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Linear algebra, help understanding cayley + jordan form
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On page 107 of evan chen's napkin, I don't understand why T is indecomposable means there is an eigenvalue.
On page 110 I don't get how $(X- lambda )^d$ is the zero map.
https://usamo.files.wordpress.com/2017/12/napkin-2017-12-11.pdf
linear-algebra
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RayOfHope
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On page 107 of evan chen's napkin, I don't understand why T is indecomposable means there is an eigenvalue.
On page 110 I don't get how $(X- lambda )^d$ is the zero map.
https://usamo.files.wordpress.com/2017/12/napkin-2017-12-11.pdf
linear-algebra
asked yesterday
RayOfHope
335
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
On page 107 of evan chen's napkin, I don't understand why T is indecomposable means there is an eigenvalue.
On page 110 I don't get how $(X- lambda )^d$ is the zero map.
https://usamo.files.wordpress.com/2017/12/napkin-2017-12-11.pdf
linear-algebra
asked yesterday
RayOfHope
335
On page 107 of evan chen's napkin, I don't understand why T is indecomposable means there is an eigenvalue.
On page 110 I don't get how $(X- lambda )^d$ is the zero map.
https://usamo.files.wordpress.com/2017/12/napkin-2017-12-11.pdf
linear-algebra
asked yesterday
RayOfHope
335
asked yesterday
RayOfHope
335
asked yesterday
RayOfHope
335
asked yesterday
RayOfHope
335
335
add a comment |Â
add a comment |Â
1 Answer
1
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1
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I don't understand why T is indecomposable means there is an eigenvalue.
Jordan form $J$ of a matrix $A$ is
$$
J = M^-1AM=beginbmatrix
J_1 & \
& . \
& & .\
& & &J_s
endbmatrixqquad(1)
$$
where each block $J_i$ is a Jordan block, given by
$$
J_i = beginbmatrix
lambda_i & 1 \
& lambda_i & . \
& & . &1\
& & &lambda_i
endbmatrixqquad(2)
$$
This Jordan block represents a single eigenvalue $lambda_i$ of the matrix $A$
So $T$ is indecomposable means that $T$ is of the form $J_i$, in this case $T:V_1 to V_1 $ , $V_1 = [J_1]$
and since $J_1$ represents a single eigenvalue of $A$, there is an eigenvalue.
I don't get how $(X−λ)^d$ is the zero map.
From Pg.110 of the pdf you linked,
$$
p_T(X) = (X−λ_1I)^d_1(X−λ_2I)^d_2...(X−λ_mI)^d_m
$$
Writing each $X$ in its Jordan form, $X = MJM^-1$,
$$
p_T(J) = (MJM^-1−λ_1MM^-1)^d_1(MJM^-1−λ_2MM^-1)^d_2...(MJM^-1−λ_mMM^-1)^d_m
$$
$$
p_T(J) = bigl(M(J-λ_1I)M^-1bigr)^d_1bigl(M(J-λ_2I)M^-1bigr)^d_2...bigl(M(J-λ_mI)M^-1bigr)^d_m
$$
Now concentrating on each $J-λ_iI$, from eq(1) and eq(2), this produces an upper triangular block matrix with $0$s on diagonal. For each term in the product, there are $0$s on corresponding diagonals and their product will result in a Zero matrix.
answered yesterday
artha
264
Thx, another question, on page 105, they said ''T acts on each of these subspaces independently. These subspaces correspond to the blocks in the matrix above''. What exactly do they mean by this?
– RayOfHope
23 hours ago
Each block corresponds to a single eigenvector. All these eigenvectors are independent and so T acts on each of them independently. Does this answer your question? If not I will add a detailed explanation.
– artha
22 hours ago
Could you add a detailed explanation please?
– RayOfHope
33 mins ago
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
I don't understand why T is indecomposable means there is an eigenvalue.
Jordan form $J$ of a matrix $A$ is
$$
J = M^-1AM=beginbmatrix
J_1 & \
& . \
& & .\
& & &J_s
endbmatrixqquad(1)
$$
where each block $J_i$ is a Jordan block, given by
$$
J_i = beginbmatrix
lambda_i & 1 \
& lambda_i & . \
& & . &1\
& & &lambda_i
endbmatrixqquad(2)
$$
This Jordan block represents a single eigenvalue $lambda_i$ of the matrix $A$
So $T$ is indecomposable means that $T$ is of the form $J_i$, in this case $T:V_1 to V_1 $ , $V_1 = [J_1]$
and since $J_1$ represents a single eigenvalue of $A$, there is an eigenvalue.
I don't get how $(X−λ)^d$ is the zero map.
