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

Gastric acid as a weapon

What LEGO pieces have "real-world" functionality?

iPhone Wallpaper?

Did Xerox really develop the first LAN?

Using et al. for a last / senior author rather than for a first author

What causes the vertical darker bands in my photo?

Why don't the Weasley twins use magic outside of school if the Trace can only find the location of spells cast?

Right-skewed distribution with mean equals to mode?

How can I make names more distinctive without making them longer?

I am not a queen, who am I?

Dominant seventh chord in the major scale contains diminished triad of the seventh?

I need to find the potential function of a vector field.

Why is "Consequences inflicted." not a sentence?

How can I fade player character when he goes inside or outside of the area?

What happens to sewage if there is no river near by?

Is the address of a local variable a constexpr?

How widely used is the term Treppenwitz? Is it something that most Germans know?

When -s is used with third person singular. What's its use in this context?

Withdrew £2800, but only £2000 shows as withdrawn on online banking; what are my obligations?

Sorting numerically

List *all* the tuples!

Is there a documented rationale why the House Ways and Means chairman can demand tax info?

What would be the ideal power source for a cybernetic eye?

Storing hydrofluoric acid before the invention of plastics



Proof involving the spectral radius and the 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 operatorSpectral radius, second induced normContinuity of the spectral radiusSpectral radius and matrix norm inequality as its consequenceConfusion between spectral radius of matrix and spectral radius of the operatorComputing the Jordan Form of a MatrixProof of Gelfand's formula without using $rho(A) < 1$ iff $lim A^n = 0$Spectral radius of a matrixLower bound spectral radius of matrixFinding the Jordan Form of a matrix…












4












$begingroup$



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



Hint: Use the Jordan canonical form.




I am 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 matrices jordan-normal-form spectral-radius







share|cite|improve this question











$endgroup$












  • $begingroup$
    This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
    $endgroup$
    – copper.hat
    1 hour ago
















4












$begingroup$



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



Hint: Use the Jordan canonical form.




I am 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 matrices jordan-normal-form spectral-radius







share|cite|improve this question











$endgroup$












  • $begingroup$
    This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
    $endgroup$
    – copper.hat
    1 hour ago














4












4








4





$begingroup$



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



Hint: Use the Jordan canonical form.




I am 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 matrices jordan-normal-form 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 $rho(A) < 1$, where $rho(A)$ denotes the spectral radius of $A$.



Hint: Use the Jordan canonical form.




I am 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 matrices jordan-normal-form spectral-radius




linear-algebra matrices jordan-normal-form spectral-radius






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Rodrigo de Azevedo

13.2k41961




13.2k41961










asked 4 hours ago









mXdXmXdX

1168




1168












  • $begingroup$
    This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
    $endgroup$
    – copper.hat
    1 hour ago


















  • $begingroup$
    This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
    $endgroup$
    – copper.hat
    1 hour ago
















$begingroup$
This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
$endgroup$
– copper.hat
1 hour ago




$begingroup$
This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
$endgroup$
– copper.hat
1 hour ago










2 Answers
2






active

oldest

votes


















8












$begingroup$

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












    $begingroup$

    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
      3 hours 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.
      3 hours 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
      3 hours 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-the-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









    8












    $begingroup$

    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









    $endgroup$


















      8












      $begingroup$

      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









      $endgroup$
















        8












        8








        8





        $begingroup$

        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 3 hours ago









        Robert IsraelRobert Israel

        332k23221478




        332k23221478























            4












            $begingroup$

            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
              3 hours 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.
              3 hours 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
              3 hours ago
















            4












            $begingroup$

            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
              3 hours 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.
              3 hours 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
              3 hours ago














            4












            4








            4





            $begingroup$

            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 3 hours 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
              3 hours 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.
              3 hours 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
              3 hours 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
              3 hours 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.
              3 hours 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
              3 hours 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
            3 hours 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
            3 hours 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.
            3 hours 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.
            3 hours 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
            3 hours 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
            3 hours 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-the-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

            Taj Mahal Inhaltsverzeichnis Aufbau | Geschichte | 350-Jahr-Feier | Heutige Bedeutung | Siehe auch |...

