Kernel of polynomial of matrixLinear Algebra: Minimum PolynomialCharacteristic Polynomial of a Linear MapPolynomial of matrixImage and Kernel of a MatrixKernel of a polynomialA positive definite matrix A can be express as a polynomial of $ A^2 $How does minimal polynomial divide characteristic polynomial of matrix?Kernel and Image of matricesFinding a matrix given its characteristic polynomialChange of basis of the kernel of a rectangular matrix

Constructing Group Divisible Designs - Algorithms?

Proving a function is onto where f(x)=|x|.

What's the difference between 違法 and 不法?

What (else) happened July 1st 1858 in London?

Does the Mind Blank spell prevent the target from being frightened?

Has Darkwing Duck ever met Scrooge McDuck?

Is camera lens focus an exact point or a range?

Can someone explain how this makes sense electrically?

Drawing ramified coverings with tikz

Is it possible to have a strip of cold climate in the middle of a planet?

Can somebody explain Brexit in a few child-proof sentences?

Customize circled numbers

Two-sided logarithm inequality

Is a model fitted to data or is data fitted to a model?

List of people who lose a child in תנ"ך

Folder comparison

Can the Supreme Court overturn an impeachment?

Why has "pence" been used in this sentence, not "pences"?

Can I use my Chinese passport to enter China after I acquired another citizenship?

MAXDOP Settings for SQL Server 2014

How do ground effect vehicles perform turns?

Is there a word to describe the feeling of being transfixed out of horror?

Have I saved too much for retirement so far?

Do Legal Documents Require Signing In Standard Pen Colors?



Kernel of polynomial of matrix


Linear Algebra: Minimum PolynomialCharacteristic Polynomial of a Linear MapPolynomial of matrixImage and Kernel of a MatrixKernel of a polynomialA positive definite matrix A can be express as a polynomial of $ A^2 $How does minimal polynomial divide characteristic polynomial of matrix?Kernel and Image of matricesFinding a matrix given its characteristic polynomialChange of basis of the kernel of a rectangular matrix













1












$begingroup$


I am given that $f(x)$, $g(x)$ and $h(x)$ are polynomials such that $f(x)=g(x)h(x)$.



I am then asked to show using induction on the degree of $f$ that $f(A)$ would be an $n times n$-matrix if $A$ is an $n times n$ matrix. I am told any constant term of $f(x), c,$ will be replaced by $cI$ in the matrix polynomial $f(A)$, where $I$ is the $n times n$ identity matrix - same size as A.



Wouldn’t this be trivial as if $I$ is the $n times n$ identity matrix, A would need to be the same size in order to add to I and thus any linear combination of powers of A would be the same size? I don’t know what induction has to do with this.



I am also asked to show that if $f(x)=g(x) h(x)$ and $f(A)=0$, then for any vector $v$ and any polynomial $b(A)$, then $b(A) h(A) v inker(g(A))$ and, similarly, $b(A) g(A) v inker(h(A))$. I don’t know how to start because I think that $b(A)h(A)v$ and $b(A)g(A)v$ are both vectors and we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.



Any hints or help would be greatly appreciated!










share|cite|improve this question











$endgroup$
















    1












    $begingroup$


    I am given that $f(x)$, $g(x)$ and $h(x)$ are polynomials such that $f(x)=g(x)h(x)$.



    I am then asked to show using induction on the degree of $f$ that $f(A)$ would be an $n times n$-matrix if $A$ is an $n times n$ matrix. I am told any constant term of $f(x), c,$ will be replaced by $cI$ in the matrix polynomial $f(A)$, where $I$ is the $n times n$ identity matrix - same size as A.



    Wouldn’t this be trivial as if $I$ is the $n times n$ identity matrix, A would need to be the same size in order to add to I and thus any linear combination of powers of A would be the same size? I don’t know what induction has to do with this.



    I am also asked to show that if $f(x)=g(x) h(x)$ and $f(A)=0$, then for any vector $v$ and any polynomial $b(A)$, then $b(A) h(A) v inker(g(A))$ and, similarly, $b(A) g(A) v inker(h(A))$. I don’t know how to start because I think that $b(A)h(A)v$ and $b(A)g(A)v$ are both vectors and we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.



