If and only if d negationslash 0 inserting the values

• Notes
• 7

This preview shows page 6 - 7 out of 7 pages.

if (and only if) d negationslash = 0. Inserting the values of x 2 and x 3 in the first equation, it can always be solved for x 1 if (and only if) a negationslash = 0. Summary: An upper triangular matrix A is invertible if and only if none of its diagonal elements are 0. b) If A is invertible, is its inverse also upper triangular? Solution: In the above computation, notice that x 3 only depends on y 3 . Then x 2 only depends on y 2 and y 3 . Finally, x 1 depends on y 1 , y 2 , and y 3 . Thus the inverse matrix is also upper triangular. c) Show that the product of two n × n upper triangular matrices is also upper trian- gular. Solution: Try the 3 × 3 case. The general case is the same – but takes some thought to write-out clearly and briefly. It is a consequence of three observations: 1. A matrix C := c 11 · · · c 1 n . . . . . . . . . c n 1 · · · c nn is upper-triangular if all the elements below the main diagonal are zero, that is, c jk = 0 for all j > k . 2. For any matrices, to compute the product AB , the jk element is the dot product of the j th row of A with the k th column of B . 3. For upper-triangular matrices: 6

Subscribe to view the full document.

the j th row of A is (0 , ... 0 , a jj ,...,a jn ) while the k th column of B is b 1 k . . . b kk 0 . . . 0 . For j > k , take the dot product of these vectors. The result is now obvious. d) Show that an upper triangular matrix is invertible if none of the elements a 11 , a 22 , a 33 , . . . , a nn , on the main diagonal are zero. Solution: This is the same as part a). The equations Avectorx = vector y are a 11 x 1 + a 12 x 2 + · · · + a 1 n - 1 x n - 1 + a 1 n x n = y 1 a 22 x 2 + · · · + a 2 n - 1 x n - 1 + a 2 n x n = y 2 . . . . . . . . . a n - 1 n - 1 x n - 1 + a n - 1 n x n = y n - 1 a nn x n = y n . To begin, solve the last equation for x n . This can always be done if (and only if) a nn negationslash = 0. Then solve the second from the last for x n - 1 , etc. This computation also proves the converse (below). As in part b), the inverse, if it exists, is also upper triangular. e) Conversely, if an upper triangular matrix is invertible show that none of the ele- ments on the main diagonal can be zero. Bonus Problem [Please give these directly to Professor Kazdan] 1-B Let L : V V be a linear map on a linear space V . a) Show that ker L ker L 2 and, more generally, ker L k ker L k +1 for all k 1. b) If ker L j = ker L j +1 for some integer j , show that ker L k = ker L k +1 for all k j .
• Fall '12
• JerryKazdan

{[ snackBarMessage ]}

Get FREE access by uploading your study materials

Upload your study materials now and get free access to over 25 million documents.

Upload now for FREE access Or pay now for instant access
"Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

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