QS-025 · Linear Algebra · Medium
NPTEL GATE DA MOCK | Week 1 - 57
Consider a set $V={(x,y)\mid x,y\in\mathbb{R}}$ with the usual rule of addition borrowed from $\mathbb{R}^2$ and scalar multiplication defined as
$$ c(x,y)= \begin{cases} (0,0), & c=0,\\ (x,y), & c\neq0. \end{cases} $$
Consider the following statements:
P: $V$ is closed under addition.
Q: $V$ has a zero element with respect to addition.
R: $1\cdot v=v$, where $1\in\mathbb{R}$ and $v\in V$.
S: $a(v_1+v_2)=av_1+av_2$, where $v_1,v_2\in V$ and $a\in\mathbb{R}$.
T: $(a+b)v=av+bv$, where $a,b\in\mathbb{R}$ and $v\in V$.
Which of the following options are correct?
A. Only P is true
B. Only Q is true
C. P, Q and S are true
D. Both R and T are not true
Consider a set $V={(x,y)\mid x,y\in\mathbb{R}}$ with the usual rule of addition borrowed from $\mathbb{R}^2$ and scalar multiplication defined as
$$ c(x,y)= \begin{cases} (0,0), & c=0,\\ (x,y), & c\neq0. \end{cases} $$
Consider the following statements:
P: $V$ is closed under addition.
Q: $V$ has a zero element with respect to addition.
R: $1\cdot v=v$, where $1\in\mathbb{R}$ and $v\in V$.
S: $a(v_1+v_2)=av_1+av_2$, where $v_1,v_2\in V$ and $a\in\mathbb{R}$.
T: $(a+b)v=av+bv$, where $a,b\in\mathbb{R}$ and $v\in V$.
Which of the following options are correct?
A. Only P is true
B. Only Q is true
C. P, Q and S are true
D. Both R and T are not true
Solution
The addition is the usual addition in $\mathbb{R}^2$, so $V$ is closed under addition. Hence, P is true.
The zero element is $(0,0)$ because
$$ (x,y)+(0,0)=(x,y). $$
Hence, Q is true.
For R, take any nonzero $v\in V$. Since $1\neq0$,
$$ 1\cdot v=v. $$
Hence, R is true.
For S, consider $a=0$. Then
$$ 0(v_1+v_2)=(0,0), $$
and
$$ 0v_1+0v_2=(0,0)+(0,0)=(0,0). $$
For $a\neq0$,
$$ a(v_1+v_2)=v_1+v_2=av_1+av_2. $$
Hence, S is true.
For T, take $a=1$ and $b=-1$. Then
$$ (a+b)v=0v=(0,0), $$
but
$$ av+bv=v+v=2v, $$
which is not $(0,0)$ for $v\neq(0,0)$. Hence, T is false. Therefore, P,Q,R,S are true and T is false.
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