¢¸
  • ¿¬¼¼´ë ¼±Çü´ë¼öÇÐ 3Â÷½ÃÇè Á·º¸.pdf   (1 ÆäÀÌÁö)
    1

  • ¿¬¼¼´ë ¼±Çü´ë¼öÇÐ 3Â÷½ÃÇè Á·º¸.pdf   (2 ÆäÀÌÁö)
    2

  • ¿¬¼¼´ë ¼±Çü´ë¼öÇÐ 3Â÷½ÃÇè Á·º¸.pdf   (3 ÆäÀÌÁö)
    3

  • ¿¬¼¼´ë ¼±Çü´ë¼öÇÐ 3Â÷½ÃÇè Á·º¸.pdf   (4 ÆäÀÌÁö)
    4


  • º» ¹®¼­ÀÇ
    ¹Ì¸®º¸±â´Â
    4 Pg ±îÁö¸¸
    °¡´ÉÇÕ´Ï´Ù.
¢º
Ŭ¸¯ : ´õ Å©°Ôº¸±â
  • ¿¬¼¼´ë ¼±Çü´ë¼öÇÐ 3Â÷½ÃÇè Á·º¸.pdf   (1 ÆäÀÌÁö)
    1

  • ¿¬¼¼´ë ¼±Çü´ë¼öÇÐ 3Â÷½ÃÇè Á·º¸.pdf   (2 ÆäÀÌÁö)
    2

  • ¿¬¼¼´ë ¼±Çü´ë¼öÇÐ 3Â÷½ÃÇè Á·º¸.pdf   (3 ÆäÀÌÁö)
    3

  • ¿¬¼¼´ë ¼±Çü´ë¼öÇÐ 3Â÷½ÃÇè Á·º¸.pdf   (4 ÆäÀÌÁö)
    4



  • º» ¹®¼­ÀÇ
    (Å« À̹ÌÁö)
    ¹Ì¸®º¸±â´Â
    4 Page ±îÁö¸¸
    °¡´ÉÇÕ´Ï´Ù.
´õºíŬ¸¯ : ´Ý±â
X ´Ý±â
µå·¡±× : Á¿ìÀ̵¿

¿¬¼¼´ë ¼±Çü´ë¼öÇÐ 3Â÷½ÃÇè Á·º¸.pdf

½ÃÇèÁ·º¸ > Á·º¸³ëÆ® ÀÎ ¼â ¹Ù·Î°¡±âÀúÀå
Áñ°Üã±â
Å°º¸µå¸¦ ´­·¯ÁÖ¼¼¿ä
( Ctrl + D )
¸µÅ©º¹»ç
Ŭ¸³º¸µå¿¡ º¹»ç µÇ¾ú½À´Ï´Ù.
¿øÇÏ´Â °÷¿¡ ºÙÇô³Ö±â Çϼ¼¿ä
( Ctrl + V )
ÆÄÀÏ : ¿¬¼¼´ë ¼±Çü´ë¼öÇÐ 3Â÷½ÃÇè Á·º¸.pdf.pdf   [Size : 118 Kbyte ]
ºÐ·®   4 Page
°¡°Ý  1,000 ¿ø

Ä«Ä«¿À ID·Î
´Ù¿î ¹Þ±â
±¸±Û ID·Î
´Ù¿î ¹Þ±â
ÆäÀ̽ººÏ ID·Î
´Ù¿î ¹Þ±â


¸ñÂ÷/Â÷·Ê
Problem 1. Indicate whether the statement is true(T) or (5) If A is a symmetric matrix, then eigenvectors from dierent eigenspaces are orthogonal. (T) false(F). Justify your answer. [each 3pt] (1) If T : Rn ¡æ Rn is a linear operator, and if [T ]B = [T ]B with respect to two bases B and B for Rn , then B = B . (F) Solve If T is a zero operator, then [T ]B = O for any basis for R . So [T ]B = [T ]B but B = B . So (¥ë1 ¥ë2 )(x1 x2 ) = 0 and thus x1 x2 = 0.
n
solve Suppose that x1 ¡ô E¥ë1 and x...
º»¹®/³»¿ë

