Proof involving the spectral radius and Jordan Canonical form Announcing the arrival of Valued...

What are the motives behind Cersei's orders given to Bronn?

ListPlot join points by nearest neighbor rather than order

How do I keep my slimes from escaping their pens?

Is there a Spanish version of "dot your i's and cross your t's" that includes the letter 'ñ'?

Storing hydrofluoric acid before the invention of plastics

What is a Meta algorithm?

Proof involving the spectral radius and Jordan Canonical form

Does accepting a pardon have any bearing on trying that person for the same crime in a sovereign jurisdiction?

Did Xerox really develop the first LAN?

Why is black pepper both grey and black?

How discoverable are IPv6 addresses and AAAA names by potential attackers?

How to find all the available tools in macOS terminal?

Is the Standard Deduction better than Itemized when both are the same amount?

Is it true that "carbohydrates are of no use for the basal metabolic need"?

Antler Helmet: Can it work?

Do I really need recursive chmod to restrict access to a folder?

Doubts about chords

"Seemed to had" is it correct?

Should I discuss the type of campaign with my players?

3 doors, three guards, one stone

How to deal with a team lead who never gives me credit?

What is this single-engine low-wing propeller plane?

Is a manifold-with-boundary with given interior and non-empty boundary essentially unique?

Is there a service that would inform me whenever a new direct route is scheduled from a given airport?



Proof involving the spectral radius and Jordan Canonical form



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Spectral radius of the Volterra operatorExample that the Jordan canonical form is not “robust.”The unit vector in the direction of uWhat is the purpose of Jordan Canonical Form?Confusion between spectral radius of matrix and spectral radius of the operatorComputing the Jordan Form of a MatrixSpectral radius of perturbed bipartite graphsA proof involving invertible $ntimes n$ matricesProof of Gelfand's formula without using $rho(A) < 1$ iff $lim A^n = 0$Computing Canonical Jordan Form over a field $mathbb{Q}$












1












$begingroup$


Let $A$ be a square matrix. Show that if $lim_{n to infty} A^{n} = 0$, then $rho(A) < 1$. Hint: Use the Jordan Canonical form. Here, $rho(A)$ denotes the spectral radius of $A$.



I'm self-studying and have been working through a few linear algebra exercises. I'm struggling a bit in applying the hint to this problem -- I don't know where to start. Any help appreciated.






linear-algebra spectral-radius







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $A$ be a square matrix. Show that if $lim_{n to infty} A^{n} = 0$, then $rho(A) < 1$. Hint: Use the Jordan Canonical form. Here, $rho(A)$ denotes the spectral radius of $A$.



    I'm self-studying and have been working through a few linear algebra exercises. I'm struggling a bit in applying the hint to this problem -- I don't know where to start. Any help appreciated.






    linear-algebra spectral-radius







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $A$ be a square matrix. Show that if $lim_{n to infty} A^{n} = 0$, then $rho(A) < 1$. Hint: Use the Jordan Canonical form. Here, $rho(A)$ denotes the spectral radius of $A$.



      I'm self-studying and have been working through a few linear algebra exercises. I'm struggling a bit in applying the hint to this problem -- I don't know where to start. Any help appreciated.






      linear-algebra spectral-radius







      share|cite|improve this question









      $endgroup$




      Let $A$ be a square matrix. Show that if $lim_{n to infty} A^{n} = 0$, then $rho(A) < 1$. Hint: Use the Jordan Canonical form. Here, $rho(A)$ denotes the spectral radius of $A$.



      I'm self-studying and have been working through a few linear algebra exercises. I'm struggling a bit in applying the hint to this problem -- I don't know where to start. Any help appreciated.






      linear-algebra spectral-radius




      linear-algebra spectral-radius






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 58 mins ago









      mXdXmXdX

      1018




      1018






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          You don't really need Jordan canonical form. If $rho(A) ge 1$, $A$ has an eigenvalue $lambda$ with $|lambda| ge 1$. That eigenvalue has an eigenvector $v$. Then $A^n v = lambda^n v$, so $|A^n v| = |lambda|^n |v| ge |v|$ does not go to $0$ as $n to infty$, which is impossible if $A^n to 0$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Hint



            $$A=PJP^{-1} \
            J=begin{bmatrix}
            lambda_1 & * & 0 & 0 & 0 & ... & 0 \
            0& lambda_2 & * & 0 & 0 & ... & 0 \
            ...&...&...&...&....&....&....\
            0 & 0 & 0 & 0&0&...&lambda_n \
            end{bmatrix}$$
            where each $*$ is either $0$ or $1$.



