Distinct eigenvalues implies $A in mathbb{R}^{n times n}$ is diagonalisable
$begingroup$
Theorem:
If an $n times n$ matrix has n distinct eigenvalues then A is diagonalisable.
Proof:
Let $A in mathbb{R}^{n times n}$. Suppose A is not diagonalisable.
Then, by definition, for a given $D in mathbb{R}^{n times n}$ there exists no invertible matrix $P in mathbb{R}^{n times n}$ such that $P^{-1}AP = D$.
Any hint(s) to assist me would be helpful.
Thanks in advance.
linear-algebra proof-verification eigenvalues-eigenvectors eigenfunctions
edited Dec 16 '18 at 2:55
JimmyK4542
41.1k245107
asked Dec 16 '18 at 2:52
MathematicingMathematicing
2,45521856
$endgroup$
add a comment |
$begingroup$
Theorem:
If an $n times n$ matrix has n distinct eigenvalues then A is diagonalisable.
Proof:
Let $A in mathbb{R}^{n times n}$. Suppose A is not diagonalisable.
Then, by definition, for a given $D in mathbb{R}^{n times n}$ there exists no invertible matrix $P in mathbb{R}^{n times n}$ such that $P^{-1}AP = D$.
Any hint(s) to assist me would be helpful.
Thanks in advance.
linear-algebra proof-verification eigenvalues-eigenvectors eigenfunctions
edited Dec 16 '18 at 2:55
JimmyK4542
41.1k245107
asked Dec 16 '18 at 2:52
MathematicingMathematicing
2,45521856
$endgroup$
add a comment |
$begingroup$
Theorem:
If an $n times n$ matrix has n distinct eigenvalues then A is diagonalisable.
Proof:
Let $A in mathbb{R}^{n times n}$. Suppose A is not diagonalisable.
Then, by definition, for a given $D in mathbb{R}^{n times n}$ there exists no invertible matrix $P in mathbb{R}^{n times n}$ such that $P^{-1}AP = D$.
Any hint(s) to assist me would be helpful.
Thanks in advance.
linear-algebra proof-verification eigenvalues-eigenvectors eigenfunctions
edited Dec 16 '18 at 2:55
JimmyK4542
41.1k245107
asked Dec 16 '18 at 2:52
MathematicingMathematicing
2,45521856
$endgroup$
Theorem:
If an $n times n$ matrix has n distinct eigenvalues then A is diagonalisable.
Proof:
Let $A in mathbb{R}^{n times n}$. Suppose A is not diagonalisable.
Then, by definition, for a given $D in mathbb{R}^{n times n}$ there exists no invertible matrix $P in mathbb{R}^{n times n}$ such that $P^{-1}AP = D$.
Any hint(s) to assist me would be helpful.
Thanks in advance.
linear-algebra proof-verification eigenvalues-eigenvectors eigenfunctions
linear-algebra proof-verification eigenvalues-eigenvectors eigenfunctions
edited Dec 16 '18 at 2:55
JimmyK4542
41.1k245107
asked Dec 16 '18 at 2:52
MathematicingMathematicing
2,45521856
edited Dec 16 '18 at 2:55
JimmyK4542
41.1k245107
asked Dec 16 '18 at 2:52
MathematicingMathematicing
2,45521856
edited Dec 16 '18 at 2:55
JimmyK4542
41.1k245107
edited Dec 16 '18 at 2:55
JimmyK4542
41.1k245107
edited Dec 16 '18 at 2:55
JimmyK4542
41.1k245107
41.1k245107
asked Dec 16 '18 at 2:52
MathematicingMathematicing
2,45521856
asked Dec 16 '18 at 2:52
MathematicingMathematicing
2,45521856
asked Dec 16 '18 at 2:52
MathematicingMathematicing
2,45521856
2,45521856
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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Hint: I would prefer to prove it directly.
Eigenvectors for distinct eigenvalues are linearly independent. Thus we have a basis for $V$ consisting of eigenvectors. Simply let $P$ be the matrix whose columns are the basis vectors.
Then $P^{-1}AP=D$, where $D$ is diagonal, and the entries on the diagonal are the eigenvalues.
answered Dec 16 '18 at 3:04
Chris CusterChris Custer
13.7k3827
$endgroup$
$begingroup$
don't you need symmetry for linear independent eigenvectors?