    Any hints or help would be greatly appreciated!










    share|cite|improve this question











    $endgroup$














      1












      1








      1





      $begingroup$


      I am given that $f(x)$, $g(x)$ and $h(x)$ are polynomials such that $f(x)=g(x)h(x)$.



      I am then asked to show using induction on the degree of $f$ that $f(A)$ would be an $n times n$-matrix if $A$ is an $n times n$ matrix. I am told any constant term of $f(x), c,$ will be replaced by $cI$ in the matrix polynomial $f(A)$, where $I$ is the $n times n$ identity matrix - same size as A.



      Wouldn’t this be trivial as if $I$ is the $n times n$ identity matrix, A would need to be the same size in order to add to I and thus any linear combination of powers of A would be the same size? I don’t know what induction has to do with this.



      I am also asked to show that if $f(x)=g(x) h(x)$ and $f(A)=0$, then for any vector $v$ and any polynomial $b(A)$, then $b(A) h(A) v inker(g(A))$ and, similarly, $b(A) g(A) v inker(h(A))$. I don’t know how to start because I think that $b(A)h(A)v$ and $b(A)g(A)v$ are both vectors and we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.



      Any hints or help would be greatly appreciated!










      share|cite|improve this question











      $endgroup$




      I am given that $f(x)$, $g(x)$ and $h(x)$ are polynomials such that $f(x)=g(x)h(x)$.



      I am then asked to show using induction on the degree of $f$ that $f(A)$ would be an $n times n$-matrix if $A$ is an $n times n$ matrix. I am told any constant term of $f(x), c,$ will be replaced by $cI$ in the matrix polynomial $f(A)$, where $I$ is the $n times n$ identity matrix - same size as A.



      Wouldn’t this be trivial as if $I$ is the $n times n$ identity matrix, A would need to be the same size in order to add to I and thus any linear combination of powers of A would be the same size? I don’t know what induction has to do with this.



      I am also asked to show that if $f(x)=g(x) h(x)$ and $f(A)=0$, then for any vector $v$ and any polynomial $b(A)$, then $b(A) h(A) v inker(g(A))$ and, similarly, $b(A) g(A) v inker(h(A))$. I don’t know how to start because I think that $b(A)h(A)v$ and $b(A)g(A)v$ are both vectors and we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.



      Any hints or help would be greatly appreciated!







      linear-algebra matrices polynomials






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 16 at 10:32







      Andrew

















      asked Mar 16 at 10:02









      AndrewAndrew

      357213




      357213




















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          To show that $f(A)$ is an $ntimes n$-matrix, you could prove by induction that $A^k$ is an $ntimes n$-matrix for all $kinBbbN$, and then note that any linear combination of $ntimes n$-matrices is again an $ntimes n$-matrix. I agree that it's a rather trivial thing to prove though.



          To show that $b(A)h(A)vinker g(A)$ for any polynomial $b$ and any vector $v$, it suffices to show that $h(A)vinker g(A)$. You claim that




          ...we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.




          Indeed you can't take powers of vectors, but that's not what is being claimed.



          You have just shown (by induction) that $g(A)$ is an $ntimes n$-matrix. So you can multiply it by an $n$-vector. Because $h(A)$ is also an $ntimes n$-matrix $h(A)v$ is an $n$-vector. So you can multiply $g(A)$ by $h(A)v$. To show that $h(A)vinker g(A)$ it suffices to show that the product is $0$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you. I can’t use the null factor law, though, can I? That is if $f(A)=g(A)h(A)=0$ this does not imply that either $g(A)$ or $h(A)$ is equal to zero because two nonzero matrices can multiply to zero. Sorry that I’m the original question I forgot to add that $f(A)=0$ for this second part.
            $endgroup$
            – Andrew
            Mar 16 at 10:31










          • $begingroup$
            @Andrew I assumed that $f(A)=0$ for the second part as otherwise it doesn't make sense. And indeed, $g(A)h(A)=0$ does not imply that $g(A)=0$ or $h(A)=0$. But you do know that $g(A)cdot h(A)v=0$ for all $v$...
            $endgroup$
            – Servaes
            Mar 16 at 10:33











          • $begingroup$
            Of course! Thank you very much.
            $endgroup$
            – Andrew
            Mar 16 at 10:38










          Your Answer





          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "69"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader:
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          ,
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3150251%2fkernel-of-polynomial-of-matrix%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          To show that $f(A)$ is an $ntimes n$-matrix, you could prove by induction that $A^k$ is an $ntimes n$-matrix for all $kinBbbN$, and then note that any linear combination of $ntimes n$-matrices is again an $ntimes n$-matrix. I agree that it's a rather trivial thing to prove though.



