Question

PLEASE PROVE PARTS a and b by CONTRADICTION and solve for c as well! Could you explain your steps as well

2. (a) (10 marks) Suppose A is an n x n real matrix. Show that A can be written as a sum of two invertible matrices. HINT: for any lER, we can write A = XI + (A - XI) (b) (10 marks) Suppose V is a proper subspace of Mnn(R). That is to say, V is a subspace, and V + Mnn(R) (there is some Me Mnn(R) such that M&V). Show that there exists an invertible matrix M E Mn,n(R) such that M€ V. You may find it useful to use (a), along with a proof by contradiction. (c) (5 marks) Suppose T : Mnn (R) + W is linear, and T(A) + Ow for some nxn matrix A. Show that there exists an invertible matrix M such that T(M) + 0w.

Answer 1

I I. SOLUTION (a). We have that A in a mom real waths, hel de in such that dto. then A=I&CA-dI), where I is men identity Since dto we have that dI is invertible and its inverse is d Suppose that A cannot be written as hum of two invertible amatrices that is whenevere A og Dygrered Sum of two matrices e, there are of them or both of them are nou-invertible representation H by on doing at og invertible 0 ve must have that A-dI if not invertible for all der dto. theit 12 det CA-dI) =0 / der, dto where det () denotes determinant matrix = A has infinitely many eigenvalues contradition as A is a matrix proximum number of real eigenvalues it can have is m... Hope our aloumption that Andi Paveltible here we must have der snel that AdI ir invertible where da o for quick closite ofd we wolle A as we have of او رار ولري 18 not for all der d to id wrong А. di +1A-d I)

dine dto =dI is invertible Also by choice of a A-d7 is invertible Heme we have our inter seuault. (b). V is a proper Subspare of Monan (IB). we have to show that there exist au invertible matrix ME MM, CIR] Such that MEV. we alsume that this does not hold That is whenever Miq a matrix satisfying ME Mun LR) and M & w, then M is nou invertible Since. ME Mm, n (R) by part (as we have that there exists. a der ME dI +(M-di where di and M-d I are invertible, dI and M-di are invertible and Maun CIR). V dou not have a invertible matrix guch that =) aure en V d I and M- di Since Vis a subspace d1+(M-d I )'evi =) MEN which is a contradiction that M&V. Hence one alsumption was wrong must have a matt M which is enreutible gatisfying ME Main CIR] and M&V. Hence proved. and we

کیا د -1 we hewe (6). Let T: Min (IR) be a linear transformation Also there exists A E Mun CIR) such that TCA) tow where Ow denotes W. zero rector or additive identity of Since A E Man LR) by paut there are are B, ce Mum. IR) dum that B and C are invertible and A-BTC =) T(A)= T(B)+ TCC) (by linearity 2 a) TLA) = T(B+C)

Part (a) is proved by contradiction. In proving it we have used that any matrix of order nxn can have at most n real eigenvalues. Part (b) is also proved by contradiction using part (a) and V is proper subspace. For proving part (c), part (a) is used and we prove it by contradiction.

3 (2 helby If Since TCA) to w Tub) + Te & Ow. both TIB) and Toll equals On. it. TEB) = Tee = On then T16) + 7) = Owton which is a contradiction to 6 Hence, we must have that either TIB) tow. or Teej tow or both are nou- zero Alen B and Ĉ are invertible. Without loss of generality assume that TIB) tou Let Me B E ManelR). then Mis invertible and TIM) tow Heure proved.