If and only if d negationslash 0 inserting the values

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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
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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 .
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Christopher Reinemann
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