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can a function of a diagonalizable matrix be non-diagonalizable?
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Suppose we have a diagonalizable matrix $M$
$$M=P D P^-1$$
with $D$ diagonal. The usual definition of a function of this matrix, $f(M)$, is
$$f(M)=P textdiag(f(D_11),f(D_22),cdots,f(D_nn)) P^-1$$
Could it be that $f(M)$ is not diagonalizable?
To me, it initially seems that the answer is obviousely no, the diagonalisation of $f(M)$ is right there, it's $P textdiag(f(D_11),f(D_22),cdots,f(D_nn)) P^-1$. But I remain uncertain. Is this reasoning correct?
matrices matrix-decomposition
asked Jul 31 at 15:19
Wouter
5,50021334
 |Â
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up vote
0
down vote
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Suppose we have a diagonalizable matrix $M$
$$M=P D P^-1$$
with $D$ diagonal. The usual definition of a function of this matrix, $f(M)$, is
$$f(M)=P textdiag(f(D_11),f(D_22),cdots,f(D_nn)) P^-1$$
Could it be that $f(M)$ is not diagonalizable?
To me, it initially seems that the answer is obviousely no, the diagonalisation of $f(M)$ is right there, it's $P textdiag(f(D_11),f(D_22),cdots,f(D_nn)) P^-1$. But I remain uncertain. Is this reasoning correct?
matrices matrix-decomposition
asked Jul 31 at 15:19
Wouter
5,50021334
4
As you said, with that definition of $f(M)$ the statement is true by definition.
– Roberto Rastapopoulos
Jul 31 at 15:25
1
If you define it by some power series expansion then it can't. (why)
– mathreadler
Jul 31 at 15:31
Why are you uncertain?
– copper.hat
Jul 31 at 15:35
I cannot give much context. During a paper review, someone asserted that is is easy to find functions of diagonalizable matrices which are non-diagonalizable. It's probably a matter of differing definitions of "a function of a matrix".
– Wouter
Jul 31 at 15:39
I need be more specific on my comment, for power series expansions with scalar coefficients.
– mathreadler
Jul 31 at 15:47
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose we have a diagonalizable matrix $M$
$$M=P D P^-1$$
with $D$ diagonal. The usual definition of a function of this matrix, $f(M)$, is
$$f(M)=P textdiag(f(D_11),f(D_22),cdots,f(D_nn)) P^-1$$
Could it be that $f(M)$ is not diagonalizable?
To me, it initially seems that the answer is obviousely no, the diagonalisation of $f(M)$ is right there, it's $P textdiag(f(D_11),f(D_22),cdots,f(D_nn)) P^-1$. But I remain uncertain. Is this reasoning correct?
matrices matrix-decomposition
asked Jul 31 at 15:19
Wouter
5,50021334
Suppose we have a diagonalizable matrix $M$
$$M=P D P^-1$$
with $D$ diagonal. The usual definition of a function of this matrix, $f(M)$, is
$$f(M)=P textdiag(f(D_11),f(D_22),cdots,f(D_nn)) P^-1$$
Could it be that $f(M)$ is not diagonalizable?
To me, it initially seems that the answer is obviousely no, the diagonalisation of $f(M)$ is right there, it's $P textdiag(f(D_11),f(D_22),cdots,f(D_nn)) P^-1$. But I remain uncertain. Is this reasoning correct?
matrices matrix-decomposition
asked Jul 31 at 15:19
Wouter
5,50021334
asked Jul 31 at 15:19
Wouter
5,50021334
asked Jul 31 at 15:19
Wouter
5,50021334
asked Jul 31 at 15:19
Wouter
5,50021334
5,50021334
4
As you said, with that definition of $f(M)$ the statement is true by definition.
– Roberto Rastapopoulos
Jul 31 at 15:25
1
If you define it by some power series expansion then it can't. (why)
– mathreadler
Jul 31 at 15:31
Why are you uncertain?
– copper.hat
Jul 31 at 15:35
I cannot give much context. During a paper review, someone asserted that is is easy to find functions of diagonalizable matrices which are non-diagonalizable. It's probably a matter of differing definitions of "a function of a matrix".
– Wouter
Jul 31 at 15:39
I need be more specific on my comment, for power series expansions with scalar coefficients.
