$Ainmathbb R^{ntimes n}$ has a single eigenvalue over $mathbb C$, then $A$ is diagonalizable $iff$ $A$ is a...
$begingroup$
So while playing with matrices a little, I found something that I think is true but I don't know how to prove in the general case. I tried to formalize it and this is the result:
Let $Ain mathbb R^{ntimes n}$ be a square matrix such that it has a single eigenvalue over $mathbb C$. Then $A$ is diagonalizable, if and only if, $A$ is a scalar matrix.
I succeeded proving this for both of the cases where $n=2$ and $n=3$ so I believe this might be true. Couldn't find a counterexample either.
Thanks!
linear-algebra matrices
asked Jan 21 at 17:18
Amit ZachAmit Zach
595
$endgroup$
add a comment |
$begingroup$
So while playing with matrices a little, I found something that I think is true but I don't know how to prove in the general case. I tried to formalize it and this is the result:
Let $Ain mathbb R^{ntimes n}$ be a square matrix such that it has a single eigenvalue over $mathbb C$. Then $A$ is diagonalizable, if and only if, $A$ is a scalar matrix.
I succeeded proving this for both of the cases where $n=2$ and $n=3$ so I believe this might be true. Couldn't find a counterexample either.
Thanks!
linear-algebra matrices
asked Jan 21 at 17:18
Amit ZachAmit Zach
595
$endgroup$
$begingroup$
For a diagonalizable matrix, the minimal polynomial (the roots of which are exactly the complex eigenvalues) has simple roots. So if a matrix with exactly one eigenvalue is diagonalizable, the minimal polynomial must have degree $1$ thus the matrix must be scalar.
$endgroup$
– Mindlack
Jan 21 at 17:24
add a comment |
$begingroup$
So while playing with matrices a little, I found something that I think is true but I don't know how to prove in the general case. I tried to formalize it and this is the result:
Let $Ain mathbb R^{ntimes n}$ be a square matrix such that it has a single eigenvalue over $mathbb C$. Then $A$ is diagonalizable, if and only if, $A$ is a scalar matrix.
I succeeded proving this for both of the cases where $n=2$ and $n=3$ so I believe this might be true. Couldn't find a counterexample either.
Thanks!
linear-algebra matrices
asked Jan 21 at 17:18
Amit ZachAmit Zach
595
$endgroup$
So while playing with matrices a little, I found something that I think is true but I don't know how to prove in the general case. I tried to formalize it and this is the result:
Let $Ain mathbb R^{ntimes n}$ be a square matrix such that it has a single eigenvalue over $mathbb C$. Then $A$ is diagonalizable, if and only if, $A$ is a scalar matrix.
I succeeded proving this for both of the cases where $n=2$ and $n=3$ so I believe this might be true. Couldn't find a counterexample either.
Thanks!
linear-algebra matrices
linear-algebra matrices
asked Jan 21 at 17:18
Amit ZachAmit Zach
595
asked Jan 21 at 17:18
Amit ZachAmit Zach
595
asked Jan 21 at 17:18
Amit ZachAmit Zach
595
asked Jan 21 at 17:18
Amit ZachAmit Zach
595
asked Jan 21 at 17:18
Amit ZachAmit Zach
595
595
$begingroup$
For a diagonalizable matrix, the minimal polynomial (the roots of which are exactly the complex eigenvalues) has simple roots. So if a matrix with exactly one eigenvalue is diagonalizable, the minimal polynomial must have degree $1$ thus the matrix must be scalar.
$endgroup$
– Mindlack
Jan 21 at 17:24
add a comment |
$begingroup$
For a diagonalizable matrix, the minimal polynomial (the roots of which are exactly the complex eigenvalues) has simple roots. So if a matrix with exactly one eigenvalue is diagonalizable, the minimal polynomial must have degree $1$ thus the matrix must be scalar.
$endgroup$
– Mindlack
Jan 21 at 17:24
$begingroup$
For a diagonalizable matrix, the minimal polynomial (the roots of which are exactly the complex eigenvalues) has simple roots. So if a matrix with exactly one eigenvalue is diagonalizable, the minimal polynomial must have degree $1$ thus the matrix must be scalar.
$endgroup$
– Mindlack
Jan 21 at 17:24
$begingroup$
For a diagonalizable matrix, the minimal polynomial (the roots of which are exactly the complex eigenvalues) has simple roots. So if a matrix with exactly one eigenvalue is diagonalizable, the minimal polynomial must have degree $1$ thus the matrix must be scalar.
