# Matrix Analysis and Applied Linear Algebra: Solutions Manual by Carl D. Meyer

By Carl D. Meyer

This ebook avoids the normal definition-theorem-proof layout; in its place a clean procedure introduces a number of difficulties and examples all in a transparent and casual sort. The in-depth specialize in purposes separates this ebook from others, and is helping scholars to work out how linear algebra might be utilized to real-life events. the various extra modern subject matters of utilized linear algebra are integrated the following which aren't regularly present in undergraduate textbooks. Theoretical advancements are continuously observed with special examples, and every part ends with a couple of routines from which scholars can achieve additional perception. additionally, the inclusion of ancient details presents own insights into the mathematicians who built this topic. The textbook includes various examples and routines, historic notes, and reviews on numerical functionality and the prospective pitfalls of algorithms. ideas to all the workouts are supplied, in addition to a CD-ROM containing a searchable reproduction of the textbook.

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Extra info for Matrix Analysis and Applied Linear Algebra: Solutions Manual

Sample text

1. (a) Yes, because T(0) = 0. (b) Yes, because x ∈ V =⇒ T(x) ∈ V. 2. Every subspace of V is invariant under I. 3. (a) X is invariant because x ∈ X ⇐⇒ x = (α, β, 0, 0) for α, β ∈ , so T(x) = T(α, β, 0, 0) = (α + β, β, 0, 0) ∈ X . 4. (a) 1 1 ∗ ∗ 1 1 0 1 [T]B =  0 0 0 0 ∗ ∗ Q is nonsingular.  ∗ ∗  ∗ ∗ (b) X is invariant because     1 1  1  2 T(α1 Q∗1 + α2 Q∗2 ) = α1   + α2   = α1 Q∗1 + α2 (Q∗1 + Q∗2 ) −2 −2 3 2 = (α1 + α2 )Q∗1 + α2 Q∗2 ∈ span {Q∗1 , Q∗2 } . Y is invariant because     0 0 0  3 T(α3 Q∗3 + α4 Q∗4 ) = α3   + α4   = α4 Q∗3 ∈ span {Q∗3 , Q∗4 } .

B) When m = n, Y is square and (YT Y)−1 YT = Y−1 so that P = Y−1 X. 11. (a) Because B contains n vectors, you need only show that B is linearly inn−1 i n−1 dependent. To do this, suppose to both i=0 αi N (y) = 0 and apply N n−1 n−1 i sides to get α0 N (y) = 0 =⇒ α0 = 0. Now i=1 αi N (y) = 0. Apply Nn−2 to both sides of this to conclude that α1 = 0. Continue this process until you have α0 = α1 = · · · = αn−1 = 0. (b) Any n × n nilpotent matrix of index n can be viewed as a nilpotent operator of index n on n .

B) This inequality follows from the inequality of part (a) by setting α = |ˆ xi |p , β = |ˆ yi |q , λ = 1/p, (1 − λ) = 1/q. and (c) H¨older’s inequality results from part (b) by setting x ˆi = xi / x p and yˆi = yi / y q . To obtain the “vector form” of the inequality, use the triangle inequality for complex numbers to write n n |x∗ y| = i=1 = x p n |xi | |yi | = xi yi ≤ i=1 y q 1/p n |xi yi | ≤ i=1 1/q n |xi |p |yi |q i=1 i=1 . 13. For p = 1, Minkowski’s inequality is a consequence of the triangle inequality for scalars.