$endgroup$
– LinAlg
Dec 16 '18 at 3:09
1
$begingroup$
No. See this: math.stackexchange.com/a/29374
$endgroup$
– Chris Custer
Dec 16 '18 at 3:20
$begingroup$
@ChrisCuster By construction, the j column of the matrix P is the j eigenvectors in the L.I set S of eigenvectors. Clearly, set S is the column space of P. I'd like to show that every column of P has a pivot position. Can you give me a hint on this?
$endgroup$
– Mathematicing
Dec 16 '18 at 3:28
1
$begingroup$
Hmm. Clearly the columns are independent, as they are the elements of a basis. To get pivots you would have to take the transpose and row-reduce, say.
$endgroup$
– Chris Custer
Dec 16 '18 at 3:32
$begingroup$
@ChrisCuster Satisfied.
$endgroup$
– Mathematicing
Dec 16 '18 at 3:35
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
oldest
votes
$begingroup$
Hint: I would prefer to prove it directly.
Eigenvectors for distinct eigenvalues are linearly independent. Thus we have a basis for $V$ consisting of eigenvectors. Simply let $P$ be the matrix whose columns are the basis vectors.
Then $P^{-1}AP=D$, where $D$ is diagonal, and the entries on the diagonal are the eigenvalues.
answered Dec 16 '18 at 3:04
Chris CusterChris Custer
13.7k3827
$endgroup$
$begingroup$
don't you need symmetry for linear independent eigenvectors?
$endgroup$
– LinAlg
Dec 16 '18 at 3:09
1
$begingroup$
No. See this: math.stackexchange.com/a/29374
$endgroup$
– Chris Custer
Dec 16 '18 at 3:20
$begingroup$
@ChrisCuster By construction, the j column of the matrix P is the j eigenvectors in the L.I set S of eigenvectors. Clearly, set S is the column space of P. I'd like to show that every column of P has a pivot position. Can you give me a hint on this?
$endgroup$
– Mathematicing
Dec 16 '18 at 3:28
1
$begingroup$
Hmm. Clearly the columns are independent, as they are the elements of a basis. To get pivots you would have to take the transpose and row-reduce, say.
$endgroup$
– Chris Custer
Dec 16 '18 at 3:32
$begingroup$
@ChrisCuster Satisfied.
$endgroup$
– Mathematicing
Dec 16 '18 at 3:35
add a comment |
$begingroup$
Hint: I would prefer to prove it directly.
Eigenvectors for distinct eigenvalues are linearly independent. Thus we have a basis for $V$ consisting of eigenvectors. Simply let $P$ be the matrix whose columns are the basis vectors.
Then $P^{-1}AP=D$, where $D$ is diagonal, and the entries on the diagonal are the eigenvalues.
answered Dec 16 '18 at 3:04
Chris CusterChris Custer
13.7k3827
$endgroup$
$begingroup$
don't you need symmetry for linear independent eigenvectors?
$endgroup$
– LinAlg
Dec 16 '18 at 3:09
1
$begingroup$
No. See this: math.stackexchange.com/a/29374
$endgroup$
– Chris Custer
Dec 16 '18 at 3:20
$begingroup$
@ChrisCuster By construction, the j column of the matrix P is the j eigenvectors in the L.I set S of eigenvectors. Clearly, set S is the column space of P. I'd like to show that every column of P has a pivot position. Can you give me a hint on this?
$endgroup$
– Mathematicing
Dec 16 '18 at 3:28
1
$begingroup$
Hmm. Clearly the columns are independent, as they are the elements of a basis. To get pivots you would have to take the transpose and row-reduce, say.
$endgroup$
– Chris Custer
Dec 16 '18 at 3:32
$begingroup$
@ChrisCuster Satisfied.
$endgroup$
– Mathematicing
Dec 16 '18 at 3:35
add a comment |
$begingroup$
Hint: I would prefer to prove it directly.
Eigenvectors for distinct eigenvalues are linearly independent. Thus we have a basis for $V$ consisting of eigenvectors. Simply let $P$ be the matrix whose columns are the basis vectors.
Then $P^{-1}AP=D$, where $D$ is diagonal, and the entries on the diagonal are the eigenvalues.
answered Dec 16 '18 at 3:04
Chris CusterChris Custer
13.7k3827
$endgroup$
Hint: I would prefer to prove it directly.
Eigenvectors for distinct eigenvalues are linearly independent. Thus we have a basis for $V$ consisting of eigenvectors. Simply let $P$ be the matrix whose columns are the basis vectors.