– mathreadler
Jul 31 at 15:47
 |Â
show 1 more comment
4
As you said, with that definition of $f(M)$ the statement is true by definition.
– Roberto Rastapopoulos
Jul 31 at 15:25
1
If you define it by some power series expansion then it can't. (why)
– mathreadler
Jul 31 at 15:31
Why are you uncertain?
– copper.hat
Jul 31 at 15:35
I cannot give much context. During a paper review, someone asserted that is is easy to find functions of diagonalizable matrices which are non-diagonalizable. It's probably a matter of differing definitions of "a function of a matrix".
– Wouter
Jul 31 at 15:39
I need be more specific on my comment, for power series expansions with scalar coefficients.
– mathreadler
Jul 31 at 15:47
4
4
As you said, with that definition of $f(M)$ the statement is true by definition.
– Roberto Rastapopoulos
Jul 31 at 15:25
As you said, with that definition of $f(M)$ the statement is true by definition.
– Roberto Rastapopoulos
Jul 31 at 15:25
1
1
If you define it by some power series expansion then it can't. (why)
– mathreadler
Jul 31 at 15:31
If you define it by some power series expansion then it can't. (why)
– mathreadler
Jul 31 at 15:31
Why are you uncertain?
– copper.hat
Jul 31 at 15:35
Why are you uncertain?
– copper.hat
Jul 31 at 15:35
I cannot give much context. During a paper review, someone asserted that is is easy to find functions of diagonalizable matrices which are non-diagonalizable. It's probably a matter of differing definitions of "a function of a matrix".
– Wouter
Jul 31 at 15:39
I cannot give much context. During a paper review, someone asserted that is is easy to find functions of diagonalizable matrices which are non-diagonalizable. It's probably a matter of differing definitions of "a function of a matrix".
– Wouter
Jul 31 at 15:39
I need be more specific on my comment, for power series expansions with scalar coefficients.
– mathreadler
Jul 31 at 15:47
I need be more specific on my comment, for power series expansions with scalar coefficients.
– mathreadler
Jul 31 at 15:47
 |Â
show 1 more comment
1 Answer
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A function is just an assignment of a value to every argument. This assignment doesn't need to be in form of an explicit expression, and even when it is, the assignment for a diagonalizable matrix doesn't have to be diagonalizable. For example, let for any diagonalizable matrix $d$ of arbitrary quadratic size,
$$f(d) := beginpmatrix0 & 1 \ 0 & 0endpmatrix$$
This is not diagonalizable although $d$ is.
answered Jul 31 at 16:10
md2perpe
5,6391921
In this context, $f$ is a function from $mathbb C$ to $mathbb C$. This is standard operator theory jargon: en.wikipedia.org/wiki/Functional_calculus
– Giuseppe Negro
Jul 31 at 16:12
And that page says that there isn't always an obvious way to extend a function to operators (like matrices) although often $f(z)$ can be written as a polynomial or a series, and then just replacing $z$ with the matrix in the polynomial or series is the most natural.
– md2perpe
Jul 31 at 16:24
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
A function is just an assignment of a value to every argument. This assignment doesn't need to be in form of an explicit expression, and even when it is, the assignment for a diagonalizable matrix doesn't have to be diagonalizable. For example, let for any diagonalizable matrix $d$ of arbitrary quadratic size,
$$f(d) := beginpmatrix0 & 1 \ 0 & 0endpmatrix$$
This is not diagonalizable although $d$ is.
answered Jul 31 at 16:10
md2perpe
5,6391921
In this context, $f$ is a function from $mathbb C$ to $mathbb C$. This is standard operator theory jargon: en.wikipedia.org/wiki/Functional_calculus
– Giuseppe Negro
Jul 31 at 16:12
And that page says that there isn't always an obvious way to extend a function to operators (like matrices) although often $f(z)$ can be written as a polynomial or a series, and then just replacing $z$ with the matrix in the polynomial or series is the most natural.