            Prove by induction that
            $$J^m=begin{bmatrix}
            lambda_1^m & star & star & star & star & ... & star \
            0& lambda_2^m & star & star & star & ... & star \
            ...&...&...&...&....&....&....\
            0 & 0 & 0 & 0&0&...&lambda_n^m \
            end{bmatrix}$$
            where the $star$s represent numbers, that is $J^m$ is an upper triangular matrix
            with the $m$^th powers of the eigenvalues on the diagonal.



            Note The above claim for $J^m$ is not fully using that $J$ is a Jordan cannonical form. It only uses that $J$ is upper triangular.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So, $A^{m} = PJ^{m}P^{-1}$. If I can show what you're asking by induction, would the limit of $J^{m} = 0$? I'm sure it is because the diagonal entries are less than one, right?
              $endgroup$
              – mXdX
              21 mins ago










            • $begingroup$
              @mXdX Well, that is the point. First $$lim_m J^m= lim_m P^{-1} A^m P =0$$ Now, since $lim J^m=0$ you can deduce that the diagonal entries converge to zero, meaning $lambda_j^m to 0$. This implies that $|lambda_j |<1$
              $endgroup$
              – N. S.
              16 mins ago










            • $begingroup$
              I understand now. Thanks. So I would have to show, like you said, that the diagonal entries of $J^{m}$ are the $m$th powers of the eigenvalues.
              $endgroup$
              – mXdX
              10 mins ago












            Your Answer








            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%2f3189376%2fproof-involving-the-spectral-radius-and-jordan-canonical-form%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            You don't really need Jordan canonical form. If $rho(A) ge 1$, $A$ has an eigenvalue $lambda$ with $|lambda| ge 1$. That eigenvalue has an eigenvector $v$. Then $A^n v = lambda^n v$, so $|A^n v| = |lambda|^n |v| ge |v|$ does not go to $0$ as $n to infty$, which is impossible if $A^n to 0$.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              You don't really need Jordan canonical form. If $rho(A) ge 1$, $A$ has an eigenvalue $lambda$ with $|lambda| ge 1$. That eigenvalue has an eigenvector $v$. Then $A^n v = lambda^n v$, so $|A^n v| = |lambda|^n |v| ge |v|$ does not go to $0$ as $n to infty$, which is impossible if $A^n to 0$.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                You don't really need Jordan canonical form. If $rho(A) ge 1$, $A$ has an eigenvalue $lambda$ with $|lambda| ge 1$. That eigenvalue has an eigenvector $v$. Then $A^n v = lambda^n v$, so $|A^n v| = |lambda|^n |v| ge |v|$ does not go to $0$ as $n to infty$, which is impossible if $A^n to 0$.






                share|cite|improve this answer









                $endgroup$



                You don't really need Jordan canonical form. If $rho(A) ge 1$, $A$ has an eigenvalue $lambda$ with $|lambda| ge 1$. That eigenvalue has an eigenvector $v$. Then $A^n v = lambda^n v$, so $|A^n v| = |lambda|^n |v| ge |v|$ does not go to $0$ as $n to infty$, which is impossible if $A^n to 0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 27 mins ago









                Robert IsraelRobert Israel

                332k23221478




                332k23221478























                    1












                    $begingroup$

                    Hint



                    $$A=PJP^{-1} \
                    J=begin{bmatrix}
                    lambda_1 & * & 0 & 0 & 0 & ... & 0 \
                    0& lambda_2 & * & 0 & 0 & ... & 0 \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n \
                    end{bmatrix}$$
                    where each $*$ is either $0$ or $1$.



                    Prove by induction that
                    $$J^m=begin{bmatrix}
                    lambda_1^m & star & star & star & star & ... & star \
                    0& lambda_2^m & star & star & star & ... & star \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n^m \
                    end{bmatrix}$$
                    where the $star$s represent numbers, that is $J^m$ is an upper triangular matrix
                    with the $m$^th powers of the eigenvalues on the diagonal.