          To show that $b(A)h(A)vinker g(A)$ for any polynomial $b$ and any vector $v$, it suffices to show that $h(A)vinker g(A)$. You claim that




          ...we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.




          Indeed you can't take powers of vectors, but that's not what is being claimed.



          You have just shown (by induction) that $g(A)$ is an $ntimes n$-matrix. So you can multiply it by an $n$-vector. Because $h(A)$ is also an $ntimes n$-matrix $h(A)v$ is an $n$-vector. So you can multiply $g(A)$ by $h(A)v$. To show that $h(A)vinker g(A)$ it suffices to show that the product is $0$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you. I can’t use the null factor law, though, can I? That is if $f(A)=g(A)h(A)=0$ this does not imply that either $g(A)$ or $h(A)$ is equal to zero because two nonzero matrices can multiply to zero. Sorry that I’m the original question I forgot to add that $f(A)=0$ for this second part.
            $endgroup$
            – Andrew
            Mar 16 at 10:31










          • $begingroup$
            @Andrew I assumed that $f(A)=0$ for the second part as otherwise it doesn't make sense. And indeed, $g(A)h(A)=0$ does not imply that $g(A)=0$ or $h(A)=0$. But you do know that $g(A)cdot h(A)v=0$ for all $v$...
            $endgroup$
            – Servaes
            Mar 16 at 10:33











          • $begingroup$
            Of course! Thank you very much.
            $endgroup$
            – Andrew
            Mar 16 at 10:38















          1












          $begingroup$

          To show that $f(A)$ is an $ntimes n$-matrix, you could prove by induction that $A^k$ is an $ntimes n$-matrix for all $kinBbbN$, and then note that any linear combination of $ntimes n$-matrices is again an $ntimes n$-matrix. I agree that it's a rather trivial thing to prove though.



          To show that $b(A)h(A)vinker g(A)$ for any polynomial $b$ and any vector $v$, it suffices to show that $h(A)vinker g(A)$. You claim that




          ...we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.




          Indeed you can't take powers of vectors, but that's not what is being claimed.



          You have just shown (by induction) that $g(A)$ is an $ntimes n$-matrix. So you can multiply it by an $n$-vector. Because $h(A)$ is also an $ntimes n$-matrix $h(A)v$ is an $n$-vector. So you can multiply $g(A)$ by $h(A)v$. To show that $h(A)vinker g(A)$ it suffices to show that the product is $0$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you. I can’t use the null factor law, though, can I? That is if $f(A)=g(A)h(A)=0$ this does not imply that either $g(A)$ or $h(A)$ is equal to zero because two nonzero matrices can multiply to zero. Sorry that I’m the original question I forgot to add that $f(A)=0$ for this second part.
            $endgroup$
            – Andrew
            Mar 16 at 10:31










          • $begingroup$
            @Andrew I assumed that $f(A)=0$ for the second part as otherwise it doesn't make sense. And indeed, $g(A)h(A)=0$ does not imply that $g(A)=0$ or $h(A)=0$. But you do know that $g(A)cdot h(A)v=0$ for all $v$...
            $endgroup$
            – Servaes
            Mar 16 at 10:33











          • $begingroup$
            Of course! Thank you very much.
            $endgroup$
            – Andrew
            Mar 16 at 10:38













          1












          1








          1





          $begingroup$

          To show that $f(A)$ is an $ntimes n$-matrix, you could prove by induction that $A^k$ is an $ntimes n$-matrix for all $kinBbbN$, and then note that any linear combination of $ntimes n$-matrices is again an $ntimes n$-matrix. I agree that it's a rather trivial thing to prove though.



          To show that $b(A)h(A)vinker g(A)$ for any polynomial $b$ and any vector $v$, it suffices to show that $h(A)vinker g(A)$. You claim that




          ...we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.




          Indeed you can't take powers of vectors, but that's not what is being claimed.



