Solving overdetermined system by QR decomposition












3












$begingroup$


I need to solve $Ax=b$ in lots of ways using QR decomposition.



$$A = begin{bmatrix}
1 & 1 \
-1 & 1 \
1 & 2
end{bmatrix}, b = begin{bmatrix}
1 \
0 \
1
end{bmatrix}$$



This is an overdetermined system. That is, it has more equations than needed for a unique solution.



I need to find $min ||Ax-b||$. How should I solve it using QR?



I know that QR can be used to reduce the problem to
$$Vert Ax - b Vert = Vert QRx - b Vert = Vert Rx - Q^{-1}b Vert.$$



but what do I do after this?






linear-algebra numerical-methods numerical-linear-algebra







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    I need to solve $Ax=b$ in lots of ways using QR decomposition.



    $$A = begin{bmatrix}
    1 & 1 \
    -1 & 1 \
    1 & 2
    end{bmatrix}, b = begin{bmatrix}
    1 \
    0 \
    1
    end{bmatrix}$$



    This is an overdetermined system. That is, it has more equations than needed for a unique solution.



    I need to find $min ||Ax-b||$. How should I solve it using QR?



    I know that QR can be used to reduce the problem to
    $$Vert Ax - b Vert = Vert QRx - b Vert = Vert Rx - Q^{-1}b Vert.$$



    but what do I do after this?






    linear-algebra numerical-methods numerical-linear-algebra







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      I need to solve $Ax=b$ in lots of ways using QR decomposition.



      $$A = begin{bmatrix}
      1 & 1 \
      -1 & 1 \
      1 & 2
      end{bmatrix}, b = begin{bmatrix}
      1 \
      0 \
      1
      end{bmatrix}$$



      This is an overdetermined system. That is, it has more equations than needed for a unique solution.



      I need to find $min ||Ax-b||$. How should I solve it using QR?



      I know that QR can be used to reduce the problem to
      $$Vert Ax - b Vert = Vert QRx - b Vert = Vert Rx - Q^{-1}b Vert.$$



      but what do I do after this?






      linear-algebra numerical-methods numerical-linear-algebra







      share|cite|improve this question









      $endgroup$




      I need to solve $Ax=b$ in lots of ways using QR decomposition.



      $$A = begin{bmatrix}
      1 & 1 \
      -1 & 1 \
      1 & 2
      end{bmatrix}, b = begin{bmatrix}
      1 \
      0 \
      1
      end{bmatrix}$$



      This is an overdetermined system. That is, it has more equations than needed for a unique solution.



      I need to find $min ||Ax-b||$. How should I solve it using QR?



      I know that QR can be used to reduce the problem to
      $$Vert Ax - b Vert = Vert QRx - b Vert = Vert Rx - Q^{-1}b Vert.$$



      but what do I do after this?






      linear-algebra numerical-methods numerical-linear-algebra




      linear-algebra numerical-methods numerical-linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked yesterday









      Guerlando OCsGuerlando OCs

      10821856




      10821856






















          2 Answers
          2






          active

          oldest

          votes


















          7












          $begingroup$

          The most straightforward way I know is to pass through the normal equations:



          $$A^T A x = A^T b$$



          and substitute in the $QR$ decomposition of $A$ (with the convention $Q in mathbb{R}^{m times n},R in mathbb{R}^{n times n}$). Thus you get



          $$R^T Q^T Q R x = R^T Q^T b.$$



          But $Q^T Q=I_n$. (Note that in this convention $Q$ isn't an orthogonal matrix, so $Q Q^T neq I_m$, but this doesn't matter here.) Thus:



          $$R^T R x = R^T Q^T b.$$



          If $A$ has linearly independent columns (as is usually the case with overdetermined systems), then $R^T$ is injective, so by multiplying both sides by the left inverse of $R^T$ you get



          $$Rx=Q^T b.$$



          This system is now easy to solve numerically.