Then $P^{-1}AP=D$, where $D$ is diagonal, and the entries on the diagonal are the eigenvalues.
answered Dec 16 '18 at 3:04
Chris CusterChris Custer
13.7k3827
answered Dec 16 '18 at 3:04
Chris CusterChris Custer
13.7k3827
answered Dec 16 '18 at 3:04
Chris CusterChris Custer
13.7k3827
answered Dec 16 '18 at 3:04
Chris CusterChris Custer
13.7k3827
13.7k3827
$begingroup$
don't you need symmetry for linear independent eigenvectors?
$endgroup$
– LinAlg
Dec 16 '18 at 3:09
1
$begingroup$
No. See this: math.stackexchange.com/a/29374
$endgroup$
– Chris Custer
Dec 16 '18 at 3:20
$begingroup$
@ChrisCuster By construction, the j column of the matrix P is the j eigenvectors in the L.I set S of eigenvectors. Clearly, set S is the column space of P. I'd like to show that every column of P has a pivot position. Can you give me a hint on this?
$endgroup$
– Mathematicing
Dec 16 '18 at 3:28
1
$begingroup$
Hmm. Clearly the columns are independent, as they are the elements of a basis. To get pivots you would have to take the transpose and row-reduce, say.
$endgroup$
– Chris Custer
Dec 16 '18 at 3:32
$begingroup$
@ChrisCuster Satisfied.
$endgroup$
– Mathematicing
Dec 16 '18 at 3:35
add a comment |
$begingroup$
don't you need symmetry for linear independent eigenvectors?
$endgroup$
– LinAlg
Dec 16 '18 at 3:09
1
$begingroup$
No. See this: math.stackexchange.com/a/29374
$endgroup$
– Chris Custer
Dec 16 '18 at 3:20
$begingroup$
@ChrisCuster By construction, the j column of the matrix P is the j eigenvectors in the L.I set S of eigenvectors. Clearly, set S is the column space of P. I'd like to show that every column of P has a pivot position. Can you give me a hint on this?
$endgroup$
– Mathematicing
Dec 16 '18 at 3:28
1
$begingroup$
Hmm. Clearly the columns are independent, as they are the elements of a basis. To get pivots you would have to take the transpose and row-reduce, say.
$endgroup$
– Chris Custer
Dec 16 '18 at 3:32
$begingroup$
@ChrisCuster Satisfied.
$endgroup$
– Mathematicing
Dec 16 '18 at 3:35
$begingroup$
don't you need symmetry for linear independent eigenvectors?
$endgroup$
– LinAlg
Dec 16 '18 at 3:09
$begingroup$
don't you need symmetry for linear independent eigenvectors?
$endgroup$
– LinAlg
Dec 16 '18 at 3:09
1
1
$begingroup$
No. See this: math.stackexchange.com/a/29374
$endgroup$
– Chris Custer
Dec 16 '18 at 3:20
$begingroup$
No. See this: math.stackexchange.com/a/29374
$endgroup$
– Chris Custer
Dec 16 '18 at 3:20
$begingroup$
@ChrisCuster By construction, the j column of the matrix P is the j eigenvectors in the L.I set S of eigenvectors. Clearly, set S is the column space of P. I'd like to show that every column of P has a pivot position. Can you give me a hint on this?
$endgroup$
– Mathematicing
Dec 16 '18 at 3:28
$begingroup$
@ChrisCuster By construction, the j column of the matrix P is the j eigenvectors in the L.I set S of eigenvectors. Clearly, set S is the column space of P. I'd like to show that every column of P has a pivot position. Can you give me a hint on this?
$endgroup$
– Mathematicing
Dec 16 '18 at 3:28
1
1
$begingroup$
Hmm. Clearly the columns are independent, as they are the elements of a basis. To get pivots you would have to take the transpose and row-reduce, say.
$endgroup$
– Chris Custer
Dec 16 '18 at 3:32
$begingroup$
Hmm. Clearly the columns are independent, as they are the elements of a basis. To get pivots you would have to take the transpose and row-reduce, say.
$endgroup$
– Chris Custer
Dec 16 '18 at 3:32
$begingroup$
@ChrisCuster Satisfied.
$endgroup$
– Mathematicing
Dec 16 '18 at 3:35
$begingroup$
@ChrisCuster Satisfied.
$endgroup$
– Mathematicing
Dec 16 '18 at 3:35
add a comment |
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