is the set of all points Its image is the set of all points XÐ: Ñ œ XÐ: Ñ Ñ in 7

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1 For a linear transformation X À Ä parallelogram determined by two vectors? 8 7 : what is the image of a line segment? of a Á! and : 8 À The set of points : are a line in 8 Þ The line contains the point :Ð let> œ!ñ The set of points : where > is restricted, say! Ÿ > Ÿ "ß are a line segment in 8 Þ The segment starts at : (when > œ!ñ and ends at Ð when > œ "Ñ Its image is the set of all points XÐ: Ñ œ XÐ: Ñ >XÐ@ Ñ in 7 These points form Ú a line segment in from XÐ: Ñ to XÐ@Ñ in 7 if XÐ@ Ñ Á! Û Ð this line segment goes from XÐ: Ñ to Ñ Ü the point XÐ: Ñ if it happens that XÐ@ Ñ œ! Loosely speaking, if we think of a single point as a degenerate line or line segment, (length!), then we could say: for a linear transformation X À the image of a line segment is a line segment the image of a line (no restriction on > ) is a line. The image of a parallelogram Suppose? ß@ are linearly independent (so neither is a scalar multiple of the other). Then: the set of all linear combinations =? (! Ÿ = Ÿ " and! Ÿ > Ÿ " ) is the parallelogram determined by!,? Its image in 7 is the set of all points =XÐ? Ñ >XÐ@ Ñ (! Ÿ = Ÿ " and! Ÿ > Ÿ " ) These form a parallelogram in 7 Ð possibly degenerate" with area!). Ú parallelogram if XÐ? ÑßXÐ@ Ñ are linearly independent, or Û line segment if XÐ? Ñ, XÐ@ Ñ are linearly dependent and not both! Ü point if both XÐ? Ñ and XÐ@ Ñ are! See picture for an illustration with X À # # Ä Þ

2 In the picture the red parallelogram (upper right) is determined by!,?, A typical point in the parallelogram is =? (the corner of the green parallelogram). Its image (drawn on the same coordinate grid) is the red parallelogram (lower left) determined by! ß XÐ? Ñ and XÐ@ ÑÞ Every point inside the original parallelogram Ð see green parallelogram, =? has an image inside the other parallelogram (see green parallelogram at lower left, =XÐ? Ñ >XÐ@ ÑÑ We can say that the image of a parallelogram under a linear transformation is a parallelogram provided we remember that the image might be a degenerate parallelogram (a line segment, or a point). This is true even if the parallelogram doesn't have a vertex at! œ see below.

3 Linear transformation applied to a parallelogram with one vertex at : Adding : to each point in a parallelogram simply translates the parallelogram by : The image is the set of all points : =? >@ Å Å translate by : parallelogram vertex at! XÐ:Ñ =XÐ?Ñ >XÐ@Ñ Å Å translate by XÐ: Ñ parallelogram containing! Å # # See picture for an illustration with X À Ä Þ the same figure. The image parallelogram is drawn in

4 ExamplesX À # # Ä linear, " i) XÐB Ñ œ reflection across the B -axis. Here XÐ/ Ñ œ œ /! 0 XÐ/ # Ñ œ œ /#. The standard matrix is ÒXÐ/ " Ñ XÐ/ # ÑÓ À 1 # " " 0 ii) XÐB Ñ œ reflection across the line B# œ B", Here XÐ/ " Ñ œ œ / 1 " XÐ/ # Ñ œ œ /". The standard matrix is ÒXÐ/ " Ñ XÐ/ # ÑÓ! See the pictures below. In each case, the parallelogram ( œ the unit square ) formed by / and / are transformed by X into the blue parallelogram in the figure. 2 and and " #

5 " 5 Example Suppose XÐBÑ œ EB œ! " B, where 5!Þ Show the effect of this transformation on the unit square. We know that the unit square with two sides /" and /# is mapped to the parallelogram with " 5 sides XÐ/ " Ñ= and XÐ/ # Ñ œ! " Þ ( See the right-hand picture below)). this transformation is called a horizontal shear to the right. A formula is XÐ B " 1 Ñ œ 5 B " B " 5B # = B 0 1 B B The case 5 Ÿ! is also illustrated below. # # #

6 For X À 8 Ä 7 Å Å domain of X codomain of X the range of is the set of all those,'s in that are the image of some B from More precisely 7 8 range of X œ Ö, in :, œ XÐBÑ for some B in The range is a subset of the codomain but might or might not equal 7 Þ Ô B" Ô B" $ $ For example, if X À Ä is projection onto the B" - B# plane: XÐ B# Ñ œ B# ÕB Ø Õ! Ø Ô 1 then (for example) 1 is in the codomain $ but not in the range of X ( why?) Õ 1Ø Definition A linear transformation X À Ä is called onto if the range œ all of Ðthe codomain). $

7 8 7 Suppose X À Ä linear, with standard matrix E, so that XÐBÑ œ EB The following are equivalent (all true or all false) 7 X is called onto if for every, 7 For every, there is at least oneb 8 for which Í there is at least one XÐB ) œ, B 8 for which EB œ, Ô Ô ( see Theorem 4, p. 37) 7 7 for every,, the equation every vector, is XÐB Ñ œ, has at least one solution a linear combo of the columns of E Ô ( see Theorem 4, p. 37) the columns of E span 7 Ô ( see Theorem 4, p. 37) E has a pivot position in every row Å This version allows us to actually figure out whether a specific XÐBÑ œ E is onto (by using row reduction) Examples: Ô #B" "%B# $B$ B% &B& B" #B# B% %B& & & 1) Suppose X À Ä ß where XÐBÑ œ $B" B# B$ B&. I= T onto? Ö Ù #B" #B# )B$ )B% B& Õ #B %B $B B B Ø " # $ % &

8 Ô # "% $ " & " #! " % Write XÐBÑ œ EB œ $ " "! " B Ö Ù # # ) ) " Õ # % $ " " Ø Ô # "% $ " & Ô "!!!! " #! " %! "!!! Row reducing E œ $ " "! " µ ÞÞÞ µ!! "!! Ö Ù Ö Ù # # ) ) "!!! "! Õ # % $ " " Ø Õ!!!! " Ø There is a pivot in every row of E so the linear transformation X is onto 2) Can there be an onto linear transformation X À % Ä '? % ' If X À Ä and XÐBÑ œ EB, then E is a ' % matrix. Since there are more rows that columns in E, there cannot be a pivot position in every row. There for a linear transformation X À % Ä ' cannot be onto.