          For numerical purposes it's important that the removal of $Q^T Q$ and $R^T$ from the problem is done analytically, and in particular $A^T A$ is never constructed numerically.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Is there any reason to make this so convoluted? From $Ax=b$ you have $QRx=b$, multiply by $Q^T$ on the left.
            $endgroup$
            – Martin Argerami
            yesterday










          • $begingroup$
            @MartinArgerami Because actually the least squares solution usually does not satisfy $Ax=b$. This simple perspective only shows you that this approach gives you a solution when a solution exists. Now you could argue directly that multiplying both sides by $Q^T$ furnishes an equation whose solution is the least squares solution. (Such an argument would resemble the usual geometric argument for deriving the normal equations.) This would make a good alternative answer to mine.
            $endgroup$
            – Ian
            yesterday





















          2












          $begingroup$

          Note that $Rx$ has the form
          $$Rx = begin{bmatrix} y_1 \ y_2 \ 0end{bmatrix} $$
          , so if $$ Q^{-1}b = begin{bmatrix} z_1 \ z_2 \ z_3end{bmatrix}$$
          then $|| Rx - Q^{-1}b||$ will be minimal for $y_1 = z_1$, $y_2=z_2$. This set of equation is no longer overdetermined.



          Using matrix notation, if tou write $R = begin{bmatrix} R_1 \ 0end{bmatrix}$ and intoduce $P=begin{bmatrix}1 & 0 & 0 \ 0 & 1& 0end{bmatrix}$, then you have
          $$ R_1x = PQ^{-1}b$$
          $$ x = (R_1)^{-1}PQ^{-1}b$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            The key trick in this answer is that by the orthogonality, $| Ax - b | = | Rx - Q^T b |$.
            $endgroup$
            – Ian
            yesterday










          • $begingroup$
            @Ian, That's something that OP has alredy obtained on his own (since $Q$ is orthogonal, $Q^{-1}=Q^T$).
            $endgroup$
            – Adam Latosiński
            14 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%2f3185239%2fsolving-overdetermined-system-by-qr-decomposition%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









          7












          $begingroup$

          The most straightforward way I know is to pass through the normal equations:



          $$A^T A x = A^T b$$



          and substitute in the $QR$ decomposition of $A$ (with the convention $Q in mathbb{R}^{m times n},R in mathbb{R}^{n times n}$). Thus you get



          $$R^T Q^T Q R x = R^T Q^T b.$$



          But $Q^T Q=I_n$. (Note that in this convention $Q$ isn't an orthogonal matrix, so $Q Q^T neq I_m$, but this doesn't matter here.) Thus:



          $$R^T R x = R^T Q^T b.$$



          If $A$ has linearly independent columns (as is usually the case with overdetermined systems), then $R^T$ is injective, so by multiplying both sides by the left inverse of $R^T$ you get



          $$Rx=Q^T b.$$



          This system is now easy to solve numerically.



          For numerical purposes it's important that the removal of $Q^T Q$ and $R^T$ from the problem is done analytically, and in particular $A^T A$ is never constructed numerically.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Is there any reason to make this so convoluted? From $Ax=b$ you have $QRx=b$, multiply by $Q^T$ on the left.
            $endgroup$
            – Martin Argerami
            yesterday










          • $begingroup$
            @MartinArgerami Because actually the least squares solution usually does not satisfy $Ax=b$. This simple perspective only shows you that this approach gives you a solution when a solution exists. Now you could argue directly that multiplying both sides by $Q^T$ furnishes an equation whose solution is the least squares solution. (Such an argument would resemble the usual geometric argument for deriving the normal equations.) This would make a good alternative answer to mine.
            $endgroup$
            – Ian
            yesterday


















          7












          $begingroup$

          The most straightforward way I know is to pass through the normal equations:



          $$A^T A x = A^T b$$



          and substitute in the $QR$ decomposition of $A$ (with the convention $Q in mathbb{R}^{m times n},R in mathbb{R}^{n times n}$). Thus you get



          $$R^T Q^T Q R x = R^T Q^T b.$$



          But $Q^T Q=I_n$. (Note that in this convention $Q$ isn't an orthogonal matrix, so $Q Q^T neq I_m$, but this doesn't matter here.) Thus:



          $$R^T R x = R^T Q^T b.$$



          If $A$ has linearly independent columns (as is usually the case with overdetermined systems), then $R^T$ is injective, so by multiplying both sides by the left inverse of $R^T$ you get



          $$Rx=Q^T b.$$



          This system is now easy to solve numerically.