                    Note The above claim for $J^m$ is not fully using that $J$ is a Jordan cannonical form. It only uses that $J$ is upper triangular.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      So, $A^{m} = PJ^{m}P^{-1}$. If I can show what you're asking by induction, would the limit of $J^{m} = 0$? I'm sure it is because the diagonal entries are less than one, right?
                      $endgroup$
                      – mXdX
                      21 mins ago










                    • $begingroup$
                      @mXdX Well, that is the point. First $$lim_m J^m= lim_m P^{-1} A^m P =0$$ Now, since $lim J^m=0$ you can deduce that the diagonal entries converge to zero, meaning $lambda_j^m to 0$. This implies that $|lambda_j |<1$
                      $endgroup$
                      – N. S.
                      16 mins ago










                    • $begingroup$
                      I understand now. Thanks. So I would have to show, like you said, that the diagonal entries of $J^{m}$ are the $m$th powers of the eigenvalues.
                      $endgroup$
                      – mXdX
                      10 mins ago
















                    1












                    $begingroup$

                    Hint



                    $$A=PJP^{-1} \
                    J=begin{bmatrix}
                    lambda_1 & * & 0 & 0 & 0 & ... & 0 \
                    0& lambda_2 & * & 0 & 0 & ... & 0 \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n \
                    end{bmatrix}$$
                    where each $*$ is either $0$ or $1$.



                    Prove by induction that
                    $$J^m=begin{bmatrix}
                    lambda_1^m & star & star & star & star & ... & star \
                    0& lambda_2^m & star & star & star & ... & star \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n^m \
                    end{bmatrix}$$
                    where the $star$s represent numbers, that is $J^m$ is an upper triangular matrix
                    with the $m$^th powers of the eigenvalues on the diagonal.



                    Note The above claim for $J^m$ is not fully using that $J$ is a Jordan cannonical form. It only uses that $J$ is upper triangular.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      So, $A^{m} = PJ^{m}P^{-1}$. If I can show what you're asking by induction, would the limit of $J^{m} = 0$? I'm sure it is because the diagonal entries are less than one, right?
                      $endgroup$
                      – mXdX
                      21 mins ago










                    • $begingroup$
                      @mXdX Well, that is the point. First $$lim_m J^m= lim_m P^{-1} A^m P =0$$ Now, since $lim J^m=0$ you can deduce that the diagonal entries converge to zero, meaning $lambda_j^m to 0$. This implies that $|lambda_j |<1$
                      $endgroup$
                      – N. S.
                      16 mins ago










                    • $begingroup$
                      I understand now. Thanks. So I would have to show, like you said, that the diagonal entries of $J^{m}$ are the $m$th powers of the eigenvalues.
                      $endgroup$
                      – mXdX
                      10 mins ago














                    1












                    1








                    1





                    $begingroup$

                    Hint



                    $$A=PJP^{-1} \
                    J=begin{bmatrix}
                    lambda_1 & * & 0 & 0 & 0 & ... & 0 \
                    0& lambda_2 & * & 0 & 0 & ... & 0 \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n \
                    end{bmatrix}$$
                    where each $*$ is either $0$ or $1$.



                    Prove by induction that
                    $$J^m=begin{bmatrix}
                    lambda_1^m & star & star & star & star & ... & star \
                    0& lambda_2^m & star & star & star & ... & star \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n^m \
                    end{bmatrix}$$
                    where the $star$s represent numbers, that is $J^m$ is an upper triangular matrix
                    with the $m$^th powers of the eigenvalues on the diagonal.



                    Note The above claim for $J^m$ is not fully using that $J$ is a Jordan cannonical form. It only uses that $J$ is upper triangular.






                    share|cite|improve this answer









                    $endgroup$



                    Hint



                    $$A=PJP^{-1} \
                    J=begin{bmatrix}
                    lambda_1 & * & 0 & 0 & 0 & ... & 0 \
                    0& lambda_2 & * & 0 & 0 & ... & 0 \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n \
                    end{bmatrix}$$
                    where each $*$ is either $0$ or $1$.



                    Prove by induction that
                    $$J^m=begin{bmatrix}
                    lambda_1^m & star & star & star & star & ... & star \
                    0& lambda_2^m & star & star & star & ... & star \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n^m \
                    end{bmatrix}$$
                    where the $star$s represent numbers, that is $J^m$ is an upper triangular matrix
                    with the $m$^th powers of the eigenvalues on the diagonal.