– md2perpe
Jul 31 at 16:24
add a comment |Â
up vote
0
down vote
A function is just an assignment of a value to every argument. This assignment doesn't need to be in form of an explicit expression, and even when it is, the assignment for a diagonalizable matrix doesn't have to be diagonalizable. For example, let for any diagonalizable matrix $d$ of arbitrary quadratic size,
$$f(d) := beginpmatrix0 & 1 \ 0 & 0endpmatrix$$
This is not diagonalizable although $d$ is.
answered Jul 31 at 16:10
md2perpe
5,6391921
In this context, $f$ is a function from $mathbb C$ to $mathbb C$. This is standard operator theory jargon: en.wikipedia.org/wiki/Functional_calculus
– Giuseppe Negro
Jul 31 at 16:12
And that page says that there isn't always an obvious way to extend a function to operators (like matrices) although often $f(z)$ can be written as a polynomial or a series, and then just replacing $z$ with the matrix in the polynomial or series is the most natural.
– md2perpe
Jul 31 at 16:24
add a comment |Â
up vote
0
down vote
up vote
0
down vote
A function is just an assignment of a value to every argument. This assignment doesn't need to be in form of an explicit expression, and even when it is, the assignment for a diagonalizable matrix doesn't have to be diagonalizable. For example, let for any diagonalizable matrix $d$ of arbitrary quadratic size,
$$f(d) := beginpmatrix0 & 1 \ 0 & 0endpmatrix$$
This is not diagonalizable although $d$ is.
answered Jul 31 at 16:10
md2perpe
5,6391921
A function is just an assignment of a value to every argument. This assignment doesn't need to be in form of an explicit expression, and even when it is, the assignment for a diagonalizable matrix doesn't have to be diagonalizable. For example, let for any diagonalizable matrix $d$ of arbitrary quadratic size,
$$f(d) := beginpmatrix0 & 1 \ 0 & 0endpmatrix$$
This is not diagonalizable although $d$ is.
answered Jul 31 at 16:10
md2perpe
5,6391921
answered Jul 31 at 16:10
md2perpe
5,6391921
answered Jul 31 at 16:10
md2perpe
5,6391921
answered Jul 31 at 16:10
md2perpe
5,6391921
5,6391921
In this context, $f$ is a function from $mathbb C$ to $mathbb C$. This is standard operator theory jargon: en.wikipedia.org/wiki/Functional_calculus
– Giuseppe Negro
Jul 31 at 16:12
And that page says that there isn't always an obvious way to extend a function to operators (like matrices) although often $f(z)$ can be written as a polynomial or a series, and then just replacing $z$ with the matrix in the polynomial or series is the most natural.
– md2perpe
Jul 31 at 16:24
add a comment |Â
In this context, $f$ is a function from $mathbb C$ to $mathbb C$. This is standard operator theory jargon: en.wikipedia.org/wiki/Functional_calculus
– Giuseppe Negro
Jul 31 at 16:12
And that page says that there isn't always an obvious way to extend a function to operators (like matrices) although often $f(z)$ can be written as a polynomial or a series, and then just replacing $z$ with the matrix in the polynomial or series is the most natural.
– md2perpe
Jul 31 at 16:24
In this context, $f$ is a function from $mathbb C$ to $mathbb C$. This is standard operator theory jargon: en.wikipedia.org/wiki/Functional_calculus
– Giuseppe Negro
Jul 31 at 16:12
In this context, $f$ is a function from $mathbb C$ to $mathbb C$. This is standard operator theory jargon: en.wikipedia.org/wiki/Functional_calculus
– Giuseppe Negro
Jul 31 at 16:12
And that page says that there isn't always an obvious way to extend a function to operators (like matrices) although often $f(z)$ can be written as a polynomial or a series, and then just replacing $z$ with the matrix in the polynomial or series is the most natural.
– md2perpe
Jul 31 at 16:24
And that page says that there isn't always an obvious way to extend a function to operators (like matrices) although often $f(z)$ can be written as a polynomial or a series, and then just replacing $z$ with the matrix in the polynomial or series is the most natural.
– md2perpe
Jul 31 at 16:24
add a comment |Â
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4
As you said, with that definition of $f(M)$ the statement is true by definition.
– Roberto Rastapopoulos
Jul 31 at 15:25
1
If you define it by some power series expansion then it can't. (why)
– mathreadler
Jul 31 at 15:31
Why are you uncertain?
– copper.hat
Jul 31 at 15:35
I cannot give much context. During a paper review, someone asserted that is is easy to find functions of diagonalizable matrices which are non-diagonalizable. It's probably a matter of differing definitions of "a function of a matrix".
– Wouter
Jul 31 at 15:39
I need be more specific on my comment, for power series expansions with scalar coefficients.
– mathreadler
Jul 31 at 15:47