          For numerical purposes it's important that the removal of $Q^T Q$ and $R^T$ from the problem is done analytically, and in particular $A^T A$ is never constructed numerically.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Is there any reason to make this so convoluted? From $Ax=b$ you have $QRx=b$, multiply by $Q^T$ on the left.
            $endgroup$
            – Martin Argerami
            yesterday










          • $begingroup$
            @MartinArgerami Because actually the least squares solution usually does not satisfy $Ax=b$. This simple perspective only shows you that this approach gives you a solution when a solution exists. Now you could argue directly that multiplying both sides by $Q^T$ furnishes an equation whose solution is the least squares solution. (Such an argument would resemble the usual geometric argument for deriving the normal equations.) This would make a good alternative answer to mine.
            $endgroup$
            – Ian
            yesterday
















          7












          7








          7





          $begingroup$

          The most straightforward way I know is to pass through the normal equations:



          $$A^T A x = A^T b$$



          and substitute in the $QR$ decomposition of $A$ (with the convention $Q in mathbb{R}^{m times n},R in mathbb{R}^{n times n}$). Thus you get



          $$R^T Q^T Q R x = R^T Q^T b.$$



          But $Q^T Q=I_n$. (Note that in this convention $Q$ isn't an orthogonal matrix, so $Q Q^T neq I_m$, but this doesn't matter here.) Thus:



          $$R^T R x = R^T Q^T b.$$



          If $A$ has linearly independent columns (as is usually the case with overdetermined systems), then $R^T$ is injective, so by multiplying both sides by the left inverse of $R^T$ you get



          $$Rx=Q^T b.$$



          This system is now easy to solve numerically.



          For numerical purposes it's important that the removal of $Q^T Q$ and $R^T$ from the problem is done analytically, and in particular $A^T A$ is never constructed numerically.






          share|cite|improve this answer











          $endgroup$



          The most straightforward way I know is to pass through the normal equations:



          $$A^T A x = A^T b$$



          and substitute in the $QR$ decomposition of $A$ (with the convention $Q in mathbb{R}^{m times n},R in mathbb{R}^{n times n}$). Thus you get



          $$R^T Q^T Q R x = R^T Q^T b.$$



          But $Q^T Q=I_n$. (Note that in this convention $Q$ isn't an orthogonal matrix, so $Q Q^T neq I_m$, but this doesn't matter here.) Thus:



          $$R^T R x = R^T Q^T b.$$



          If $A$ has linearly independent columns (as is usually the case with overdetermined systems), then $R^T$ is injective, so by multiplying both sides by the left inverse of $R^T$ you get



          $$Rx=Q^T b.$$



          This system is now easy to solve numerically.



          For numerical purposes it's important that the removal of $Q^T Q$ and $R^T$ from the problem is done analytically, and in particular $A^T A$ is never constructed numerically.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered yesterday









          IanIan

          69.2k25392




          69.2k25392








          • 1




            $begingroup$
            Is there any reason to make this so convoluted? From $Ax=b$ you have $QRx=b$, multiply by $Q^T$ on the left.
            $endgroup$
            – Martin Argerami
            yesterday










          • $begingroup$
            @MartinArgerami Because actually the least squares solution usually does not satisfy $Ax=b$. This simple perspective only shows you that this approach gives you a solution when a solution exists. Now you could argue directly that multiplying both sides by $Q^T$ furnishes an equation whose solution is the least squares solution. (Such an argument would resemble the usual geometric argument for deriving the normal equations.) This would make a good alternative answer to mine.
            $endgroup$
            – Ian
            yesterday
















          • 1




            $begingroup$
            Is there any reason to make this so convoluted? From $Ax=b$ you have $QRx=b$, multiply by $Q^T$ on the left.
            $endgroup$
            – Martin Argerami
            yesterday










          • $begingroup$
            @MartinArgerami Because actually the least squares solution usually does not satisfy $Ax=b$. This simple perspective only shows you that this approach gives you a solution when a solution exists. Now you could argue directly that multiplying both sides by $Q^T$ furnishes an equation whose solution is the least squares solution. (Such an argument would resemble the usual geometric argument for deriving the normal equations.) This would make a good alternative answer to mine.
            $endgroup$
            – Ian
            yesterday










          1




          1




          $begingroup$
          Is there any reason to make this so convoluted? From $Ax=b$ you have $QRx=b$, multiply by $Q^T$ on the left.
          $endgroup$
          – Martin Argerami
          yesterday




          $begingroup$
          Is there any reason to make this so convoluted? From $Ax=b$ you have $QRx=b$, multiply by $Q^T$ on the left.
          $endgroup$
          – Martin Argerami
          yesterday












