GO Classes DA 2025 | Weekly Quiz 6 | Change of Basis & Linear Transformation | Question: 12

Question

Consider two statements S1 and S2.
S1: If $\left\{v_{1}, \ldots, v_{n}\right\}$ are linearly INDEPENDENT vectors in $V$, then $\left\{T\left(v_{1}\right), \ldots, T\left(v_{n}\right)\right\}$ are linearly INDEPENDENT in $W$.
S2: If $\left\{v_{1}, \ldots, v_{n}\right\}$ are linearly DEPENDENT vectors in $V$, then $\left\{T\left(v_{1}\right), \ldots, T\left(v_{n}\right)\right\}$ are linearly DEPENDENT in $W$.
Which of the following is true?

• A. $\mathrm{S} 1$ is true, $\mathrm{S} 2$ is false.
• B. $\mathrm{S} 1$ is false, $\mathrm{S} 2$ is true.
• C. Both S1 and S2 are true.
• D. Both S1 and S2 are false.

Answer 1

(i) If $\left\{v_{1}, \ldots, v_{n}\right\}$ are linearly INDEPENDENT vectors in $V$, then $\left\{T\left(v_{1}\right), \ldots, T\left(v_{n}\right)\right\}$ are linearly INDEPENDENT in $W$.

False. In general, any non-injective map $T$ will yield a counterexample.

For instance, take $T$ to be the $O$-map. $T(\vec{v})=\overrightarrow{0} \quad \forall \vec{v} \in V$.

(ii) If $\left\{v_{1}, \ldots, v_{n}\right\}$ are linearly DEPENDENT vectors in $V$, then $\left\{T\left(v_{1}\right), \ldots, T\left(v_{n}\right)\right\}$ are linearly DEPENDENT in $W$

True. If $c_{1} \vec{v}_{1}+\cdots+c_{n} \vec{v}_{n}=\overrightarrow{0}$ and not all $c_{i}=0$, then

$$
\begin{aligned}
& T\left(c_{1} \vec{v}_{1}+\cdots+c_{n} \vec{v}_{n}\right)=T(\overrightarrow{0}) \\
\Rightarrow \quad & c_{1} T\left(\vec{v}_{1}\right)+\cdots+c_{n} T\left(\vec{v}_{n}\right)=\overrightarrow{0} \text { by linearity } \\
\Rightarrow & \left\{T\left(\vec{v}_{1}\right), \ldots, T\left(\vec{v}_{n}\right)\right\} \text { are Linearly } \\
& \text { Dependent. }
\end{aligned}
$$

Answer 2

Hint:
$rank(A,B) \leq min\left \{rank(A), rank(B\right \}$

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Statement 1: FALSE

Vectors of V are LI, that is, V is full rank
T(v) = Av

So
rank(AV) <= min{rank(A), rank(V)}
It means that rank of {Tv1, Tv2...Tvn} will also depend on A. So it is not always LI

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Statement 2: TRUE

{a, b, c} are LI iff c1a + c2b + c3c = 0 and c1 = c2 = c3 = 0

If even 1 constant is non-zero, then that vector can be written in the form of linear equation of other vectors, that is,

x = Summation(constant . vector) / constant of x

and the set of vectors will be LD.

Now, vectors of V are LD, that means there is atleast 1 vector which can be written as linear form of other vectors
and rank of V < n and nullity of V >= 1

Rank(AV) cant be more than rank of (V) 
It means Rank(AV) != n (impossible)
So, it is impossible that AV can be LI