MATH 122: Matrixology (Linear Algebra) Solutions to Level Tetris (1984), 10 of 10 University of Vermont, Fall 2016

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1 MATH : Matrixology (Linear Algebra) Solutions to Level Tetris (984), 0 of 0 University of Vermont, Fall 0 (Q 4, 5) Show that the function f (x, x ) x + 4x x + x does not have a minimum at (0, 0) even though it has positive coefficients x Do this by rewriting f (x, x ) as x x A and finding the pivots of A x and noting their signs (and explaining why the signs of the pivots matter) Write f as a difference of squares and find a point (x, x ) where f is negative Note of caution: All of this signs matching for pivots and eigenvalues falls apart if we have to do row swaps in our reduction First, we can rewrite our function as x x f (x, x ) x x A x x x x We need to do one step of row reduction to reveal the pivots: ; A R R - R 0 The pivots are and - so we must have one positive and one negative eigenvalue: f is therefore not positive definite Completing the square: f (x, x ) x + 4x x + x (x + x ) 4x + x (x + x ) (x ) Note the appearance of the pivots and - in front of the squares As we saw in class, the LU factorization of symmetric matrices, A LDLT, is behind all of this (Q 9, 5) Find the by matrix A and its pivots, rank, eigenvalues, and determinant: x x x x A x 4(x x + x ) x

2 Is this matrix positive definite, semi-positive definite, or neither? Expanding 4(x x + x ) we have 4x + 4x + x 8x x x x + x x and this can be written as x x x We can now find the pivots of A (much easier than finding the eigenvalues): R R R - R R 8 8 R The pivots are 4, 0, 0 and our matrix is therefore semi-positive definite Some bonus sneaky grooviness: we can see straight away that A is a rank one matrix: A 4 / 4 / 4 / / / / We now have A in its spectral decomposition form: A n λ nˆv iˆv i T i So the eigenvalues are 4, 0, and 0, which means that A is semi-positive definite Another way to see this: we know from the pivots that two of the eigenvalues are 0 Since the trace of A is the sum of the eigenvalues, we have that the trace of A must be λ λ Checking A, we have λ 4 The determinant of A is zilch since we have 0 eigenvalues (following set of questions based on Q 7, Section 7) Singular Value Decomposition Happiness Consider A 0 0 x x x

3 (a) What are m, n, and r for this matrix? (b) What are the dimensions of U, Σ, and V? (c) Calculate A T A and AA T (a) m, n, and r (b) U is x, Σ is x, and V is x (c) and A T A AA T For the matrix A given above, find the eigenvalues and eigenvectors of A T A, and thereby construct V and Σ See this tweet for some post-it based help: Okay, we have to solve A λi 0 Using the big formula and going across the top row (to take advantage of the 0 in the (,) entry), we have: 0 λ 0 λ ( λ) λ 0 λ λ 0 λ ( λ)( λ)( λ) ()() ()( λ) (0)() λ 4λ λ λ(λ )(λ ) Our eigenvalues are λ, λ, and λ 0 Ordering for largest to smallest is important here We notice a couple of things: () The eigenvalues are all 0 This is good as these are the squares of our singular values, the σ i () One eigenvalue is 0 This

4 makes sense as the rank r which means that we have two non-zero singular values Our singular values are the square roots of the eigenvalues: σ and σ Note that there are only two singular values as A is x Next task: find the eigenvectors (a) For λ, we solve (A T A I) v v 0 You can do this be inspection, or by systematically finding the nullspace vector, or however you please By inspection: v Normalizing, we have ˆv (b) For λ, we solve (A T A I) v 0: By inspection: v ˆv (c) For λ 0, solve (A T A 0I) v 0: v 0 and the normalized eigenvector is 0 0 v 0 4

5 By inspection: v ˆv We can now write down V ˆv ˆv ˆv : and the normalized eigenvector is V 0 And the Σ matrix is For the same A, now find the basis {û i } using the essential connection Aˆv i σ i û i Construct U from the basis you find Again see this tweet for some post-it based help: We multiply the ˆv i for which σ i > 0 by A to find the û i We ll need to pull the σ i out to find the û i Recall that σ and σ First off: 0 Aˆv 0 σ û 5

6 Notice how when we pull out σ, we (almost magically) end up with a happy little unit vector Second vector: Aˆv 0 0 σ û 0 Smashing Note that û T û 0 and we have an orthonormal basis for R Finally, U Next find the {û i } in a different way by finding the eigenvalues and eigenvectors of AA T Eigenvalues: 0 AA T λi ( λ) λ λ ( λ )( λ + ) ( λ)( λ), where we have used the difference of perfect squares So λ and λ which again gives σ and σ Eigenvector time (la-la-la-la) for λ : 0 (AA T λ I) u I

7 By inspection, we have û Next, This gives û the negative of the one we have above 0 (AA T λ I) u I Note that we could have chosen û In fact, we always need to compute Aˆv i to find out which direction û i should take Beyond this, we don t need to compute the û i directly ever as once we have v i we need only multiply by A (as per the previous question) We found the u s directly here to () see that both ways give the same thing and () punish ourselves just a little more 7 (a) Put everything together and show that A UΣV T (b) Draw the big picture for this A showing which ˆv i s are mapped to which û i s (c) Which basis vectors, if any, belong to the two nullspaces?, (a) UΣV T / + / + 0 / / / / + 0 / + / 0 0 7

8 (b) (c) Left nullspace is just { 0} A s nullspace has dimension and has the basis vector ˆv 8 Finally, for this same A, perform the following calculation: where r is the rank of A σ û ˆv T + σ û ˆv T + + σ r û rˆv T r 8

9 You should obtain A σ û ˆv T +σ û ˆv T A Matlab question Verify the signs you found for the pivots of A in question by using Matlab to find A s eigenvalues Using Matlab, we find λ 0 and λ 4: >> eig( ; ) ans -0 4 One positive and one negative, matching the signs of the pivots 0 Matlab question Use Matlab to compute the SVD for the matrix A you explored in questions 8 >> U,Sigma,V svd( 0 ; 0 ) U Sigma 9

10 V (The bonus one pointer) Where does the fearsome kiwi rank among among rattites and what s unusual about the kiwi egg? The kiwi is the smallest of all struthious birds A kiwi egg can weight up to /4 of the mother s own weight, which is believed to be the highest ratio of all birds 0