Problem 1. Indicate whether the statement is true(T) or (5) If A is a symmetric matrix, then eigenvectors from dierent eigenspaces are orthogonal. (T) false(F). Justify your answer. [each 3pt] (1) If T : Rn ¡æ Rn is a linear operator, and if [T ]B = [T ]B with respect to two bases B and B for Rn , then B = B . (F) Solve If T is a zero operator, then [T ]B = O for any basis for R . So [T ]B = [T ]B but B = B . So (¥ë1 ¥ë2 )(x1 x2 ) = 0 and thus x1 x2 = 0.
n

solve Suppose that x1 ¡ô E¥ë1 and x2 ¡ô E¥ë2 are eigenvectors from dierent eigenspaces. Then, (¥ë1 x1 ) x2 = (Ax1 ) x2 = x1 (AT x2 )

= x1 (Ax2 ) = x1 (¥ë2 x2 )

(2) If V and W are distinct subspaces of Rn with the same dimension, then neither V nor W is a subspace of the other. (T) Solve With out loss of generality, if V is a subspace of W . Since dim(V ) = dim(W ) and V is a subspace of W , V = W . It is a contradiction. Problem 2. Indicate whether the statement is true(T) or false(F). [each 2pt] (1) If A = U ¥ÒV ¡¦(»ý·«)


ÀúÀÛ±ÇÁ¤º¸
À§ Á¤º¸ ¹× °Ô½Ã¹° ³»¿ëÀÇ Áø½Ç¼º¿¡ ´ëÇÏ¿© ȸ»ç´Â º¸ÁõÇÏÁö ¾Æ´ÏÇϸç, ÇØ´ç Á¤º¸ ¹× °Ô½Ã¹° ÀúÀ۱ǰú ±âŸ ¹ýÀû Ã¥ÀÓÀº ÀÚ·á µî·ÏÀÚ¿¡°Ô ÀÖ½À´Ï´Ù. À§ Á¤º¸ ¹× °Ô½Ã¹° ³»¿ëÀÇ ºÒ¹ýÀû ÀÌ¿ë, ¹«´Ü ÀüÀ硤¹èÆ÷´Â ±ÝÁöµÇ¾î ÀÖ½À´Ï´Ù. ÀúÀÛ±ÇħÇØ, ¸í¿¹ÈÑ¼Õ µî ºÐÀï¿ä¼Ò ¹ß°ß½Ã °í°´¼¾ÅÍÀÇ ÀúÀÛ±ÇħÇØ½Å°í ¸¦ ÀÌ¿ëÇØ Áֽñ⠹ٶø´Ï´Ù.
ÀÚ·áÁ¤º¸
ID : comm***
Regist : 2017-04-03
Update : 2017-04-03
FileNo : 17042092

Àå¹Ù±¸´Ï

¿¬°ü°Ë»ö(#)
¿¬¼¼´ë   ¼±Çü´ë¼öÇÐ   Â÷½ÃÇè   Á·º¸   3Â÷½ÃÇè  


ȸ»ç¼Ò°³ | ÀÌ¿ë¾à°ü | °³ÀÎÁ¤º¸Ãë±Þ¹æħ | °í°´¼¾ÅÍ ¤Ó olle@olleSoft.co.kr
¿Ã·¹¼ÒÇÁÆ® | »ç¾÷ÀÚ : 408-04-51642 ¤Ó ±¤ÁÖ±¤¿ª½Ã ±¤»ê±¸ ¹«Áø´ë·Î 326-6, 201È£ | äÈñÁØ | Åë½Å : ±¤»ê0561È£
Copyright¨Ï ¿Ã·¹¼ÒÇÁÆ® All rights reserved | Tel.070-8744-9518
ÀÌ¿ë¾à°ü | °³ÀÎÁ¤º¸Ãë±Þ¹æħ ¤Ó °í°´¼¾ÅÍ ¤Ó olle@olleSoft.co.kr
¿Ã·¹¼ÒÇÁÆ® | »ç¾÷ÀÚ : 408-04-51642 | Tel.070-8744-9518