Question

6. True or False: (a) An eigenvalue of the matrix A is a non-zero vector y such that Ac = Xū. (b) Let A be a 3 x 4 matrix. Then ker A is non-trivial. (e) Let A be an n x n matrix. Ta is injective (i.e. one-to-one) if and only if TA is surjective (i.e. onto). (d) If A is a singular matrix, then A must have an eigenvalue. (e) The set {A € M,(F): det(A) = +1} is closed under matrix multiplication.

Answer 1

Since A has 4 coluning, nullity (A) =dim (kerty 6. (a) Let A be a nth square nan square matrix. Then eigen vector of A is a non-zero vector e in Rh such that Anado for some sealand. Since it A is not square then the equation A = de does not fulfill the equality of dimensions. a eigen value of a metrix Aų non-zero vector ao such that Au=dza. This statement is Thee. (6) Let A bg a from Rank- nullity Theorem Rank(Alt Nullity (A) = Number of or, dim (ker A) so a 3 X4 matrix. columns in a =4- Rank (A) Since maximum possible Rank of A is 3. Also we get the inequality os Rank (A) < 3 - Rank (A) >,-3 so o

OV or 4-07 4 - Rank (A) 74-3 , 47, dim(ker A) 711 So Is dimension of Ker I <3 so dimension of ker A not equal to o, So, dimlker A) 70 so, ker a to so, ker A is non-trivial, Then ker A ; non So, Let A be trivial 3x4 matrix We 3 this statement ý Trus. le) know the theorem: It T is a linear oferator on finite dimension vectors space" u. dim (") =n, the [+] = A, where A is nxn square matrix. Then these are equivalent. ① T is infective (one-one) T is onto T e bisective. so Let A be nyn matrix. Ta is inseetine (i.e. one-one) if and only if Ta is Surseetine (ie, onto). so the statement (d) Since A is a singular metrix it and only it o į an eigen value of A. so, Singular metrices has at least one value that is O. it A is singular matrix. Then A must have an eigen so the statenient ý tree is true eigen vålue that so eigen value.

set LA (e) Let The At Mnl F): det (A) = H is denoted by S. Then, sa At Mn(F); det (Al = 11 4) = +1] dit (B) = ܓ Let AES and BES Then det (A) = is equal to either for- Also det (B) is equal to either for- Then det (AB) det (A) det (B) Here are four eases are arrived ① If det (A) =1, det (B)=1 Them det (AB) = 1 ① If det (A)=1, det (B) = 1 then det (AB) = 4. ③ af det (A) =-1 Then det (AB) =-| 6 If det (Al =-, det (1) End Then det (AB) = 1 sol det (AR) is either 1 oral. So, ABES. Hence the set s į closed under metrix multiplication, so the set saa. Mulif) : det (A) = +1] ý closed under matrix multiplication. This statement ý true /