          $begingroup$
          @MartinArgerami Because actually the least squares solution usually does not satisfy $Ax=b$. This simple perspective only shows you that this approach gives you a solution when a solution exists. Now you could argue directly that multiplying both sides by $Q^T$ furnishes an equation whose solution is the least squares solution. (Such an argument would resemble the usual geometric argument for deriving the normal equations.) This would make a good alternative answer to mine.
          $endgroup$
          – Ian
          yesterday






          $begingroup$
          @MartinArgerami Because actually the least squares solution usually does not satisfy $Ax=b$. This simple perspective only shows you that this approach gives you a solution when a solution exists. Now you could argue directly that multiplying both sides by $Q^T$ furnishes an equation whose solution is the least squares solution. (Such an argument would resemble the usual geometric argument for deriving the normal equations.) This would make a good alternative answer to mine.
          $endgroup$
          – Ian
          yesterday













          2












          $begingroup$

          Note that $Rx$ has the form
          $$Rx = begin{bmatrix} y_1 \ y_2 \ 0end{bmatrix} $$
          , so if $$ Q^{-1}b = begin{bmatrix} z_1 \ z_2 \ z_3end{bmatrix}$$
          then $|| Rx - Q^{-1}b||$ will be minimal for $y_1 = z_1$, $y_2=z_2$. This set of equation is no longer overdetermined.



          Using matrix notation, if tou write $R = begin{bmatrix} R_1 \ 0end{bmatrix}$ and intoduce $P=begin{bmatrix}1 & 0 & 0 \ 0 & 1& 0end{bmatrix}$, then you have
          $$ R_1x = PQ^{-1}b$$
          $$ x = (R_1)^{-1}PQ^{-1}b$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            The key trick in this answer is that by the orthogonality, $| Ax - b | = | Rx - Q^T b |$.
            $endgroup$
            – Ian
            yesterday










          • $begingroup$
            @Ian, That's something that OP has alredy obtained on his own (since $Q$ is orthogonal, $Q^{-1}=Q^T$).
            $endgroup$
            – Adam Latosiński
            14 hours ago


















          2












          $begingroup$

          Note that $Rx$ has the form
          $$Rx = begin{bmatrix} y_1 \ y_2 \ 0end{bmatrix} $$
          , so if $$ Q^{-1}b = begin{bmatrix} z_1 \ z_2 \ z_3end{bmatrix}$$
          then $|| Rx - Q^{-1}b||$ will be minimal for $y_1 = z_1$, $y_2=z_2$. This set of equation is no longer overdetermined.



          Using matrix notation, if tou write $R = begin{bmatrix} R_1 \ 0end{bmatrix}$ and intoduce $P=begin{bmatrix}1 & 0 & 0 \ 0 & 1& 0end{bmatrix}$, then you have
          $$ R_1x = PQ^{-1}b$$
          $$ x = (R_1)^{-1}PQ^{-1}b$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            The key trick in this answer is that by the orthogonality, $| Ax - b | = | Rx - Q^T b |$.
            $endgroup$
            – Ian
            yesterday










          • $begingroup$
            @Ian, That's something that OP has alredy obtained on his own (since $Q$ is orthogonal, $Q^{-1}=Q^T$).
            $endgroup$
            – Adam Latosiński
            14 hours ago
















          2












          2








          2





          $begingroup$

          Note that $Rx$ has the form
          $$Rx = begin{bmatrix} y_1 \ y_2 \ 0end{bmatrix} $$
          , so if $$ Q^{-1}b = begin{bmatrix} z_1 \ z_2 \ z_3end{bmatrix}$$
          then $|| Rx - Q^{-1}b||$ will be minimal for $y_1 = z_1$, $y_2=z_2$. This set of equation is no longer overdetermined.



          Using matrix notation, if tou write $R = begin{bmatrix} R_1 \ 0end{bmatrix}$ and intoduce $P=begin{bmatrix}1 & 0 & 0 \ 0 & 1& 0end{bmatrix}$, then you have
          $$ R_1x = PQ^{-1}b$$
          $$ x = (R_1)^{-1}PQ^{-1}b$$






          share|cite|improve this answer









          $endgroup$



          Note that $Rx$ has the form
          $$Rx = begin{bmatrix} y_1 \ y_2 \ 0end{bmatrix} $$
          , so if $$ Q^{-1}b = begin{bmatrix} z_1 \ z_2 \ z_3end{bmatrix}$$
          then $|| Rx - Q^{-1}b||$ will be minimal for $y_1 = z_1$, $y_2=z_2$. This set of equation is no longer overdetermined.



