If $tilde{lambda}$ is an eigenvalue of $A$ of multiplicity $n$, then $A$ is a scalar matrix, where $A$ is $n...
If $tilde{lambda}$ is an eigenvalue of $A$ of multiplicity $n$, then $A$ is a scalar matrix, where $A$ is $n times n$ real symmetric matrix.
So I know that $A = begin{bmatrix}a&b\b&cend{bmatrix} $
and that $$det(A-lambda I_2) = (a-λ)(c-λ) - b^2$$
However I don't really know how to proceed from there.
linear-algebra
edited Dec 2 '18 at 5:00
Brahadeesh
6,11742361
asked Dec 2 '18 at 1:43
Zhe Tian
112
add a comment |
If $tilde{lambda}$ is an eigenvalue of $A$ of multiplicity $n$, then $A$ is a scalar matrix, where $A$ is $n times n$ real symmetric matrix.
So I know that $A = begin{bmatrix}a&b\b&cend{bmatrix} $
and that $$det(A-lambda I_2) = (a-λ)(c-λ) - b^2$$
However I don't really know how to proceed from there.
linear-algebra
edited Dec 2 '18 at 5:00
Brahadeesh
6,11742361
asked Dec 2 '18 at 1:43
Zhe Tian
112
1. Dont assume A is 2$times$2. Also you know that det(A-$lambda$I) = $(x-lambda)^n$.
– Joel Pereira
Dec 2 '18 at 1:49
add a comment |
If $tilde{lambda}$ is an eigenvalue of $A$ of multiplicity $n$, then $A$ is a scalar matrix, where $A$ is $n times n$ real symmetric matrix.
So I know that $A = begin{bmatrix}a&b\b&cend{bmatrix} $
and that $$det(A-lambda I_2) = (a-λ)(c-λ) - b^2$$
However I don't really know how to proceed from there.
linear-algebra
edited Dec 2 '18 at 5:00
Brahadeesh
6,11742361
asked Dec 2 '18 at 1:43
Zhe Tian
112
If $tilde{lambda}$ is an eigenvalue of $A$ of multiplicity $n$, then $A$ is a scalar matrix, where $A$ is $n times n$ real symmetric matrix.
So I know that $A = begin{bmatrix}a&b\b&cend{bmatrix} $
and that $$det(A-lambda I_2) = (a-λ)(c-λ) - b^2$$
However I don't really know how to proceed from there.
linear-algebra
linear-algebra
edited Dec 2 '18 at 5:00
Brahadeesh
6,11742361
asked Dec 2 '18 at 1:43
Zhe Tian
112
edited Dec 2 '18 at 5:00
Brahadeesh
6,11742361
asked Dec 2 '18 at 1:43
Zhe Tian
112
edited Dec 2 '18 at 5:00
Brahadeesh
6,11742361
edited Dec 2 '18 at 5:00
Brahadeesh
6,11742361
edited Dec 2 '18 at 5:00
Brahadeesh
6,11742361
6,11742361
asked Dec 2 '18 at 1:43
Zhe Tian
112
asked Dec 2 '18 at 1:43
Zhe Tian
112
asked Dec 2 '18 at 1:43
Zhe Tian
112
112
1. Dont assume A is 2$times$2. Also you know that det(A-$lambda$I) = $(x-lambda)^n$.
– Joel Pereira
Dec 2 '18 at 1:49
add a comment |
1. Dont assume A is 2$times$2. Also you know that det(A-$lambda$I) = $(x-lambda)^n$.
– Joel Pereira
Dec 2 '18 at 1:49
1. Dont assume A is 2$times$2. Also you know that det(A-$lambda$I) = $(x-lambda)^n$.
– Joel Pereira
Dec 2 '18 at 1:49
1. Dont assume A is 2$times$2. Also you know that det(A-$lambda$I) = $(x-lambda)^n$.
– Joel Pereira
Dec 2 '18 at 1:49
add a comment |
2 Answers
2
active
oldest
votes
Hint:
If $A$ has an eigenvalue of multiplicity $n$, it is the only eigenvalue of $A$, so. Furthermore real symmetric matrices are diagonalisable, so its diagonal form is $lambda I$. Can you deduce that $A$ is $lambda I$?.
