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Prove that if symmetric $A$ and skew-symmetric $B$ are similar, then $A = B = O$
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Let $A$, $B$ and $Q$ be square matrices with real entries. Show that if $A$ is symmetric, $B$ is skew-symmetric, and $Q$ is invertible such that $Q^-1 A Q=B$, then $A=B=0$.
linear-algebra matrices
edited Jul 21 at 11:30
Rodrigo de Azevedo
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éƒÂJOE
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Let $A$, $B$ and $Q$ be square matrices with real entries. Show that if $A$ is symmetric, $B$ is skew-symmetric, and $Q$ is invertible such that $Q^-1 A Q=B$, then $A=B=0$.
linear-algebra matrices
edited Jul 21 at 11:30
Rodrigo de Azevedo
12.6k41751
asked Jul 21 at 8:16
éƒÂJOE
61
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $A$, $B$ and $Q$ be square matrices with real entries. Show that if $A$ is symmetric, $B$ is skew-symmetric, and $Q$ is invertible such that $Q^-1 A Q=B$, then $A=B=0$.
linear-algebra matrices
edited Jul 21 at 11:30
Rodrigo de Azevedo
12.6k41751
asked Jul 21 at 8:16
éƒÂJOE
61
Let $A$, $B$ and $Q$ be square matrices with real entries. Show that if $A$ is symmetric, $B$ is skew-symmetric, and $Q$ is invertible such that $Q^-1 A Q=B$, then $A=B=0$.
linear-algebra matrices
edited Jul 21 at 11:30
Rodrigo de Azevedo
12.6k41751
asked Jul 21 at 8:16
éƒÂJOE
61
edited Jul 21 at 11:30
Rodrigo de Azevedo
12.6k41751
edited Jul 21 at 11:30
Rodrigo de Azevedo
12.6k41751
edited Jul 21 at 11:30
Rodrigo de Azevedo
12.6k41751
12.6k41751
asked Jul 21 at 8:16
éƒÂJOE
61
asked Jul 21 at 8:16
éƒÂJOE
61
asked Jul 21 at 8:16
éƒÂJOE
61
61
add a comment |Â
add a comment |Â
1 Answer
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From the condition $Q^-1AQ = B$ we have $A = QBQ^-1$ which can be viewed as normal Jordan form of matrix $A$, however the Jordan blocks are upper triangular matrices and the only possibility for them to be skew-symmetric is that matrix $A$ is diagonolizable and each Jordan block has size of $1times 1$. Hence, $B$ is diagonal and from the skew-symmetry it follows that the diagonal elements are zeros as well. Hence, matrix $A$ is also zero-matrix.
answered Jul 21 at 8:30
pointguard0
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
From the condition $Q^-1AQ = B$ we have $A = QBQ^-1$ which can be viewed as normal Jordan form of matrix $A$, however the Jordan blocks are upper triangular matrices and the only possibility for them to be skew-symmetric is that matrix $A$ is diagonolizable and each Jordan block has size of $1times 1$. Hence, $B$ is diagonal and from the skew-symmetry it follows that the diagonal elements are zeros as well. Hence, matrix $A$ is also zero-matrix.
answered Jul 21 at 8:30
pointguard0
1,027718
add a comment |Â
up vote
0
down vote
From the condition $Q^-1AQ = B$ we have $A = QBQ^-1$ which can be viewed as normal Jordan form of matrix $A$, however the Jordan blocks are upper triangular matrices and the only possibility for them to be skew-symmetric is that matrix $A$ is diagonolizable and each Jordan block has size of $1times 1$. Hence, $B$ is diagonal and from the skew-symmetry it follows that the diagonal elements are zeros as well. Hence, matrix $A$ is also zero-matrix.
answered Jul 21 at 8:30
pointguard0
1,027718
add a comment |Â
up vote
0
down vote
up vote
0
down vote
From the condition $Q^-1AQ = B$ we have $A = QBQ^-1$ which can be viewed as normal Jordan form of matrix $A$, however the Jordan blocks are upper triangular matrices and the only possibility for them to be skew-symmetric is that matrix $A$ is diagonolizable and each Jordan block has size of $1times 1$. Hence, $B$ is diagonal and from the skew-symmetry it follows that the diagonal elements are zeros as well. Hence, matrix $A$ is also zero-matrix.
answered Jul 21 at 8:30
pointguard0
1,027718
From the condition $Q^-1AQ = B$ we have $A = QBQ^-1$ which can be viewed as normal Jordan form of matrix $A$, however the Jordan blocks are upper triangular matrices and the only possibility for them to be skew-symmetric is that matrix $A$ is diagonolizable and each Jordan block has size of $1times 1$. Hence, $B$ is diagonal and from the skew-symmetry it follows that the diagonal elements are zeros as well. Hence, matrix $A$ is also zero-matrix.
answered Jul 21 at 8:30
pointguard0
1,027718
answered Jul 21 at 8:30
pointguard0
1,027718
answered Jul 21 at 8:30
pointguard0
1,027718
answered Jul 21 at 8:30
pointguard0
1,027718
1,027718
add a comment |Â
add a comment |Â
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