Exercise 1: Computing µ1 PurposeCompute upper and lower bounds for the structured singular value of constant matrices. Notethe dependence of µ∆(M) on the particular structure ∆, and verify the correctness of thebounds produced by the calculation.2 Commandsmussv upper and lower bounds for µ∆(M)mussvextract extract verification (D, G scalings, perturbation)Calling Synta x for mussv:≫ [bnds,muinfo] = mussv(mat,blk);≫ [bnds,muinfo] = mussv(mat,blk,opt);≫ [VDelta,VSigma,VLmi] = mussvextract(muinfo);Descriptionmat Matrix to calculate µ ofblk block structure information about the set ∆; the number of perturbation blocks, theirsizes and types. For example:The block s tr ucture∆ :=δ10 0 00 δ20 00 0 δ300 0 0 δ4: δi∈ Cis represented by the array≫ blk = [1 1;1 1;1 1;1 1];The block s tr ucture∆ :=ndiag [∆1∆2δ3] : ∆1∈ C3×2, ∆2∈ C4×5, δ3∈ Co,is represented by the array1≫ blk = [3 2;4 5;1 1];The block s tr ucture∆ :=δ1I30 00 ∆200 0 δ3I2: δ1, δ3∈ C, ∆2∈ C2×2is represented by the array≫ blk = [3 0;2 2;2 0];The block s tr ucture∆ :=δ10 0 00 δ20 00 0 δ300 0 0 δ4: δi∈ C, i = 1, 2, δj∈ R, j = 3, 4.is represented by the array≫ blk = [1 1;1 1;-1 0;-1 0];The block s tr ucture∆ :=ndiag [∆1∆2δ3I3×3] : ∆1∈ C3×2, ∆2∈ C4×5, δ3∈ Ro,is represented by the array≫ blk = [3 2;4 5;-3 0];Finally, the block structure∆ :=δ1I3×30 00 ∆200 0 δ3I2×2: δ1∈ C, δ3∈ R, ∆2∈ C2×2is represented by the array≫ blk = [3 0;2 2;-2 0];23 Upper BoundFor easier math equations below, use notation as:• let β := bnd(1, 1), the r eturned u pper bound ;• let ∆ denote the block structure as defined by blk;• let M denote the matrix in mat.The upper bound satisfies µ∆(M) ≤ β. The fields of VLmi are Dr, Dc, Grc and Gcr. Theywill satisfy:1. Dr= D∗r≻ 0, Dc= D∗c≻ 02. D1/2c∆ = ∆ D1/2rfor every ∆ ∈ ∆3. Gcr= G∗cr, Grc= G∗rc4. Grc∆ = ∆∗Gcrfor all ∆ ∈ ∆5. M∗DrM − β2Dc+ j (GcrM − M∗Grc)  0Together, these imply µ∆(M) ≤ β, as was derived in the handouts.Of course, if there are no real blocks, then Gcr= Grc= 0, and equivalently¯σD1/2rMD−1/2c= β4 Lower BoundFor easier math equations below, use notation as:• let α := bnd(1, 2), the returned lower bound;• let ∆ denote the returned perturbation, VDelta.The lower bound satisfies α ≤ µ∆(M). This is verified by checkin g 3 things:1. ∆ ∈ ∆2. ¯σ (∆) =1α3. det (I − M∆) = 0, equivalently, that M∆ has an eigenvalue at 135 ExampleVerify all of this on some random complex matrices.≫ mat = crandn(11,12);≫ blk = [-3 0;2 0;4 2;-2 0;1 2];≫ [bnds,muinfo] = mussv(mat,blk);≫ [VDelta,VSigma,VLmi] = mussvextract(muinfo);≫ % check the conditions above6 OptionsThere are a few options w hich yield better lower and upper bounds, at the expense of morecomputation. Use a 3rd argument≫ [bnds,muinfo] = mussv(mat,blk,’f’);≫ [bnds,muinfo] = mussv(mat,blk,’a’);≫ [bnds,muinfo] = mussv(mat,blk,’m2’);≫ [bnds,muinfo] = mussv(mat,blk,’m9’);≫ [bnds,muinfo] = mussv(mat,blk,’am7’); % bothThe argument ’f’ uses a fast, approximate algorithm to optimize the D and G scales. Theupper bound is usually inferior (larger) to the default.By contrast, the argument ’a’ improves the upper bound, but uses additional computationaltime.The argument ’m2’ - ’m9’ improves the lower bound, by trying the power iteration morethan once, each time starting from a new random vector. It then takes the best answer thatit gets.7 QuestionsWhich block structures seem to benefit from the extra computation that the ’a’ and ’m’options require?How much time d oes this add to the computation, as a