Question

Can you please answer questions 1-6，thank you a lot！Thumbs up for great answer，Thx！

Remember: to show that a property is true you must check every possibility (probably using variables and general vectors). To show that a property is false you only need to give one counterexample. 1. Find a set of vectors in R2 which is closed under vector addition but not scalar multiplication. 2. Find a set of vectors in R? which is closed under scalar multiplication but not vector addition, 3. Consider the set of vectors in R3 whose components have a sum of 0. Show whether this set is a subspace of Rs. 4. Consider the set of vectors in R? whose components have a product of 0. Show whether this set is a subspace of R. 5. Consider the set of 2-by-2 invertible matrices. Show whether this set is a subspace of the vector space of all 2-by-2 matrices. 6. Consider the set of 4-by-4 symmetric matrices. Show whether this set is a subspace of the vector space of all 4-by-4 matrices. (Recall: a matrix A is symmetric if A = A", that is, if ajj = a; for all i and

Answer 1

Solution with Explanation:

L = a +b7,0 17 S={(a,0) e R² 1 ano} CR?. clearely s is closed under additim. let (a,0), (6,0) ES (a,0) + (6,0) = (atb, o) ES ano, bo Now for scalare Multiplication, od dro n otoo let (2,0) ES and HER -|(2,0) = (-2,0) fs as -2<0. : S is closed cendere addition but not a under scalare multiplication.

2 S={(a, b) E IR? I ab=off R² let (a,b) Es, and a be any scalare then x (a,b) = (da, ab). La.ab = K² ab = x² 0 = 0 = (La , kb) ES, = x (a,b) € Si. Si is closed undere mentora Scalare multiplication Now, (1.0), (0.1) E SI. (,0) + (0,1)= (1.1)&s, as 1.1=1707 Siin NOT closed undere vector addition.

(3 let s be the set of vectores is R3 whoge Components have a sum of o. that u s= {(a, b, c) | atb+c= of C R² . 0+0+0=0 = (0,0,0) e S S is non empty set. Now. Let (a, b, a), (A2, bz, (2) ES (a, b, c) + (aq, bz, C2) = (a +az, bitbe, Gt (2) - at azt bit bz+G+ = (a + b + c) + (azt bat (2) Foto (a, b, G) ES out bitc=0 abz (₂) ES - azt b ₂ + (2=0 – (a, b, a) + (A2, bz, C2) ES. let (a, b, c) ES and XER be any scalar. d (a,b,c) - (da, &b, Qc) Now Ka+ & 6+&C = a(a+b+c) a,b,c) ES - atb+c=0 - x.o -0 So from the above we conclude that the set is a subspace of R. a

(4) let s denote the set of vectores in R² whose Components have a product o that n o S={(a,b,c)ER) abc=o} CR. a This NOT a subspace of Rs as the set Sis Not closed cendere aloon corcontropocation vectore addition 1.1.0=0 example: (1,1,0), (0,0,1) ES : 0.0.1 = 0 but (1,1,0) + (0,0,1)= (1,1,1) & S 11.1=1 70] on (5) Let M₂ denote the space of all ax2 matrices. W = { A E M₂ I det A to} 0=188)&W , as det (0) = 0 So, w is NOT a subspace of M₂. example: Consider A=(68) and B=(oo) det A = det B = 1 = A BEW. But A+B = ) + () - (3) d'W.

• A matrix is invertible if and only if $\det$ is not equals to zero. $\det$ denotes the determinant of the matrix.

C My . (6) Let W, ={A= (aij e Mu | A- AT} = {(aij) e. My | aij = aji} .c Mu. Clearly. Wi is non-empty. as. Of Wis 10ir o + jj (the null matrix)] do let A = (aij) and 13 = (bij) t Wi .. clearely, Aij= aji and bij = b; i tij Now. A+B = (diju(bij) = (aij + bij) - Cij , where Aijt bij = Cij For everey, i, j, we have lij = (ij)+(bij) (aij)e W, = (aji) + (b; i) = - aji + bj i Cji Cij = Cji tij => A+BEW, (A+B)"- AT+ B' = A + B . 13 other way to Hà Preoway to

XER be any scalar tet AEW, and (XAT = XTAT =&AT E XER &T=1 - A E WI; AT=A] - ВА = (XAT = LA - XAEWI.

from the above we have for any ABE W.. and XER A+B E W il . QAEWI => W, is a subspace of My Am.