          Using matrix notation, if tou write $R = begin{bmatrix} R_1 \ 0end{bmatrix}$ and intoduce $P=begin{bmatrix}1 & 0 & 0 \ 0 & 1& 0end{bmatrix}$, then you have
          $$ R_1x = PQ^{-1}b$$
          $$ x = (R_1)^{-1}PQ^{-1}b$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Adam LatosińskiAdam Latosiński

          6068




          6068








          • 1




            $begingroup$
            The key trick in this answer is that by the orthogonality, $| Ax - b | = | Rx - Q^T b |$.
            $endgroup$
            – Ian
            yesterday










          • $begingroup$
            @Ian, That's something that OP has alredy obtained on his own (since $Q$ is orthogonal, $Q^{-1}=Q^T$).
            $endgroup$
            – Adam Latosiński
            14 hours ago
















          • 1




            $begingroup$
            The key trick in this answer is that by the orthogonality, $| Ax - b | = | Rx - Q^T b |$.
            $endgroup$
            – Ian
            yesterday










          • $begingroup$
            @Ian, That's something that OP has alredy obtained on his own (since $Q$ is orthogonal, $Q^{-1}=Q^T$).
            $endgroup$
            – Adam Latosiński
            14 hours ago










          1




          1




          $begingroup$
          The key trick in this answer is that by the orthogonality, $| Ax - b | = | Rx - Q^T b |$.
          $endgroup$
          – Ian
          yesterday




          $begingroup$
          The key trick in this answer is that by the orthogonality, $| Ax - b | = | Rx - Q^T b |$.
          $endgroup$
          – Ian
          yesterday












          $begingroup$
          @Ian, That's something that OP has alredy obtained on his own (since $Q$ is orthogonal, $Q^{-1}=Q^T$).
          $endgroup$
          – Adam Latosiński
          14 hours ago






          $begingroup$
          @Ian, That's something that OP has alredy obtained on his own (since $Q$ is orthogonal, $Q^{-1}=Q^T$).
          $endgroup$
          – Adam Latosiński
          14 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%2f3185239%2fsolving-overdetermined-system-by-qr-decomposition%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

          香粉寮

          Image
          香粉寮 香粉寮 （英文： Heung Fan Liu ）位於香港新界沙田區大圍，其北面山坡為大圍村原居民用以安葬之地。 目录 1 命名 2 歷史 2.1 天體活動 3 相關參見 4 參考文獻 5 外部連結 命名 香粉寮中的「香粉」是指該地一帶所種植的香樹上的粉末，而「寮」則指以前當地居民在該地所搭建的寮屋。 歷史 以前，當地居民以出產香料為生，並在當地種植香木，靠旁邊的城門河水力磨製香粉。 天體活動 在1930年代，該地曾以舉辦天體活動而聞名。 在1932年，當時一名連伯氏（Lanepart）的拉脫維亞人成立香港裸體運動會（ Hong Kong Nudist Society ），在該處進行天體活動，引來香港傳媒廣泛報導 [1] 。當時華人社會對於裸體的觀點，與外籍人士大為有異，也出於好奇，探頭探腦令天體營不勝其擾，天體營最終搬離，遠走馬灣島中鑊底灣，直至1960年代，由於參與人數不足，於鑊底灣的天體活動亦告終止。 相關參見 美田邨 恆峰花園 翠景花園 香粉寮街 碧田街 香粉寮發展項目 參考文獻 ^ 拂袖生. <<初談香港早期天體運動>>(上). [ 2010-12-26 ] .   外部連結 爾東. 趣談新界街道. 明報出版社. 2006: 165. ISBN  9628918311.   这是一篇與香港地理相關的小作品。你可以通过编辑或修订扩充其内容。 查 论 编 This page is only for reference, If you need detailed information, please check here
          Read more

          GameSpot

          Image
          body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} GameSpot 戰地風雲：惡名昭彰2在Gamespot的評論 网站类型 新聞 持有者 CBS 创始人 Pete Deemer Vince Broady Jon Epstein 网站 http://www.gamespot.com/ 注册 Optional (free and paid) 推出时间 1996年5月1日 [1] GameSpot （中国大陆：游戏基地），於1996年5月由Pete Deemer和Vince Broady創立，是一個提供新聞、評論、預告片、下載及其他的相關資訊的電子遊戲網站。GameSpot被一間後來被CNET收購的企業ZDNet所收購。根據Alexa，GameSpot.com是200個網路擁擠最嚴重的網站之一。 除了由GameSpot員工創作的內容，網站還允許用戶寫評論、網誌、之後在網路論壇分享。一些在CNET旗下的GameFAQs分享。 2004年， GameSpot被Spike TV的觀眾選上「電子遊戲賞節目」贏得「最傑出遊戲網站。 [2] 其他的遊戲網站還有IGN、1UP.com、GameSpy是它最大的競爭對手。2008年，根據Compete.com的統計，「gamespot.com」吸引了最少6000萬人的點擊率。 [3] GameSpot的主頁鏈結了到最近新聞、評論、預告、和一些有關遊戲機的入口：Wii、任天堂DS、電腦遊戲、Xbox 360、PSP、PlayStation 2、PlayStation 3。它還有一列「最受歡迎遊戲名單」，還有給用戶快速獲得遊戲資訊的搜尋器。GameSpot 還包括一些小範圍的遊戲機：任天堂64、GameCube、Game Boy Color、Game Boy Advance、Xbox、PlayStation、SEGA Saturn、Dreamcast、Neo Geo Pocket Color、N-Gage、手機遊戲。 目录 1 歷史 1.1 國際歷史 1.2 著名的員工 2 評論和分...
          Read more

          璽

          Image
          國家象徵 事物 國號 國旗 國徽 國歌 國旗歌 國慶日 國家格言 國家代表色 國籍標誌 國家代碼 國家航空公司 國技 國樂 國璽 國石 （ 日语 ： 国石 ） 國菜 人物或拟人形象 國家元首 国父 国家化身 非人生物 國獸 （ 英语 ： List of national animals ） 國鳥 國花 國樹 查 论 编 本條目 論述以部分區域為主 ，未必具有普世通用的觀點。 請 協助補充內容 以避免偏頗，或討論本文的問題。   「 国玺 」重定向至此。關於法國法兰西岛大区上塞纳省的一个市镇，詳見「 索城 」。 太平天國玉璽 璽 是帝王用以署名的印章，武则天以后则多以 宝 名之。「璽」字的由來是秦始皇以和氏璧（一說是用藍田玉）製造玉璽，璽為印記之意。古時不少國家的君主均有專用的印章為印記，其中又以秦始皇的 傳國璽 最为知名。現代也以「璽」稱呼國家使用的印章，稱為 國璽 。 目录 1 中國 2 日本 3 參見 4 参考文献 5 外部链接 中國 秦、兩漢、新、魏、西晉、前趙、東晉、宋、南齊、梁、陳、隋、唐、後梁及後唐各朝沿用秦始皇之傳國玉璽，上面刻有「受命於天，既壽永昌」，以和氏璧製造，其中在王莽篡漢時，曾向當時的太皇太后王政君索取傳國玉璽，王政君非常憤怒，便將玉璽取出摔砸於地，為此傳國璽還碎了一角，雖然日後王莽以黃金補上缺角，卻仍看的出摔碎的痕跡。傳國玉璽在後唐末帝李從珂自焚之時失蹤。 在宋朝以前，皇帝通常也同时沿用秦朝 乘舆七玺 的制度，即在仅用于礼仪场合的传国玺外，另作皇帝日常办公用玺六方（皇帝之玺、天子之玺、皇帝行玺、天子行玺、皇帝信玺、天子信玺），重大礼仪时与传国玺一道作为礼器同列于御座御舆之前，象征皇权至上。惟唐初失传国玺，唐太宗另作一方宝玺代替，后复得传国玺，则自作宝玺易名“皇帝神玺”，传国玺称“受命玺”，与皇帝天子六玺合为乘舆八玺。五季纷扰，玺制混乱，至宋，皇帝就只用三方宝玺，一方为礼仪用玺受命宝（傳國璽此時已失踪，該玺為自作），另两方为日常办公所用的“书诏之宝”与“天下合同之宝”。宋徽宗时虽复活乘舆八玺之制，但为了强调自己的...
          Read more