Answer to Question #145412 in Linear Algebra for Sourav Mondal

Question

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).

Answer 1

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

True

Let $A+A'$ be a symmetric matrix $A-A'$ be a skew matrix

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

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

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

Now let $C=A-A'$

$C'=(A-A)'=A-(A)'\ (Why?)\\ C'=A'-A\ (Why?)\\ \implies -(A-A')=-C$

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

If $A$ is a square matrix in $Mn(C)$ then

$$A=\frac12 (A+A')+\frac12(A-A')$$

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

False

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

$$α_0x+α_1Tx+···+α_kT^kx= 0,$$

i.e.,

$$(α_0I+α_1T+···+α_kT^k)x= 0.$$

Thus,

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

where $p(t) :=α_0+α_1t+···+α_kt^k$ . By fundamental theorem of algebra, there exist $λ_1,...,λ_k$ in F such that

$$p(t) =α_k(t−λ_1)···(t−λ_k)$$

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

$$α_k(T−λ_1I)···(T−λ_kI)x.$$

This shows that at least one of $T−λ_1I,...,T−λ_kI$ 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 $v_1, v_2,...,v_n$ are vectors of V. The set of vectors $\{v_1, v_2,...,v_n\}$ is linearly independent.

If $f_1v_1+f_2v_2+...+f_nv_n=0$ for some $f_1,f_2, ..., f_n \in F$ where at least one of $f_1,f_2, ..., f_n$ 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 $m,n$ and $p$ 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

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

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

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

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