From Pg.110 of the pdf you linked,
$$
p_T(X) = (X−λ_1I)^d_1(X−λ_2I)^d_2...(X−λ_mI)^d_m
$$
Writing each $X$ in its Jordan form, $X = MJM^-1$,
$$
p_T(J) = (MJM^-1−λ_1MM^-1)^d_1(MJM^-1−λ_2MM^-1)^d_2...(MJM^-1−λ_mMM^-1)^d_m
$$
$$
p_T(J) = bigl(M(J-λ_1I)M^-1bigr)^d_1bigl(M(J-λ_2I)M^-1bigr)^d_2...bigl(M(J-λ_mI)M^-1bigr)^d_m
$$
Now concentrating on each $J-λ_iI$, from eq(1) and eq(2), this produces an upper triangular block matrix with $0$s on diagonal. For each term in the product, there are $0$s on corresponding diagonals and their product will result in a Zero matrix.
answered yesterday
artha
264
Thx, another question, on page 105, they said ''T acts on each of these subspaces independently. These subspaces correspond to the blocks in the matrix above''. What exactly do they mean by this?
– RayOfHope
23 hours ago
Each block corresponds to a single eigenvector. All these eigenvectors are independent and so T acts on each of them independently. Does this answer your question? If not I will add a detailed explanation.
– artha
22 hours ago
Could you add a detailed explanation please?
– RayOfHope
33 mins ago
add a comment |Â
up vote
1
down vote
I don't understand why T is indecomposable means there is an eigenvalue.
Jordan form $J$ of a matrix $A$ is
$$
J = M^-1AM=beginbmatrix
J_1 & \
& . \
& & .\
& & &J_s
endbmatrixqquad(1)
$$
where each block $J_i$ is a Jordan block, given by
$$
J_i = beginbmatrix
lambda_i & 1 \
& lambda_i & . \
& & . &1\
& & &lambda_i
endbmatrixqquad(2)
$$
This Jordan block represents a single eigenvalue $lambda_i$ of the matrix $A$
So $T$ is indecomposable means that $T$ is of the form $J_i$, in this case $T:V_1 to V_1 $ , $V_1 = [J_1]$
and since $J_1$ represents a single eigenvalue of $A$, there is an eigenvalue.
I don't get how $(X−λ)^d$ is the zero map.
From Pg.110 of the pdf you linked,
$$
p_T(X) = (X−λ_1I)^d_1(X−λ_2I)^d_2...(X−λ_mI)^d_m
$$
Writing each $X$ in its Jordan form, $X = MJM^-1$,
$$
p_T(J) = (MJM^-1−λ_1MM^-1)^d_1(MJM^-1−λ_2MM^-1)^d_2...(MJM^-1−λ_mMM^-1)^d_m
$$
$$
p_T(J) = bigl(M(J-λ_1I)M^-1bigr)^d_1bigl(M(J-λ_2I)M^-1bigr)^d_2...bigl(M(J-λ_mI)M^-1bigr)^d_m
$$
Now concentrating on each $J-λ_iI$, from eq(1) and eq(2), this produces an upper triangular block matrix with $0$s on diagonal. For each term in the product, there are $0$s on corresponding diagonals and their product will result in a Zero matrix.
answered yesterday
artha
264
Thx, another question, on page 105, they said ''T acts on each of these subspaces independently. These subspaces correspond to the blocks in the matrix above''. What exactly do they mean by this?
– RayOfHope
23 hours ago
Each block corresponds to a single eigenvector. All these eigenvectors are independent and so T acts on each of them independently. Does this answer your question? If not I will add a detailed explanation.
– artha
22 hours ago
Could you add a detailed explanation please?
– RayOfHope
33 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I don't understand why T is indecomposable means there is an eigenvalue.
Jordan form $J$ of a matrix $A$ is
$$
J = M^-1AM=beginbmatrix
J_1 & \
& . \
& & .\
& & &J_s
endbmatrixqquad(1)
$$
where each block $J_i$ is a Jordan block, given by
$$
J_i = beginbmatrix
lambda_i & 1 \
& lambda_i & . \
& & . &1\
& & &lambda_i
endbmatrixqquad(2)
$$
This Jordan block represents a single eigenvalue $lambda_i$ of the matrix $A$
So $T$ is indecomposable means that $T$ is of the form $J_i$, in this case $T:V_1 to V_1 $ , $V_1 = [J_1]$
and since $J_1$ represents a single eigenvalue of $A$, there is an eigenvalue.
I don't get how $(X−λ)^d$ is the zero map.
