Maple Worksheet #3Math 314, Section 6Fall 2006Maple Knows Vector Spaces!We are going to study the subspaces associated with matrices. First, let's load up the necessary package for doing linear algebra. Be sure to execute this command:with(LinearAlgebra);Construct a matrix using shortcut notation and 2D input:B := < < 1 | 2 | 3 | 4> , < 5 | 6 | 7 | 8 >, < 9 | 10 | 11 | 12> >;A := Matrix([[1,1,3,3,2], [0,2,2,4,0], [1,0,2,1,2],[2,1,5,4,4]]);Now let's apply some Maple commands:LUk2UmVkdWNlZFJvd0VjaGVsb25Gb3JtRzYiNiNJIkFHRiQ=What does this tell you about the row, column and null spaces of A?
Let's confirm our answers:Performing calculations in a casual usage scenario accepts optional arguments in any order. RowSpace(A);Hmmm... We'd rather our row spaces live in standard spaces, so they have to be columns. Let's fix this:PkkiVkc2Ii1JLENvbHVtblNwYWNlR0YkNiMtSSpUcmFuc3Bvc2VHRiQ2I0kiQUdGJA==That's odd. So what algorithm do you think that Maple is using for both, our "row space algorithm" or our "column space algorithm"?
Answer: (you fill it in!)O.K., let's move on. Now find the column space of A.LUksQ29sdW1uU3BhY2VHNiI2I0kiQUdGJA==Now how did Maple get that? Confirm your suspicion with an alternate calculation in the next cell:LUkpUm93U3BhY2VHNiI2Iy1JKlRyYW5zcG9zZUdGJDYjSSJBR0YkNow let's calculate the null space of A. Go back to the RREF above and confirm that you would have obtained the following:(Note: this was also one of the offending commands. It should look 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If it doesn't, go to the line below, make sure you are in 2D Input mode, and retype it.)PkkiVUc2Ii1JKk51bGxTcGFjZUdGJDYjSSJBR0YkLet U be the null space of A and V the row space. Of course, they both live in the standard space \342\204\2355. Let's confirm the formula
dim(U+V) = dim(U) + dim(V) - dim(U\342\213\202V).
For starters, we know dim(U) and dim(V) (what are they?). How do we find dim(U+V)? Let's build the matrices with column space U+V. Then calculate its rank with the Rank command.Note: We do this listing the columns from the two "lists" U and V. Actually, when Maple returns a basis for row space, column space or null space, what is is returning really is a list of rows or columns. The Maple command "Matrix" will accept a list of columns for building a matrix as well as a list of rows. Try the following:(This command has appeared munged for some people. It should look likeLUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2JVEiQ0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRKiZjb2xvbmVxO0YnL0YzUSdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGPS8lKXN0cmV0Y2h5R0Y9LyUqc3ltbWV0cmljR0Y9LyUobGFyZ2VvcEdGPS8lLm1vdmFibGVsaW1pdHNHRj0vJSdhY2NlbnRHRj0vJSVmb3JtR1EmaW5maXhGJy8lJ2xzcGFjZUdRL3RoaWNrbWF0aHNwYWNlRicvJSdyc3BhY2VHRk8vJShtaW5zaXplR1EiMUYnLyUobWF4c2l6ZUdRKWluZmluaXR5RictRiw2JVEnTWF0cml4RidGL0YyLUkobWZlbmNlZEdGJDYkLUYjNiMtRmZuNiYtRiM2MC1GLDYlUSJVRidGL0YyLUZmbjYmLUYjNiMtSSNtbkdGJDYkRlRGOUY5LyUlb3BlbkdRIltGJy8lJmNsb3NlR1EiXUYnLUY2NjBRIixGJ0Y5RjsvRj9GMUZARkJGREZGRkhGSi9GTlEkMGVtRicvRlFRM3Zlcnl0aGlja21hdGhzcGFjZUYnRlJGVUZeby1GZm42Ji1GIzYjLUZmbzYkUSIyRidGOUY5RmhvRltwRl5wRl5vLUZmbjYmLUYjNiMtRmZvNiRRIjNGJ0Y5RjlGaG9GW3BGXnAtRiw2JVEiVkYnRi9GMkZhb0ZecEZkcUZmcEY5RmhvRltwRjk=If it doesn't, go to the line below, make sure you are in 2D Input mode, and retype it.)PkkiQ0c2Ii1JJ01hdHJpeEc2JCUqcHJvdGVjdGVkR0koX3N5c2xpYkdGJDYjNycmSSJVR0YkNiMiIiImRi02IyIiIyZGLTYjIiIkJkkiVkdGJEYuJkY3RjE=LUk2UmVkdWNlZFJvd0VjaGVsb25Gb3JtRzYiNiNJIkNHRiQ=Conclusions? The rank of C is 5, so its columns are linearly independent, and the dimension of U+V is 5. Since dimension of U is 3 and of V is 2, we conclude from the dimension formula above that the dimension of dim(U\342\213\202V) must be 0. Hence dim(U\342\213\202V) = {0}.Finally, you might try the same questions with the matrix B that we introduced above by replacing the matrix A by B in the preceding commands. Better yet, just copy and paste everything we did with A down below and do the replacements.print();Assignment 2Due: Tuesday, November 7Points: 10Let subspaces U and V of \342\204\2356 by given by U = span{u1, u2, u3, u4, u5} = span{(1,3,2,-1,4,1), (1,3,0,1,1,1), (0,2,1,4,0,-3), (-1,3,1,0,4,0), (5,3,2,-1,1,3)} V = span{v1, v2, v3, v4} = span{(3,1,0,0,1,2), (3,3,-3,3,5,1), (2,2,-2,2,4,0), (1,3,0,1,1,1)}Use Maple to help you find bases for U, V, U+V, and U\342\213\202V. Use these to verify the dimension formula for these subspaces of \342\204\2356. You will find Exercise 3.6.11 and some of the discussion in this notebook helpful. Turn in a notebook with your work or printed copy. Note: Do NOT include irrelevant output. For example, if you make a copy of this notebook for your homework, delete all irrelevant material and put a colon after commands whose output you don't want to display, such as the "with(LinearAlgebra)" command. Do include relevant output, such as your name.LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEhRicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0Yn