More details:
Since is real symmetric, there exists an orthogonal matrix $P$ and a diagonal matrix $D$ such that
$$A=PDP^{-1}=P(lambda I)P^{-1}=lambda(PIP^{-1})=lambda I.$$
answered Dec 2 '18 at 2:00
Bernard
118k639112
Sorry I'm still confused on what you mean?
– Zhe Tian
Dec 2 '18 at 2:14
@ZheTian: I've added details. Is that clearer?
– Bernard
Dec 2 '18 at 12:20
add a comment |
Your definition of scalar matrix is a diagonal matrix which every element in the diagonal are the same, right?
I assumed that the question should be "... then $A$ is a matrix similar to a scalar matrix". But you also can abuse notation and say that, in fact, $A$ "is'' a scalar matrix by the fact that similiar matrices can represent the same linear transformation...
But, if $A$ is a $ntimes n$ symmetric matrix, then $A$ is diagonalizable and there is a invertible $P$ such that $A=P DP^{-1}$, where $D=diag(lambda,cdots,lambda)$.
answered Dec 2 '18 at 4:43
Robson
771221
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint:
If $A$ has an eigenvalue of multiplicity $n$, it is the only eigenvalue of $A$, so. Furthermore real symmetric matrices are diagonalisable, so its diagonal form is $lambda I$. Can you deduce that $A$ is $lambda I$?.
More details:
Since is real symmetric, there exists an orthogonal matrix $P$ and a diagonal matrix $D$ such that
$$A=PDP^{-1}=P(lambda I)P^{-1}=lambda(PIP^{-1})=lambda I.$$
answered Dec 2 '18 at 2:00
Bernard
118k639112
Sorry I'm still confused on what you mean?
– Zhe Tian
Dec 2 '18 at 2:14
@ZheTian: I've added details. Is that clearer?
– Bernard
Dec 2 '18 at 12:20
add a comment |
Hint:
If $A$ has an eigenvalue of multiplicity $n$, it is the only eigenvalue of $A$, so. Furthermore real symmetric matrices are diagonalisable, so its diagonal form is $lambda I$. Can you deduce that $A$ is $lambda I$?.
More details:
Since is real symmetric, there exists an orthogonal matrix $P$ and a diagonal matrix $D$ such that
$$A=PDP^{-1}=P(lambda I)P^{-1}=lambda(PIP^{-1})=lambda I.$$
answered Dec 2 '18 at 2:00
Bernard
118k639112
Sorry I'm still confused on what you mean?
– Zhe Tian
Dec 2 '18 at 2:14
@ZheTian: I've added details. Is that clearer?
– Bernard
Dec 2 '18 at 12:20
add a comment |
Hint:
If $A$ has an eigenvalue of multiplicity $n$, it is the only eigenvalue of $A$, so. Furthermore real symmetric matrices are diagonalisable, so its diagonal form is $lambda I$. Can you deduce that $A$ is $lambda I$?.
More details:
Since is real symmetric, there exists an orthogonal matrix $P$ and a diagonal matrix $D$ such that
$$A=PDP^{-1}=P(lambda I)P^{-1}=lambda(PIP^{-1})=lambda I.$$
answered Dec 2 '18 at 2:00
Bernard
118k639112
Hint:
If $A$ has an eigenvalue of multiplicity $n$, it is the only eigenvalue of $A$, so. Furthermore real symmetric matrices are diagonalisable, so its diagonal form is $lambda I$. Can you deduce that $A$ is $lambda I$?.
