Characteristic polynomial roots of $ 3times 3$ orthogonal matrix.
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How would you approach this problem?
Let $A$ be an orthogonal $3$ by $3$ matrix. That is, $A^TA = AA^T=I_3$. Prove that the characteristic polynomial $textitp_A$ has a real root.
I am not familiar with how to prove a third degree polynomial has a real root. I started the problem by noticing that $det(A-tI) = det(A-tAA^T)=det(A)det(I-tA^T)$, but this isn't getting me anywhere.
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
edited Aug 30 at 0:04
Bernard
112k635102
asked Aug 30 at 0:03
b_chao
111
 |Â
show 2 more comments
up vote
2
down vote
favorite
How would you approach this problem?
Let $A$ be an orthogonal $3$ by $3$ matrix. That is, $A^TA = AA^T=I_3$. Prove that the characteristic polynomial $textitp_A$ has a real root.
I am not familiar with how to prove a third degree polynomial has a real root. I started the problem by noticing that $det(A-tI) = det(A-tAA^T)=det(A)det(I-tA^T)$, but this isn't getting me anywhere.
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
edited Aug 30 at 0:04
Bernard
112k635102
asked Aug 30 at 0:03
b_chao
111
Welcome to Maths SX! All cubic polynomials have (at least) one real root, and this has nothing to do with orthogonal matrices: it relies on the Intermediate value theorem.
– Bernard
Aug 30 at 0:06
@Bernard: then the claim holds for all 3x3 matrices?
– b_chao
Aug 30 at 0:09
Over the field $mathbf R$, yes.
– Bernard
Aug 30 at 0:10
@Bernard: I see, thanks!
– b_chao
Aug 30 at 0:12
1
@TheoreticalEconomist: That's a nicer, but more sophisticated argument: it supposes you've proved $mathbf C$ is algebraically closed – which isn't exactly trivial.
– Bernard
Aug 30 at 8:23
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
How would you approach this problem?
Let $A$ be an orthogonal $3$ by $3$ matrix. That is, $A^TA = AA^T=I_3$. Prove that the characteristic polynomial $textitp_A$ has a real root.
I am not familiar with how to prove a third degree polynomial has a real root. I started the problem by noticing that $det(A-tI) = det(A-tAA^T)=det(A)det(I-tA^T)$, but this isn't getting me anywhere.
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
edited Aug 30 at 0:04
Bernard
112k635102
asked Aug 30 at 0:03
b_chao
111
How would you approach this problem?
Let $A$ be an orthogonal $3$ by $3$ matrix. That is, $A^TA = AA^T=I_3$. Prove that the characteristic polynomial $textitp_A$ has a real root.
I am not familiar with how to prove a third degree polynomial has a real root. I started the problem by noticing that $det(A-tI) = det(A-tAA^T)=det(A)det(I-tA^T)$, but this isn't getting me anywhere.
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
edited Aug 30 at 0:04
Bernard
112k635102
asked Aug 30 at 0:03
b_chao
111
edited Aug 30 at 0:04
Bernard
112k635102
asked Aug 30 at 0:03
b_chao
111
edited Aug 30 at 0:04
Bernard
112k635102
edited Aug 30 at 0:04
Bernard
112k635102
edited Aug 30 at 0:04
Bernard
112k635102
112k635102
asked Aug 30 at 0:03
b_chao
111
asked Aug 30 at 0:03
b_chao
111
asked Aug 30 at 0:03
b_chao
111
111
Welcome to Maths SX! All cubic polynomials have (at least) one real root, and this has nothing to do with orthogonal matrices: it relies on the Intermediate value theorem.
– Bernard
Aug 30 at 0:06
@Bernard: then the claim holds for all 3x3 matrices?
– b_chao
Aug 30 at 0:09
Over the field $mathbf R$, yes.
– Bernard
Aug 30 at 0:10
@Bernard: I see, thanks!
– b_chao
Aug 30 at 0:12
1
@TheoreticalEconomist: That's a nicer, but more sophisticated argument: it supposes you've proved $mathbf C$ is algebraically closed – which isn't exactly trivial.
– Bernard
Aug 30 at 8:23
 |Â
show 2 more comments
Welcome to Maths SX! All cubic polynomials have (at least) one real root, and this has nothing to do with orthogonal matrices: it relies on the Intermediate value theorem.
– Bernard
Aug 30 at 0:06
@Bernard: then the claim holds for all 3x3 matrices?
