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?







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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














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?







share|cite|improve this question















  • 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












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?







share|cite|improve this question











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?









share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









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












  • 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










1 Answer
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0
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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.






share|cite|improve this answer





















  • 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










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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.






share|cite|improve this answer





















  • 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














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.






share|cite|improve this answer





















  • 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












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.






share|cite|improve this answer













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.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











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
















  • 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












 

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