Question

Which of the following statements are True and
which are False ? Justify your answer with a
short proof or by a counter-example.

(a) The operation *, defined by x * y = log (xy) is
a binary operation on S, where
S={xER x>0}.

(b) If a and b are eigenvalues of two n x n
matrices A and B respectively, then a + b is
an eigenvalue of A + B.

(c) If S and T are linear transformations such
that SoT is defined and is 1 — 1, then S is
1 — 1.

(d) T : R³- R³: T((x1, x2, x3), (y 1 , y2, y3)) =
(x1 + x2 + x3) . (y1 + y2 + y3)
is an inner product on R³.

(e) {India, — 5, Jamila} is a set.

Answer 1

(a) The operation $*$, defined by $x * y = \log (xy)$ is not a binary operation on

$S=\{x\in \mathbb R \ |\ x>0\},$ bacause $1*1=\log(1\cdot 1)=\log 1=0\notin S.$

Answer: false

(b) Let us consider two $2\times 2$ matrices $A=\left[\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}\right]$ and $B=\left[\begin{array}{cc} -1 & 0 \\ 3 & -1 \end{array}\right]$. Let us find their eigenvalues:

$\det(A-aE)=\left|\begin{array}{cc} 1-a & 1 \\ 1 & 1-a \end{array}\right|=(1-a)^2-1=0$ implies $1-a=\pm 1$, and therefore, $a=0$ or $a=2;$

$\det(B-bE)=\left[\begin{array}{cc} -1-b & 0 \\ 3 & -1-b \end{array}\right]=(-1-b)^2=0$ implies $b=-1;$

$\det(A+B-cE)=\left|\begin{array}{cc} -c & 1 \\ 4 & -c \end{array}\right|=c^2-4=0$ implies $c=2$ or $c=-2$.

It follows that $a=0$ and $b=-1$ are the eigenvalues of the matrices $A$ and $B$ respectively, but $a+b=-1$ is not the eigenvalue of the matrix $A+B.$

Answer: false

(c) Let define the linear transformations $T:\mathbb R^2\to\mathbb R^4,\ \ T(x_1,x_2)=(x_1, x_2, 0, 0)$ and $S:\mathbb R^4\to\mathbb R^2,\ \ S(x_1,x_2,x_3,x_4)=(x_1, x_2)$. Then $S\circ T:\mathbb R^2\to\mathbb R^2.$ Since $S\circ T(x_1,x_2)=S(x_1, x_2, 0, 0)=(x_1,x_2)$, $S\circ T$ is an identity map of $\mathbb R^2$, and thus is 1 — 1. On the other hand, $S(1,1,0,0)=(1, 1)=S(1,1,1,1)$, and consequently, $S$ is not 1 — 1.

Answer: false

(d) Let $T :\mathbb R^3\to\mathbb R^3,\ \ T((x_1, x_2, x_3), (y_1 , y_2, y_3)) =(x_1 + x_2 + x_3)\cdot(y_1 + y_2 + y_3)$

Let $u=(x_1,x_2,x_3), \ v=(y_1,y_2,y_3),$ and $w=(z_1,z_2,z_3)$ be vectors and $\alpha$  be a scalar, then:

1. $T(u+v, w) =((x_1+y_1) + (x_2+y_2) + (x_3+y_3))\cdot(z_1 + z_2 + z_3)=(x_1+ x_2+ x_3+y_1+y_2 +y_3)\cdot(z_1 + z_2 + z_3)=(x_1+ x_2+ x_3)\cdot(z_1 + z_2 + z_3)+(y_1+y_2 +y_3)\cdot(z_1 + z_2 + z_3)=T(u, w)+T(v, w)$

2. $T(\alpha u, v)=T(\alpha(x_1, x_2, x_3), (y_1 , y_2, y_3)) =(\alpha x_1 + \alpha x_2 + \alpha x_3)\cdot(y_1 + y_2 + y_3)=\alpha (x_1 + x_2 + x_3)\cdot(y_1 + y_2 + y_3)=\alpha T( u, v)$

3. $T( u, v) =(x_1 + x_2 + x_3)\cdot(y_1 + y_2 + y_3)=(y_1 + y_2 + y_3)\cdot (x_1 + x_2 + x_3)=T( v, u)$

4.  $T( u, u) =(x_1 + x_2 + x_3)\cdot(x_1 + x_2 + x_3)=(x_1 + x_2 + x_3)^2\ge0$. On the other hand, for $(-1,1,0)\ne(0,0,0)$ we have $T((-1,1,0),(-1,1,0))=(-1+1+0)(-1+1+0)=0$.

Therefore, $T((x_1, x_2, x_3), (y_1 , y_2, y_3)) =(x_1 + x_2 + x_3)\cdot(y_1 + y_2 + y_3)$ is not an inner product on $\mathbb R^3$.

Answer: false

(e) {India, — 5, Jamila} is a set with three elements: India, — 5 and Jamila.

Answer: true