                    Note The above claim for $J^m$ is not fully using that $J$ is a Jordan cannonical form. It only uses that $J$ is upper triangular.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 43 mins ago









                    N. S.N. S.

                    105k7115210




                    105k7115210












                    • $begingroup$
                      So, $A^{m} = PJ^{m}P^{-1}$. If I can show what you're asking by induction, would the limit of $J^{m} = 0$? I'm sure it is because the diagonal entries are less than one, right?
                      $endgroup$
                      – mXdX
                      21 mins ago










                    • $begingroup$
                      @mXdX Well, that is the point. First $$lim_m J^m= lim_m P^{-1} A^m P =0$$ Now, since $lim J^m=0$ you can deduce that the diagonal entries converge to zero, meaning $lambda_j^m to 0$. This implies that $|lambda_j |<1$
                      $endgroup$
                      – N. S.
                      16 mins ago










                    • $begingroup$
                      I understand now. Thanks. So I would have to show, like you said, that the diagonal entries of $J^{m}$ are the $m$th powers of the eigenvalues.
                      $endgroup$
                      – mXdX
                      10 mins ago


















                    • $begingroup$
                      So, $A^{m} = PJ^{m}P^{-1}$. If I can show what you're asking by induction, would the limit of $J^{m} = 0$? I'm sure it is because the diagonal entries are less than one, right?
                      $endgroup$
                      – mXdX
                      21 mins ago










                    • $begingroup$
                      @mXdX Well, that is the point. First $$lim_m J^m= lim_m P^{-1} A^m P =0$$ Now, since $lim J^m=0$ you can deduce that the diagonal entries converge to zero, meaning $lambda_j^m to 0$. This implies that $|lambda_j |<1$
                      $endgroup$
                      – N. S.
                      16 mins ago










                    • $begingroup$
                      I understand now. Thanks. So I would have to show, like you said, that the diagonal entries of $J^{m}$ are the $m$th powers of the eigenvalues.
                      $endgroup$
                      – mXdX
                      10 mins ago
















                    $begingroup$
                    So, $A^{m} = PJ^{m}P^{-1}$. If I can show what you're asking by induction, would the limit of $J^{m} = 0$? I'm sure it is because the diagonal entries are less than one, right?
                    $endgroup$
                    – mXdX
                    21 mins ago




                    $begingroup$
                    So, $A^{m} = PJ^{m}P^{-1}$. If I can show what you're asking by induction, would the limit of $J^{m} = 0$? I'm sure it is because the diagonal entries are less than one, right?
                    $endgroup$
                    – mXdX
                    21 mins ago












                    $begingroup$
                    @mXdX Well, that is the point. First $$lim_m J^m= lim_m P^{-1} A^m P =0$$ Now, since $lim J^m=0$ you can deduce that the diagonal entries converge to zero, meaning $lambda_j^m to 0$. This implies that $|lambda_j |<1$
                    $endgroup$
                    – N. S.
                    16 mins ago




                    $begingroup$
                    @mXdX Well, that is the point. First $$lim_m J^m= lim_m P^{-1} A^m P =0$$ Now, since $lim J^m=0$ you can deduce that the diagonal entries converge to zero, meaning $lambda_j^m to 0$. This implies that $|lambda_j |<1$
                    $endgroup$
                    – N. S.
                    16 mins ago












                    $begingroup$
                    I understand now. Thanks. So I would have to show, like you said, that the diagonal entries of $J^{m}$ are the $m$th powers of the eigenvalues.
                    $endgroup$
                    – mXdX
                    10 mins ago




                    $begingroup$
                    I understand now. Thanks. So I would have to show, like you said, that the diagonal entries of $J^{m}$ are the $m$th powers of the eigenvalues.
                    $endgroup$
                    – mXdX
                    10 mins ago


















                    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%2f3189376%2fproof-involving-the-spectral-radius-and-jordan-canonical-form%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

                    Fairchild Swearingen Metro Inhaltsverzeichnis Geschichte | Innenausstattung | Nutzung | Zwischenfälle...