From Pg.110 of the pdf you linked,
$$
p_T(X) = (X−λ_1I)^d_1(X−λ_2I)^d_2...(X−λ_mI)^d_m
$$
Writing each $X$ in its Jordan form, $X = MJM^-1$,
$$
p_T(J) = (MJM^-1−λ_1MM^-1)^d_1(MJM^-1−λ_2MM^-1)^d_2...(MJM^-1−λ_mMM^-1)^d_m
$$
$$
p_T(J) = bigl(M(J-λ_1I)M^-1bigr)^d_1bigl(M(J-λ_2I)M^-1bigr)^d_2...bigl(M(J-λ_mI)M^-1bigr)^d_m
$$
Now concentrating on each $J-λ_iI$, from eq(1) and eq(2), this produces an upper triangular block matrix with $0$s on diagonal. For each term in the product, there are $0$s on corresponding diagonals and their product will result in a Zero matrix.
answered yesterday
artha
264
I don't understand why T is indecomposable means there is an eigenvalue.
Jordan form $J$ of a matrix $A$ is
$$
J = M^-1AM=beginbmatrix
J_1 & \
& . \
& & .\
& & &J_s
endbmatrixqquad(1)
$$
where each block $J_i$ is a Jordan block, given by
$$
J_i = beginbmatrix
lambda_i & 1 \
& lambda_i & . \
& & . &1\
& & &lambda_i
endbmatrixqquad(2)
$$
This Jordan block represents a single eigenvalue $lambda_i$ of the matrix $A$
So $T$ is indecomposable means that $T$ is of the form $J_i$, in this case $T:V_1 to V_1 $ , $V_1 = [J_1]$
and since $J_1$ represents a single eigenvalue of $A$, there is an eigenvalue.
I don't get how $(X−λ)^d$ is the zero map.
From Pg.110 of the pdf you linked,
$$
p_T(X) = (X−λ_1I)^d_1(X−λ_2I)^d_2...(X−λ_mI)^d_m
$$
Writing each $X$ in its Jordan form, $X = MJM^-1$,
$$
p_T(J) = (MJM^-1−λ_1MM^-1)^d_1(MJM^-1−λ_2MM^-1)^d_2...(MJM^-1−λ_mMM^-1)^d_m
$$
$$
p_T(J) = bigl(M(J-λ_1I)M^-1bigr)^d_1bigl(M(J-λ_2I)M^-1bigr)^d_2...bigl(M(J-λ_mI)M^-1bigr)^d_m
$$
Now concentrating on each $J-λ_iI$, from eq(1) and eq(2), this produces an upper triangular block matrix with $0$s on diagonal. For each term in the product, there are $0$s on corresponding diagonals and their product will result in a Zero matrix.
answered yesterday
artha
264
answered yesterday
artha
264
answered yesterday
artha
264
answered yesterday
artha
264
264
Thx, another question, on page 105, they said ''T acts on each of these subspaces independently. These subspaces correspond to the blocks in the matrix above''. What exactly do they mean by this?
– RayOfHope
23 hours ago
Each block corresponds to a single eigenvector. All these eigenvectors are independent and so T acts on each of them independently. Does this answer your question? If not I will add a detailed explanation.
– artha
22 hours ago
Could you add a detailed explanation please?
– RayOfHope
33 mins ago
add a comment |Â
Thx, another question, on page 105, they said ''T acts on each of these subspaces independently. These subspaces correspond to the blocks in the matrix above''. What exactly do they mean by this?
– RayOfHope
23 hours ago
Each block corresponds to a single eigenvector. All these eigenvectors are independent and so T acts on each of them independently. Does this answer your question? If not I will add a detailed explanation.
– artha
22 hours ago
Could you add a detailed explanation please?
– RayOfHope
33 mins ago
Thx, another question, on page 105, they said ''T acts on each of these subspaces independently. These subspaces correspond to the blocks in the matrix above''. What exactly do they mean by this?
– RayOfHope
23 hours ago
Thx, another question, on page 105, they said ''T acts on each of these subspaces independently. These subspaces correspond to the blocks in the matrix above''. What exactly do they mean by this?
– RayOfHope
23 hours ago
Each block corresponds to a single eigenvector. All these eigenvectors are independent and so T acts on each of them independently. Does this answer your question? If not I will add a detailed explanation.
– artha
22 hours ago
Each block corresponds to a single eigenvector. All these eigenvectors are independent and so T acts on each of them independently. Does this answer your question? If not I will add a detailed explanation.
– artha
22 hours ago
Could you add a detailed explanation please?
– RayOfHope
33 mins ago
Could you add a detailed explanation please?
– RayOfHope
33 mins ago
add a comment |Â
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