Mixing
Power of a Jordan Normal Form
$begingroup$
In my notes I have that the Jordan normal form of $B^2$ is $$begin{bmatrix}
0 & 1 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 &1 \
0 & 0 & 0 & 0
end{bmatrix}$$ and the notes say that because of this, the only possibility for $B$ is
$$begin{bmatrix}
0 & 1 & 0 & 0\
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0
end{bmatrix}$$
Can anybody explain why this is the case? How can we conclude the jordan form of B from the jordan form of B^2? Thanks for your help!
jordan-normal-form
asked Jan 7 at 5:33
BOlivianoperuano84BOlivianoperuano84
1778
$endgroup$
add a comment |
$begingroup$
In my notes I have that the Jordan normal form of $B^2$ is $$begin{bmatrix}
0 & 1 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 &1 \
0 & 0 & 0 & 0
end{bmatrix}$$ and the notes say that because of this, the only possibility for $B$ is
$$begin{bmatrix}
0 & 1 & 0 & 0\
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0
end{bmatrix}$$
Can anybody explain why this is the case? How can we conclude the jordan form of B from the jordan form of B^2? Thanks for your help!
jordan-normal-form
asked Jan 7 at 5:33
BOlivianoperuano84BOlivianoperuano84
1778
$endgroup$
add a comment |
$begingroup$
In my notes I have that the Jordan normal form of $B^2$ is $$begin{bmatrix}
0 & 1 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 &1 \
0 & 0 & 0 & 0
end{bmatrix}$$ and the notes say that because of this, the only possibility for $B$ is
$$begin{bmatrix}
0 & 1 & 0 & 0\
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0
end{bmatrix}$$
Can anybody explain why this is the case? How can we conclude the jordan form of B from the jordan form of B^2? Thanks for your help!
jordan-normal-form
asked Jan 7 at 5:33
BOlivianoperuano84BOlivianoperuano84
1778
$endgroup$
In my notes I have that the Jordan normal form of $B^2$ is $$begin{bmatrix}
0 & 1 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 &1 \
0 & 0 & 0 & 0
end{bmatrix}$$ and the notes say that because of this, the only possibility for $B$ is
$$begin{bmatrix}
0 & 1 & 0 & 0\
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0
end{bmatrix}$$
Can anybody explain why this is the case? How can we conclude the jordan form of B from the jordan form of B^2? Thanks for your help!
jordan-normal-form
jordan-normal-form
asked Jan 7 at 5:33
BOlivianoperuano84BOlivianoperuano84
1778
asked Jan 7 at 5:33
BOlivianoperuano84BOlivianoperuano84
1778
asked Jan 7 at 5:33
BOlivianoperuano84BOlivianoperuano84
1778
asked Jan 7 at 5:33
BOlivianoperuano84BOlivianoperuano84
1778
asked Jan 7 at 5:33
BOlivianoperuano84BOlivianoperuano84
1778
1778
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
If $lambda$ is an eigenvalue of $B$, then $lambda^2$ is an eigenvalue of $B^2$. In the Jordan Normal Form, the eigenvalues correspond to entries on the main diagonal. Since all such entries in $B^2$ are $0$, this means that all eigenvalues are $0$ for both $B$ and $B^2$. That takes care of the diagonal entries for the Jordan form of $B$.
We know that the entries above a Jordan block are all equal to $1$. Since we have only one block of a single eigenvalue, namely $0$, then the above-diagonal entries for the Jordan form of $B$ will be all $1$ and that gives the desired form.
answered Jan 7 at 7:18
EuxhenHEuxhenH
484210
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
If $lambda$ is an eigenvalue of $B$, then $lambda^2$ is an eigenvalue of $B^2$. In the Jordan Normal Form, the eigenvalues correspond to entries on the main diagonal. Since all such entries in $B^2$ are $0$, this means that all eigenvalues are $0$ for both $B$ and $B^2$. That takes care of the diagonal entries for the Jordan form of $B$.
We know that the entries above a Jordan block are all equal to $1$. Since we have only one block of a single eigenvalue, namely $0$, then the above-diagonal entries for the Jordan form of $B$ will be all $1$ and that gives the desired form.
answered Jan 7 at 7:18
EuxhenHEuxhenH
484210
$endgroup$
add a comment |
$begingroup$
If $lambda$ is an eigenvalue of $B$, then $lambda^2$ is an eigenvalue of $B^2$. In the Jordan Normal Form, the eigenvalues correspond to entries on the main diagonal. Since all such entries in $B^2$ are $0$, this means that all eigenvalues are $0$ for both $B$ and $B^2$. That takes care of the diagonal entries for the Jordan form of $B$.
We know that the entries above a Jordan block are all equal to $1$. Since we have only one block of a single eigenvalue, namely $0$, then the above-diagonal entries for the Jordan form of $B$ will be all $1$ and that gives the desired form.
answered Jan 7 at 7:18
EuxhenHEuxhenH
484210
$endgroup$
add a comment |
$begingroup$
If $lambda$ is an eigenvalue of $B$, then $lambda^2$ is an eigenvalue of $B^2$. In the Jordan Normal Form, the eigenvalues correspond to entries on the main diagonal. Since all such entries in $B^2$ are $0$, this means that all eigenvalues are $0$ for both $B$ and $B^2$. That takes care of the diagonal entries for the Jordan form of $B$.
We know that the entries above a Jordan block are all equal to $1$. Since we have only one block of a single eigenvalue, namely $0$, then the above-diagonal entries for the Jordan form of $B$ will be all $1$ and that gives the desired form.
answered Jan 7 at 7:18
EuxhenHEuxhenH
484210
$endgroup$
If $lambda$ is an eigenvalue of $B$, then $lambda^2$ is an eigenvalue of $B^2$. In the Jordan Normal Form, the eigenvalues correspond to entries on the main diagonal. Since all such entries in $B^2$ are $0$, this means that all eigenvalues are $0$ for both $B$ and $B^2$. That takes care of the diagonal entries for the Jordan form of $B$.
We know that the entries above a Jordan block are all equal to $1$. Since we have only one block of a single eigenvalue, namely $0$, then the above-diagonal entries for the Jordan form of $B$ will be all $1$ and that gives the desired form.
answered Jan 7 at 7:18
EuxhenHEuxhenH
484210
answered Jan 7 at 7:18
EuxhenHEuxhenH
484210
answered Jan 7 at 7:18
EuxhenHEuxhenH
484210
answered Jan 7 at 7:18
EuxhenHEuxhenH
484210
484210
add a comment |
add a comment |
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