Question #145412
Which of the following statements are true and
which are false ? Give reasons for your
answers.
(a) Any square matrix with real entries is either symmetric or skew-symmetric or a linear combination of such matrices.
(b) If a linear operator has an eigenvalue 0,
then it cannot be one-one.
(c) If f : V --> K is a non-zero linear functional
and V a vector space of dimension n, then
there are n - 1 linearly independent vectors
v belongs to V such that f(v) = 0.
(d) Every binary operation on Rn is
commutative, for all n belongs to N.
(e) IAdj (A)I = IAI for all A belongs Mn(R).
1
Expert's answer
2020-11-25T16:33:45-0500

(a) Any square matrix with real entries is either symmetric or skew-symmetric or a linear combination of such matrices.

True


Let A+AA+A' be a symmetric matrix AAA-A' be a skew matrix

Let B=A+AB=A+A' then B=(A+A)B'=(A+A')'

    A+(A) (as (A+B)=A+B)    A+A(as (A)=A)    A+A(as (A+B=B+A)+B\implies A'+(A')'\ (as\ (A+B)'=A'+B')\\ \implies A'+A (as\ (A')'=A)\\ \implies A+A' (as\ (A+B= B+A)+B\\

Therefore B=A+AB=A+A' is a symmetric matrix


Now let C=AAC=A-A'

C=(AA)=A(A) (Why?)C=AA (Why?)    (AA)=CC'=(A-A)'=A-(A)'\ (Why?)\\ C'=A'-A\ (Why?)\\ \implies -(A-A')=-C


Hence, AAA-A' is a skew-symmetric matrix

If AA is a square matrix in Mn(C)Mn(C) then


A=12(A+A)+12(AA)A=\frac12 (A+A')+\frac12(A-A')

(b) If a linear operator has an eigenvalue 0, then it cannot be one-one.

False


Let VV be an nn dimensional vector space over an algebraically closed field F. Let x be a non-zero vector in V. If Tx=0Tx= 0 , then 0 is an eigenvalue. Assume that Tx0Tx \ne 0 . Then we known that {x,Tx,...,Tnx}\{x,Tx,...,Tnx\} is linearly dependent, so that there exist α0,α1,...,αkα0,α1,...,αk in F with k{1,...,n}k\{1,...,n\} such that ak0ak \neq 0 and



α0x+α1Tx++αkTkx=0,α_0x+α_1Tx+···+α_kT^kx= 0,

i.e.,


(α0I+α1T++αkTk)x=0.(α_0I+α_1T+···+α_kT^k)x= 0.

Thus,


p(T)x=0,p(T)x= 0,

where p(t):=α0+α1t++αktkp(t) :=α_0+α_1t+···+α_kt^k . By fundamental theorem of algebra, there exist λ1,...,λkλ_1,...,λ_k in F such that


p(t)=αk(tλ1)(tλk)p(t) =α_k(t−λ_1)···(t−λ_k)

Since p(T)x=0p(T)x= 0 , we have


αk(Tλ1I)(TλkI)x.α_k(T−λ_1I)···(T−λ_kI)x.

This shows that at least one of Tλ1I,...,TλkIT−λ_1I,...,T−λ_kI is not one-one.

(c) If f : V --> K is a non-zero linear functional and V a vector space of dimension n, then there are n - 1 linearly independent vectors v belongs to V such that f(v) = 0.

False


By definition,

Suppose V is a vector space of dimension n, and that v1,v2,...,vnv_1, v_2,...,v_n are vectors of V. The set of vectors {v1,v2,...,vn}\{v_1, v_2,...,v_n\} is linearly independent.


If f1v1+f2v2+...+fnvn=0f_1v_1+f_2v_2+...+f_nv_n=0 for some f1,f2,...,fnFf_1,f_2, ..., f_n \in F where at least one of f1,f2,...,fnf_1,f_2, ..., f_n is a non-zero. This implies that there are n linearly independent vectors of V as opposed to n-1


(d) Every binary operation on Rn is commutative, for all n belongs to N.

True


Let∗ be a binary operation on a set Rn. Given any three elements m,nm,n and pp of a set Rn, the binary operation, applied to the elements m∗ n and p of Rn ,yields an element (m∗n)∗p of Rn, and, applied to the elements m and n∗p of Rn, yields an element m∗(n∗p) of Rn.


(e) IAdj (A)I = IAI for all A belongs Mn(R).

False


Every invertible matrix is a product of some elementary matrices. This

is seen by applications of a series of elementary row operations to the matrix

to get the identity matrix I. When matrix A is invertible, the inverse can

be found by the adjoint, the formula


A1=1Aadj(A)A^{-1}=\frac{1}{|A|}adj(A)

It is easy to see that kA=(kn)A|kA| = (k^n)|A| where k is a constant and n is the order of the square matrix A.


We know that A1 is1A|A^{-1}|\ is \frac{1}{|A|} .Since A1A^{-1} is 1Aadj(A)\frac{1}{|A|} adj (A), we get A1=(1A)nadj(A)|A^{-1}| = (\frac{1}{|A|})^n |adj(A)| which implies A(n)A\frac{|A|^{(n)}}{|A|} is adj(A)|adj(A)| .


So A(n1)=adj(A)|A|^{(n-1)} = |adj(A)| .



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