          You have just shown (by induction) that $g(A)$ is an $ntimes n$-matrix. So you can multiply it by an $n$-vector. Because $h(A)$ is also an $ntimes n$-matrix $h(A)v$ is an $n$-vector. So you can multiply $g(A)$ by $h(A)v$. To show that $h(A)vinker g(A)$ it suffices to show that the product is $0$.






          share|cite|improve this answer









          $endgroup$



          To show that $f(A)$ is an $ntimes n$-matrix, you could prove by induction that $A^k$ is an $ntimes n$-matrix for all $kinBbbN$, and then note that any linear combination of $ntimes n$-matrices is again an $ntimes n$-matrix. I agree that it's a rather trivial thing to prove though.



          To show that $b(A)h(A)vinker g(A)$ for any polynomial $b$ and any vector $v$, it suffices to show that $h(A)vinker g(A)$. You claim that




          ...we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.




          Indeed you can't take powers of vectors, but that's not what is being claimed.



          You have just shown (by induction) that $g(A)$ is an $ntimes n$-matrix. So you can multiply it by an $n$-vector. Because $h(A)$ is also an $ntimes n$-matrix $h(A)v$ is an $n$-vector. So you can multiply $g(A)$ by $h(A)v$. To show that $h(A)vinker g(A)$ it suffices to show that the product is $0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 16 at 10:23









          ServaesServaes

          28.5k34099




          28.5k34099











          • $begingroup$
            Thank you. I can’t use the null factor law, though, can I? That is if $f(A)=g(A)h(A)=0$ this does not imply that either $g(A)$ or $h(A)$ is equal to zero because two nonzero matrices can multiply to zero. Sorry that I’m the original question I forgot to add that $f(A)=0$ for this second part.
            $endgroup$
            – Andrew
            Mar 16 at 10:31










          • $begingroup$
            @Andrew I assumed that $f(A)=0$ for the second part as otherwise it doesn't make sense. And indeed, $g(A)h(A)=0$ does not imply that $g(A)=0$ or $h(A)=0$. But you do know that $g(A)cdot h(A)v=0$ for all $v$...
            $endgroup$
            – Servaes
            Mar 16 at 10:33











          • $begingroup$
            Of course! Thank you very much.
            $endgroup$
            – Andrew
            Mar 16 at 10:38
















          • $begingroup$
            Thank you. I can’t use the null factor law, though, can I? That is if $f(A)=g(A)h(A)=0$ this does not imply that either $g(A)$ or $h(A)$ is equal to zero because two nonzero matrices can multiply to zero. Sorry that I’m the original question I forgot to add that $f(A)=0$ for this second part.
            $endgroup$
            – Andrew
            Mar 16 at 10:31










          • $begingroup$
            @Andrew I assumed that $f(A)=0$ for the second part as otherwise it doesn't make sense. And indeed, $g(A)h(A)=0$ does not imply that $g(A)=0$ or $h(A)=0$. But you do know that $g(A)cdot h(A)v=0$ for all $v$...
            $endgroup$
            – Servaes
            Mar 16 at 10:33











          • $begingroup$
            Of course! Thank you very much.
            $endgroup$
            – Andrew
            Mar 16 at 10:38















          $begingroup$
          Thank you. I can’t use the null factor law, though, can I? That is if $f(A)=g(A)h(A)=0$ this does not imply that either $g(A)$ or $h(A)$ is equal to zero because two nonzero matrices can multiply to zero. Sorry that I’m the original question I forgot to add that $f(A)=0$ for this second part.
          $endgroup$
          – Andrew
          Mar 16 at 10:31




          $begingroup$
          Thank you. I can’t use the null factor law, though, can I? That is if $f(A)=g(A)h(A)=0$ this does not imply that either $g(A)$ or $h(A)$ is equal to zero because two nonzero matrices can multiply to zero. Sorry that I’m the original question I forgot to add that $f(A)=0$ for this second part.
          $endgroup$
          – Andrew
          Mar 16 at 10:31












          $begingroup$
          @Andrew I assumed that $f(A)=0$ for the second part as otherwise it doesn't make sense. And indeed, $g(A)h(A)=0$ does not imply that $g(A)=0$ or $h(A)=0$. But you do know that $g(A)cdot h(A)v=0$ for all $v$...
          $endgroup$
          – Servaes
          Mar 16 at 10:33