                    Image
                    ZubringerflugzeugTransportflugzeugZweimotoriges FlugzeugFairchildMilitärluftfahrzeug (Belgien)Militärluftfahrzeug (Schweden)Militärluftfahrzeug (Südafrika)Erstflug 1965 Vereinigte StaatenIATA-CodeTurbopropflugzeugUS-amerikanischenFlugzeugherstellersSwearingenFairchildSan AntonioEd SwearingenGeschäftsreiseflugzeugsSwearingen Merlin IIFairchildUSASkywest AirlinesAir WisconsinHorizon AirlinesComairWings WestAmerican Eagle AirlinesFairchild DornierAerocondeutschsprachigen LändernC-1C-2VC-3VC-4C-5C-6C-7C-8C-9KC-10C-11C-12YC-14YC-15C-17C-18C-19C-20C-21C-22C-23C-24VC-25C-26C-27C-28C-29C-31C-32C-33UC-35C-37C-38C-39C-40C-41KC-45KC-46C-143HC-144C-145C-146KC-767UC-880 Fairchild Swearingen Metro Swearingen Metroliner der Ostfriesische Lufttransport auf dem Flughafen Bremen Typ: Turbopropflugzeug Entwurfsland: Vereinigte Staaten  Vereinigte Staaten Hersteller: Fairchild/Swearingen Erstflug: 1965 Indienststellung: 1971 Produktionszeit: ...
                    Read more

                    Pilgersdorf Inhaltsverzeichnis Geografie | Geschichte | Bevölkerungsentwicklung | Politik | Kultur...

                    Image
                    Gemeinde im BurgenlandOrt im Bezirk OberpullendorfPilgersdorfErsterwähnung 1225 ungarischkroatischBurgenlandBezirk OberpullendorfÖsterreichZöbernbachKirchschlag in der Buckligen WeltLockenhausKatastralgemeindenNoricumkeltischen Höhensiedlung BurgBurgbergPannoniaLudwigs des DeutschenDominicusRatpotsRihherisMagyarenGüssinger FehdeAlbrechtFranziskanernKanizsayKirchschlagKrumbachTürkenProtestantenSteinbachHaidukenAugustiner-EremitenKuruzenPestVolkszählungGemeinderatBürgermeisterdirektwahlErstenZweiten WeltkriegsDeutschkreutzDraßmarktFrankenau-UnterpullendorfGroßwarasdorfHoritschonKaisersdorfKobersdorfLackenbachLackendorfLockenhausLutzmannsburgMannersdorf an der RabnitzMarkt Sankt MartinNeckenmarktNeutalNikitschOberloisdorfOberpullendorfPilgersdorfPiringsdorfRaidingRitzingSteinberg-DörflStoobUnterfrauenhaidUnterrabnitz-SchwendgrabenWeingrabenWeppersdorf Dieser Artikel erläutert die burgenländische Gemeinde. Siehe auch Pilgersdorf (Gemeinde Lasberg), Oberösterreich bzw. Pilgersdorf, ...
                    Read more

                    Marineschifffahrtleitung Inhaltsverzeichnis Geschichte | Heutige Organisation der NATO | Nationale und...

                    Image
                    MarineMinenkriegsführung (See)Geleit (Schifffahrt)NATO-Organisation SeestreitkräftenHandelsschifffahrtHandelsschiffenZweiten WeltkriegKalten KriegesGeleitKriegsschiffeU-BootenErsten Weltkrieguneingeschränkten U-Boot-Kriegerhebliche Anstrengungenweit reichenden SchutzmaßnahmenSowjetunionNATOAllied Maritime Command Headquarters NorthwoodLondonOfficer in Tactical CommandKapitänePartnerschaft für den FriedenReedereienFischereiKonvoisMinenbedrohungMinenabwehrfahrzeugenReserveoffizierenautische PatenteKonvoisInternationale Seeschifffahrts-Organisationengl.SonderorganisationVereinten NationenLondonInternational Ship and Port Facility Security CodeInternationalen HandelskammerLondonPiraterieKuala LumpurOperation AtalantaAustralienNeuseelandBundesmarineEmdenFlottenkommandoBefehlshaber der Seestreitkräfte der NordseeMarinekommandoRostockHandelsschiffsoffizierenReserveoffizierePiraterie vor der Küste SomaliasOperation Allied ProtectorOperation Ocean ShieldGolfs von AdenInternational Recommended...
                    Read more