$endgroup$
– Mindlack
Jan 21 at 17:24
add a comment |
2 Answers
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$begingroup$
If $A$ is a scalar matrix, $A$ is diagonal, hence diagonalizable.
Conversely, if $A$ is diagonalizable, then it is similar to a matrix with its eigenvalues on the diagonal. If there's only one eigenvalue, $A$ is scalar.
answered Jan 21 at 17:24
Chris CusterChris Custer
13.8k3827
$endgroup$
add a comment |
$begingroup$
If $A$ is scalar then it's of course diagonalizable.
Suppose $A$ is diagonalizable; then its unique eigenvalue $lambda$ must have geometric multiplicity $n$, which means that $A-lambda I_n$ has rank $0$. Thus $A-lambda I_n=0$.
answered Jan 21 at 18:16
egregegreg
183k1486204
$endgroup$
add a comment |
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2 Answers
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$begingroup$
If $A$ is a scalar matrix, $A$ is diagonal, hence diagonalizable.
Conversely, if $A$ is diagonalizable, then it is similar to a matrix with its eigenvalues on the diagonal. If there's only one eigenvalue, $A$ is scalar.
answered Jan 21 at 17:24
Chris CusterChris Custer
13.8k3827
$endgroup$
add a comment |
$begingroup$
If $A$ is a scalar matrix, $A$ is diagonal, hence diagonalizable.
Conversely, if $A$ is diagonalizable, then it is similar to a matrix with its eigenvalues on the diagonal. If there's only one eigenvalue, $A$ is scalar.
answered Jan 21 at 17:24
Chris CusterChris Custer
13.8k3827
$endgroup$
add a comment |
$begingroup$
If $A$ is a scalar matrix, $A$ is diagonal, hence diagonalizable.
Conversely, if $A$ is diagonalizable, then it is similar to a matrix with its eigenvalues on the diagonal. If there's only one eigenvalue, $A$ is scalar.
answered Jan 21 at 17:24
Chris CusterChris Custer
13.8k3827
$endgroup$
If $A$ is a scalar matrix, $A$ is diagonal, hence diagonalizable.
Conversely, if $A$ is diagonalizable, then it is similar to a matrix with its eigenvalues on the diagonal. If there's only one eigenvalue, $A$ is scalar.
answered Jan 21 at 17:24
Chris CusterChris Custer
13.8k3827
answered Jan 21 at 17:24
Chris CusterChris Custer
13.8k3827
answered Jan 21 at 17:24
Chris CusterChris Custer
13.8k3827
answered Jan 21 at 17:24
Chris CusterChris Custer
13.8k3827
13.8k3827
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add a comment |
$begingroup$
If $A$ is scalar then it's of course diagonalizable.
Suppose $A$ is diagonalizable; then its unique eigenvalue $lambda$ must have geometric multiplicity $n$, which means that $A-lambda I_n$ has rank $0$. Thus $A-lambda I_n=0$.
answered Jan 21 at 18:16
egregegreg
183k1486204
$endgroup$
add a comment |
$begingroup$
If $A$ is scalar then it's of course diagonalizable.
Suppose $A$ is diagonalizable; then its unique eigenvalue $lambda$ must have geometric multiplicity $n$, which means that $A-lambda I_n$ has rank $0$. Thus $A-lambda I_n=0$.
answered Jan 21 at 18:16
egregegreg
183k1486204
$endgroup$
add a comment |
$begingroup$
If $A$ is scalar then it's of course diagonalizable.
Suppose $A$ is diagonalizable; then its unique eigenvalue $lambda$ must have geometric multiplicity $n$, which means that $A-lambda I_n$ has rank $0$. Thus $A-lambda I_n=0$.
answered Jan 21 at 18:16
egregegreg
183k1486204
$endgroup$
If $A$ is scalar then it's of course diagonalizable.
Suppose $A$ is diagonalizable; then its unique eigenvalue $lambda$ must have geometric multiplicity $n$, which means that $A-lambda I_n$ has rank $0$. Thus $A-lambda I_n=0$.
answered Jan 21 at 18:16
egregegreg
183k1486204
answered Jan 21 at 18:16
egregegreg
183k1486204
answered Jan 21 at 18:16
egregegreg
183k1486204
answered Jan 21 at 18:16
egregegreg
183k1486204
183k1486204
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$begingroup$
For a diagonalizable matrix, the minimal polynomial (the roots of which are exactly the complex eigenvalues) has simple roots. So if a matrix with exactly one eigenvalue is diagonalizable, the minimal polynomial must have degree $1$ thus the matrix must be scalar.
$endgroup$
– Mindlack
Jan 21 at 17:24