{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

m235notes13-page5

# m235notes13-page5 - ◆ be a basis of R 2 What is the...

This preview shows page 1. Sign up to view the full content.

We really should have subscripts that tell us what basis we are working with. We can just as well express the linear function f in terms of some other basis. Of course the entries in this new matrix will, in general, be di erent than the entries in the matrix arising from the standard basis. We now show how to compute the matrix of a linear function given its expression in terms of the standard basis in terms of an arbitrary basis. We actually do a little more than this. Example 5 . Let G be a linear transformation R 2 -! R 2 given by the matrix g = - 1 2 1 5 E E where E is the standard basis of R 2 . Let A = f 1 = 3 2 , f 2 = 4 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ◆ be a basis of R 2 . What is the matrix of G wrt the basis A ? • We Fnd the transition matrix from the A coordinates to the E coordinates. Note that we have f 1 = ✓ 1 ◆ A ! ✓ 3 2 ◆ E , f 2 = ✓ 1 ◆ A = ✓ 4 3 ◆ E . This says that the transition matrix from the basis A to the basis E is 1 E A = ✓ 3 4 2 3 ◆ . • ±rom this we conclude that the transition matrix from the basis E to the basis A is 1 A E = (1 E A )-1 = ✓ 3-4-2 3 ◆ . • The matrix for the linear map G in terms of the basis A is 1 A E · g E E · 1 E A . 5...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern