 at 7:53
$begingroup$
I see, but if $m$ if different for $x,y$, how can we prove that $x+y in mathcal{R}_{A,lambda_i}$
$endgroup$
– Minh
Jan 13 at 7:56
$begingroup$
WOLG, suppose $m_1 geqslant m_2$, then how could you find an $m$ for $x+y$?
$endgroup$
– xbh
Jan 13 at 7:57