7. Show that two 2-dimensional subspaces of a 3-dimensional subspace must have nontrivial intersection.
|(by contradiction) Suppose are both 2-dimensional subspaces of a 3-dimension vector space and assume that have trivial intersection. Then is also a subspace of , and since have a trivial intersection . But then:|
. However subspaces must have a smaller dimension than the whole vector space and . This is a contradiction and so must have trivial intersection.
8. Let be subspaces of a finite dimensional vector space . Show that .
|Define the linear map by . Then by dimension formula First note that in general . This fact I won’t prove here but is why . Now . That is, iff . But since and and they are actually the same vector, , then we must have . That says that the elements of the kernel are ordered pairs where the first and second component are equal and must be in . Then we can write . I claim that this is isomorphic to . To prove this consider the function as . This map is an isomorphism which you can check. Since we have an isomorphism, the dimensions must equal and so . Finally let us examine . I claim that . Note, this is equal and not just isomorphic. To see this, we note that if then by subspace property. So then any is also equal to . So these sets do indeed contain the exact same elements. That means . Putting this all together gives:|
16. Show that the matrix
as a linear map satisfies .
|The matrix is already in eschelon form and has one pivot in the second column. That means that a basis for the column space which is the same as the image would be the second column. In other words, . Now for the kernel space. Writing out the equation reads or in other words . Then an arbitrary element of the kernel . So again . In other words, .|
17. Show that
defines a projection for all . Compute the kernel and image.
|First I will deal with the case . In this case the matrix is and we see by the procedure in the last problem that: and .|
Now for the case . Then we still have only one pivot and either column can form a basis for the image. Using the second column makes it look nicer, and is the same as the previous case. . The difference is when we write out the equation to find the kernel, we get . With as our free variable this means so that a basis for the kernel is .