<Text-field style="Heading 2" layout="Heading 2">Maple Knows Vector Spaces!</Text-field>

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:

LUkpUm93U3BhY2VHNiI2Iy1JKlRyYW5zcG9zZUdGJDYjSSJBR0Yk

Now 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 like

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

If it doesn't, go to the line below, make sure you are in 2D Input mode, and retype it.)

PkkiVUc2Ii1JKk51bGxTcGFjZUdGJDYjSSJBR0Yk

Let 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 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

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();