More details:
Since is real symmetric, there exists an orthogonal matrix $P$ and a diagonal matrix $D$ such that
$$A=PDP^{-1}=P(lambda I)P^{-1}=lambda(PIP^{-1})=lambda I.$$
answered Dec 2 '18 at 2:00
Bernard
118k639112
edited Dec 2 '18 at 12:20
answered Dec 2 '18 at 2:00
Bernard
118k639112
answered Dec 2 '18 at 2:00
Bernard
118k639112
answered Dec 2 '18 at 2:00
Bernard
118k639112
118k639112
Sorry I'm still confused on what you mean?
– Zhe Tian
Dec 2 '18 at 2:14
@ZheTian: I've added details. Is that clearer?
– Bernard
Dec 2 '18 at 12:20
add a comment |
Sorry I'm still confused on what you mean?
– Zhe Tian
Dec 2 '18 at 2:14
@ZheTian: I've added details. Is that clearer?
– Bernard
Dec 2 '18 at 12:20
Sorry I'm still confused on what you mean?
– Zhe Tian
Dec 2 '18 at 2:14
Sorry I'm still confused on what you mean?
– Zhe Tian
Dec 2 '18 at 2:14
@ZheTian: I've added details. Is that clearer?
– Bernard
Dec 2 '18 at 12:20
@ZheTian: I've added details. Is that clearer?
– Bernard
Dec 2 '18 at 12:20
add a comment |
Your definition of scalar matrix is a diagonal matrix which every element in the diagonal are the same, right?
I assumed that the question should be "... then $A$ is a matrix similar to a scalar matrix". But you also can abuse notation and say that, in fact, $A$ "is'' a scalar matrix by the fact that similiar matrices can represent the same linear transformation...
But, if $A$ is a $ntimes n$ symmetric matrix, then $A$ is diagonalizable and there is a invertible $P$ such that $A=P DP^{-1}$, where $D=diag(lambda,cdots,lambda)$.
answered Dec 2 '18 at 4:43
Robson
771221
add a comment |
Your definition of scalar matrix is a diagonal matrix which every element in the diagonal are the same, right?
I assumed that the question should be "... then $A$ is a matrix similar to a scalar matrix". But you also can abuse notation and say that, in fact, $A$ "is'' a scalar matrix by the fact that similiar matrices can represent the same linear transformation...
But, if $A$ is a $ntimes n$ symmetric matrix, then $A$ is diagonalizable and there is a invertible $P$ such that $A=P DP^{-1}$, where $D=diag(lambda,cdots,lambda)$.
answered Dec 2 '18 at 4:43
Robson
771221
add a comment |
Your definition of scalar matrix is a diagonal matrix which every element in the diagonal are the same, right?
I assumed that the question should be "... then $A$ is a matrix similar to a scalar matrix". But you also can abuse notation and say that, in fact, $A$ "is'' a scalar matrix by the fact that similiar matrices can represent the same linear transformation...
But, if $A$ is a $ntimes n$ symmetric matrix, then $A$ is diagonalizable and there is a invertible $P$ such that $A=P DP^{-1}$, where $D=diag(lambda,cdots,lambda)$.
answered Dec 2 '18 at 4:43
Robson
771221
Your definition of scalar matrix is a diagonal matrix which every element in the diagonal are the same, right?
I assumed that the question should be "... then $A$ is a matrix similar to a scalar matrix". But you also can abuse notation and say that, in fact, $A$ "is'' a scalar matrix by the fact that similiar matrices can represent the same linear transformation...
But, if $A$ is a $ntimes n$ symmetric matrix, then $A$ is diagonalizable and there is a invertible $P$ such that $A=P DP^{-1}$, where $D=diag(lambda,cdots,lambda)$.
answered Dec 2 '18 at 4:43
Robson
771221
answered Dec 2 '18 at 4:43
Robson
771221
answered Dec 2 '18 at 4:43
Robson
771221
answered Dec 2 '18 at 4:43
Robson
771221
771221
add a comment |
add a comment |
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1. Dont assume A is 2$times$2. Also you know that det(A-$lambda$I) = $(x-lambda)^n$.
– Joel Pereira
Dec 2 '18 at 1:49