– b_chao
Aug 30 at 0:09
Over the field $mathbf R$, yes.
– Bernard
Aug 30 at 0:10
@Bernard: I see, thanks!
– b_chao
Aug 30 at 0:12
1
@TheoreticalEconomist: That's a nicer, but more sophisticated argument: it supposes you've proved $mathbf C$ is algebraically closed – which isn't exactly trivial.
– Bernard
Aug 30 at 8:23
Welcome to Maths SX! All cubic polynomials have (at least) one real root, and this has nothing to do with orthogonal matrices: it relies on the Intermediate value theorem.
– Bernard
Aug 30 at 0:06
Welcome to Maths SX! All cubic polynomials have (at least) one real root, and this has nothing to do with orthogonal matrices: it relies on the Intermediate value theorem.
– Bernard
Aug 30 at 0:06
@Bernard: then the claim holds for all 3x3 matrices?
– b_chao
Aug 30 at 0:09
@Bernard: then the claim holds for all 3x3 matrices?
– b_chao
Aug 30 at 0:09
Over the field $mathbf R$, yes.
– Bernard
Aug 30 at 0:10
Over the field $mathbf R$, yes.
– Bernard
Aug 30 at 0:10
@Bernard: I see, thanks!
– b_chao
Aug 30 at 0:12
@Bernard: I see, thanks!
– b_chao
Aug 30 at 0:12
1
1
@TheoreticalEconomist: That's a nicer, but more sophisticated argument: it supposes you've proved $mathbf C$ is algebraically closed – which isn't exactly trivial.
– Bernard
Aug 30 at 8:23
@TheoreticalEconomist: That's a nicer, but more sophisticated argument: it supposes you've proved $mathbf C$ is algebraically closed – which isn't exactly trivial.
– Bernard
Aug 30 at 8:23
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
As you mentioned the characteristic polynomial of your matrix is a third degree polynomial and every third degree polynomial has a real root.
You do not have to prove that every third degree polynomial has a real root to solve your problem because it is a well-known fact by just looking at the end behavior of the polynomial.
answered Aug 30 at 0:22
Mohammad Riazi-Kermani
31k41853
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
As you mentioned the characteristic polynomial of your matrix is a third degree polynomial and every third degree polynomial has a real root.
You do not have to prove that every third degree polynomial has a real root to solve your problem because it is a well-known fact by just looking at the end behavior of the polynomial.
answered Aug 30 at 0:22
Mohammad Riazi-Kermani
31k41853
add a comment |Â
up vote
1
down vote
As you mentioned the characteristic polynomial of your matrix is a third degree polynomial and every third degree polynomial has a real root.
You do not have to prove that every third degree polynomial has a real root to solve your problem because it is a well-known fact by just looking at the end behavior of the polynomial.
answered Aug 30 at 0:22
Mohammad Riazi-Kermani
31k41853
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As you mentioned the characteristic polynomial of your matrix is a third degree polynomial and every third degree polynomial has a real root.
You do not have to prove that every third degree polynomial has a real root to solve your problem because it is a well-known fact by just looking at the end behavior of the polynomial.
answered Aug 30 at 0:22
Mohammad Riazi-Kermani
31k41853
As you mentioned the characteristic polynomial of your matrix is a third degree polynomial and every third degree polynomial has a real root.
You do not have to prove that every third degree polynomial has a real root to solve your problem because it is a well-known fact by just looking at the end behavior of the polynomial.
answered Aug 30 at 0:22
Mohammad Riazi-Kermani
31k41853
answered Aug 30 at 0:22
Mohammad Riazi-Kermani
31k41853
answered Aug 30 at 0:22
Mohammad Riazi-Kermani
31k41853
answered Aug 30 at 0:22
Mohammad Riazi-Kermani
31k41853
31k41853
add a comment |Â
add a comment |Â
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Welcome to Maths SX! All cubic polynomials have (at least) one real root, and this has nothing to do with orthogonal matrices: it relies on the Intermediate value theorem.
– Bernard
Aug 30 at 0:06
@Bernard: then the claim holds for all 3x3 matrices?
– b_chao
Aug 30 at 0:09
Over the field $mathbf R$, yes.
– Bernard
Aug 30 at 0:10
@Bernard: I see, thanks!
– b_chao
Aug 30 at 0:12
1
@TheoreticalEconomist: That's a nicer, but more sophisticated argument: it supposes you've proved $mathbf C$ is algebraically closed – which isn't exactly trivial.
– Bernard
Aug 30 at 8:23