# Linear independency before and after Linear Transformation

If we are given some linearly dependent vectors, would the T of those vectors necessarily be dependent (given a transformation from $R^n$ to $R^p$)?

And if we are given some linearly independent vectors, would T of those vectors necessarily be independent (given a transformation from $R^n$ to $R^p$)?

The answers are presumably no and no, but I am struggling to figure out why. Thanks!

-
The only set whose linear independence is preserved under all linear transformations is the empty set. – Jonas Meyer Jan 24 '11 at 2:18

If a set of vectors is dependent, there is a non-trivial combination of some of them that equals 0: $$a_1v_1+\dots+a_n v_n=0,$$ where not all the scalars $a_i$ are $0$. Linearity of $T$ should give you at once that the $Tv_i$ are also linearly dependent (as witnessed by the same $a_i$).

Linear independence, on the other hand, does not need to be preserved. For example, consider the linear transformation that maps all the vectors to 0.

Now, under some additional conditions, a linear transformation may preserve independence. For example, suppose that $T$ is injective (i.e., the only solution to $Tv=0$ is $v=0$). Then $T$ preserves linear independence: Suppose that $Tv_1,\dots,Tv_n$ are dependent, so there are scalars $b_1,\dots,b_n$, not all zero, such that $$b_1Tv_1+\dots+b_nTv_n=0.$$ By linearity, you get $T(b_1v_1+\dots+b_nv_n)=0$ and, if $T$ is injective, then in fact $b_1v_1+\dots+b_nv_n=0$, so the $v_i$ are dependent. (Or, if you prefer: If the $v_i$ are independent, and we have $b_1Tv_1+\dots+b_nTv_n=0$, then as above we get $b_1v_1+\dots+b_nv_n=0$, and independence gives us that $b_1=\dots=b_n=0$, so the $Tv_i$ are independent.)

Injectivity of $T$ is the only way to ensure that any independent set is mapped to an independent set. However, even if $T$ is not injective, there may be some independent sets whose independence is preserved. For example, consider the linear transformation $T:{\mathbb R}^3\to{\mathbb R}^2$ given by $T(x,y,z)=(x,y)$.

Then $T$ maps both $(1,0,0)$ and $(1,0,1)$ to $(1,0)$. The original vectors are independent, the resulting vectors obviously are not (they coincide!). But, on the other hand, $T$ preserves the linear independence of any two (independent) vectors whose last coordinate is 0.

-

You want to be a bit careful with the statements; the main difficulty lies in how you deal with collections of sets that include repetitions.

Most of the time, when we think about vectors and vector spaces, a list of vectors that includes repetitions is considered to be linearly dependent, even though as a set it may technically not be. For example, in $\mathbb{R}^2$, the list $\mathbf{v}_1=(1,0)$, $\mathbf{v}_2=(1,0)$ is considered linearly dependent, because $\mathbf{v}_1 - \mathbf{v}_2 = \mathbf{0}$. But as sets, $\{\mathbf{v}_1,\mathbf{v}_2\} = \{(1,0)\} = S$, you have that $S$ is linearly independent (it contains a single element and the element is nonzero).

If you are considering lists, and repetitions are allowed (and make a list linearly dependent), then Andres's answer is completely correct. If you are considering sets, and repetitions are an issue, then the problem is much more difficult. I suspect this is not the case, since it is almost never the case; but you might want to double check (and try to figure how you can have a linearly dependent set map to a set of vectors that is linearly independent when considered as a set, where repetitions are ignored).

-

Just to add to Arturo Magidin's answer, if you are considering sets of vectors rather than lists, then both statements are false:

Consider $T:\mathbb{R}^2 \to \mathbb{R}, \: T(x,y) = x+y,$ and sets $A = \{(2,0),(0,2),(1,1)\}$ and $B = \{(1,-1)\}$. Then $A$ is linearly dependent, but $T(A) = \{2\}$ is linearly independent; and $B$ is linearly independent, but $T(B) = \{0\}$ is linearly dependent.

-

Suppose $v_1,\cdots v_n$ are linearly independent and T is a linear transformation. Suppose ker(T) only intersects linear span $W$ of $v_1, \cdots v_n$ only at $\{0\}$. Then T preserves the linear independence of $v_1, \cdots v_n$.

This condition is necessary and sufficient.

-