            Image
            Mogul-ArchitekturMausoleum in IndienBauwerk in Uttar PradeshGrabmoscheeKuppelbauwerkWeltkulturerbe in AsienWeltkulturerbe in IndienAgraMogulreichGrabbau in AsienGrabgartenErbaut in den 1640er Jahren deutschpersoarabischDMGDevanagariMausoleumYamunaAgraindischenUttar PradeshKuppelMinaretteZiegelsteinenMarmorplattenGroßmogulShah JahanMumtaz MahalOktogonTambourgebauchtenChhatriIwanMinaretteErdbebenWendeltreppeJaliKenotapheKryptaReliefsMarmorinkrustationenKoranMärtyrerinGottesSandsteinTschahār BāghBagh-e WafaDschalalabadParadiesgartenMekkaMoscheeSüd-ZentralasiensAhmad LahoriBadakhshanAbu Fazelpersische Architekturindo-islamischen BaukunstIndienAsiensElefantenMarmorJaipurRajasthanNarbadaEdelsteinenHalbedelsteinenJaspisPunjabKarneoleAchatYemenKristallChinaLapislazuliSaphireCeylonKorallenArabienRoten MeerGranateBundelkhandDiamantenPannaOnyxPersienHeliotropFatehpur SikriTürkiseJean-Baptiste TavernierFrançois BernierMullahsSeelenTeppichenYamunaMuhammad Aurangzeb AlamgirTschahār BāghHumayunsbri
            Read more

            Baia Sprie Cuprins Etimologie | Istorie | Demografie | Politică și administrație | Arii naturale...

            Image
            Localități urbane în județul MaramureșOrașe în județul MaramureșBaia Sprie maghiarăgermanăorașjudețul MaramureșTransilvaniaRomâniaChiuzbaiaTăuții de SusSatu Nou de Susminieresașilor13291329maghiarCarol RobertBaia Mare13601348139014061851latinescul1540germanBaia Mare1564Nicolae OlahusSăsarGutâiBaia Mare1411Belgradului1427Iancu de Hunedoara14461453Matei CorvinMaria Terezaslovacirecensământului efectuat în 2011recensământul anterior din 2002românimaghiariromiortodocșiromano-catolicireformațigreco-catolicipenticostalimartori ai lui IehovaSebastian Alin BârdaPartidul Național Liberal178019321960școlii băimărene de picturăAurel PoppAdy EndreNicolae TonitzaAstra19221997România1876vulcanicetectonicădaciteandezitecuarțuluiBaia SprieBorșaCavnicDragomireștiSăliștea de SusSeiniȘomcuta MareTârgu LăpușTăuții-MăgherăușUlmeniVișeu de SusArdusatArinișAsuaju de SusBăița de sub CodruBăiuțBârsanaBăseștiBicazBistraBocicoiu MareBogdan VodăBoiu MareBotizaBudeștiCălineștiCâmpulung la TisaCerneștiCicârlăuCoa
            Read more

            Nicolae Petrescu-Găină Cuprins Biografie | Opera | In memoriam | Varia | Controverse, incertitudini...

            Image
            Articole de calitateNașteri în 1871Nașteri pe 31 martieDecese în 1931Decese pe 15 februarieArtiști români din secolul al XIX-leaArtiști români din secolul al XX-leaCaricaturiști româniTinerimea artistică CraiovaRomâniaBucureștiRomâniacaricaturistȘtefan LuchianConstantin ArtachinoAlexandru Bogdan-PiteștiNicolae Vermont1898Honoré DaumierParlamentului României19081913Constantin Jiquidi1898secolului al XIX-leaGolemo KonjariBulgariaMuzeului de Artă din Craiova2007Craiova31 martie1871limba francezălimba latinăLiceul Sfântul SavaBucureștigazului metanOperei RomâneȘtefan Petrescu-MuscăLiceul „Carol I”Școlii de Belle-ArteCostin PetrescuGheorghe TattarescuTheodor AmanConstantin I. StăncescuAlexandru ObedenaruSalonului Artiștilor IndependențiExpoziția Artiștilor în ViațăNicolae VermontȘtefan LuchianConstantin ArtachinoAlexandru Bogdan-PiteștiIon AngelescuNicolae GrantIon TicuLudovic DolinskiCaricatura Eu Sunt... Școala de Belle-Arte din BucureștiAteneului RomânGeorge RanettiCafeneaua ImperialFi
            Read more