          $begingroup$
          @Andrew I assumed that $f(A)=0$ for the second part as otherwise it doesn't make sense. And indeed, $g(A)h(A)=0$ does not imply that $g(A)=0$ or $h(A)=0$. But you do know that $g(A)cdot h(A)v=0$ for all $v$...
          $endgroup$
          – Servaes
          Mar 16 at 10:33













          $begingroup$
          Of course! Thank you very much.
          $endgroup$
          – Andrew
          Mar 16 at 10:38




          $begingroup$
          Of course! Thank you very much.
          $endgroup$
          – Andrew
          Mar 16 at 10:38

















          draft saved

          draft discarded
















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid


          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.

          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3150251%2fkernel-of-polynomial-of-matrix%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Solar Wings Breeze Design and development Specifications (Breeze) References Navigation menu1368-485X"Hang glider: Breeze (Solar Wings)"e

          Kathakali Contents Etymology and nomenclature History Repertoire Songs and musical instruments Traditional plays Styles: Sampradayam Training centers and awards Relationship to other dance forms See also Notes References External links Navigation menueThe Illustrated Encyclopedia of Hinduism: A-MSouth Asian Folklore: An EncyclopediaRoutledge International Encyclopedia of Women: Global Women's Issues and KnowledgeKathakali Dance-drama: Where Gods and Demons Come to PlayKathakali Dance-drama: Where Gods and Demons Come to PlayKathakali Dance-drama: Where Gods and Demons Come to Play10.1353/atj.2005.0004The Illustrated Encyclopedia of Hinduism: A-MEncyclopedia of HinduismKathakali Dance-drama: Where Gods and Demons Come to PlaySonic Liturgy: Ritual and Music in Hindu Tradition"The Mirror of Gesture"Kathakali Dance-drama: Where Gods and Demons Come to Play"Kathakali"Indian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceMedieval Indian Literature: An AnthologyThe Oxford Companion to Indian TheatreSouth Asian Folklore: An Encyclopedia : Afghanistan, Bangladesh, India, Nepal, Pakistan, Sri LankaThe Rise of Performance Studies: Rethinking Richard Schechner's Broad SpectrumIndian Theatre: Traditions of PerformanceModern Asian Theatre and Performance 1900-2000Critical Theory and PerformanceBetween Theater and AnthropologyKathakali603847011Indian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceBetween Theater and AnthropologyBetween Theater and AnthropologyNambeesan Smaraka AwardsArchivedThe Cambridge Guide to TheatreRoutledge International Encyclopedia of Women: Global Women's Issues and KnowledgeThe Garland Encyclopedia of World Music: South Asia : the Indian subcontinentThe Ethos of Noh: Actors and Their Art10.2307/1145740By Means of Performance: Intercultural Studies of Theatre and Ritual10.1017/s204912550000100xReconceiving the Renaissance: A Critical ReaderPerformance TheoryListening to Theatre: The Aural Dimension of Beijing Opera10.2307/1146013Kathakali: The Art of the Non-WorldlyOn KathakaliKathakali, the dance theatreThe Kathakali Complex: Performance & StructureKathakali Dance-Drama: Where Gods and Demons Come to Play10.1093/obo/9780195399318-0071Drama and Ritual of Early Hinduism"In the Shadow of Hollywood Orientalism: Authentic East Indian Dancing"10.1080/08949460490274013Sanskrit Play Production in Ancient IndiaIndian Music: History and StructureBharata, the Nāṭyaśāstra233639306Table of Contents2238067286469807Dance In Indian Painting10.2307/32047833204783Kathakali Dance-Theatre: A Visual Narrative of Sacred Indian MimeIndian Classical Dance: The Renaissance and BeyondKathakali: an indigenous art-form of Keralaeee

          Urgehal History Discography Band members References External links Navigation menu"Mediateket: Urgehal""Interview with Enzifer of Urgehal, 2007""Urgehal - Interview"Urgehal"Urgehal Frontman Trondr Nefas Dies at 35"Urgehal9042691cb161873230(data)0000 0001 0669 4224no2016126817ee6ccef6-e558-44b6-b059-dbbb5b913b24145036459145036459