Let $T:V\rightarrow U$ be a linear transformation. Prove/disprove: If $dimKerT=k$ and $ B=(b_1,...,b_n)$ is a basis of $V$, then in the set $(Tb_1,...Tb_n)$ there are exactly $k$ vectors that can be expressed as linear combinations of the previous vectors.
Hey everyone. This statement seems correct. First, we note that $dimImT$ is $n-k$.
If $k=0$, then $KerT=\{0\}$, hence $T$ is an injection- which implies $(Tb_1,...,Tb_n)$ is a linear independent set (for any injective linear transformation $F:V\rightarrow U$, $S$ is a linear independent set iff $F(S)$ is linearly independent), and in particular a basis of $V$- therefore it has exactly $0$ vectors which are linear combinations of the previous vectors, just as $dimKerT$.
Else, $k\gt 0$ and we have $|(Tb_1,...,Tb_n)|\gt dimImT=n-k$ (number of vectors in $T(B)$ is greater than $n-k$), hence is a linear dependent set. I want to prove it has exactly $k$ vectors that can be expressed as linear combinations of the previous vectors, and not at least, but am not sure how. Would love to get some help or hints on that one. Thanks in advance.
By the rank-nullity theorem, $\dim\langle Tb_1,\ldots,Tb_n\rangle=n-k$. And, if you have $n$ vectors $v_1,\ldots,v_n$, asserting that they span a space with dimension $n-k$ is equivalent to the assertion that there is a set of $k$ vectors of $v_1,\ldots,v_n$ which are linear combinations of the remaining vectors and that no set with more than $k$ vectors has that property.