Let $B=\{v_1,v_2,\cdots,v_n\}$ be a basis for $V$. $R(T)=\mbox{span}\{T(v_1),T(v_2),\cdots,T(v_n)\}$. If $T$ is linear and goes from $V$ to $W$, what are the minimum requirements so that $T(B)$ is a basis for $W$. If we allow the transformation to be represented by matrix multiplication in the form $Ax=y$, What would such a matrix look like? What would its reduced form look like?

I believe that the correct answers is: The transformation must be injective. A transformation is injective if and only if The columns of the matrix are be linearly independent. The number of columns cannot exceed the number of rows. If we reduce the matrix $A$, every column will be a pivot column.

  • $\begingroup$ Injectivity will only give that $T(B)$ is a basis for $R(T)$, for as you have said, injectivity gives linear independence. However, one also needs a basis to span the space, and for that, surjectivity is needed. $\endgroup$ – QTHalfTau Jul 7 '16 at 3:26
  • $\begingroup$ You would need that the dimension of W is less or equal the dimension of V, so it looks like you would get dimension of W = dimension of V and T -- therefore - invertible. $\endgroup$ – Alex Jul 7 '16 at 3:27

Note that $\DeclareMathOperator{Span}{Span}\Span\{T(v_1),\dotsc,T(v_n)\}=\DeclareMathOperator{image}{image}\image(T)$.

The collection $\{T(v_1),\dotsc,T(v_n)\}$ is linearly independent if and only if $\dim(V)=\dim\image(T)$. But the rank-nullity theorem states that $$ \dim\ker(T)=\dim(V)-\dim\image(T) $$ Thus $\{T(v_1),\dotsc,T(v_n)\}$ is linearly independent if and only if $T$ is injective.

The collection $\{T(v_1),\dotsc,T(v_n)\}$ spans $W$ if and only if $\image(T)=W$. That is, $\{T(v_1),\dotsc,T(v_n)\}$ spans $W$ if and only if $T$ is surjective.

So, if you want $\{T(v_1),\dotsc,T(v_n)\}$ to be a basis for $W$ then $T$ must be both injective and surjective. That is, $T$ must be an isomorphism.

  • $\begingroup$ Wouldn't it be $dim(V) \le dim(im(T))$. Since you can have some creepy basis with $\{v_1, ... , v_m\}, v_i \in \mathbb{R}^n$ with $ m \le n$ and $T \in \mathbb{K}^{n,n}$ $\endgroup$ – Patrick Abraham Jul 7 '16 at 6:37
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    $\begingroup$ @PatrickAbraham The rank-nullity theorem makes $\dim(V)<\dim(im(T))$ impossible. $\endgroup$ – Brian Fitzpatrick Jul 7 '16 at 6:46
  • $\begingroup$ Let $V$ have an basis $B = {e_1, e_2, ... , e_m}$ with $e_i \in \mathbb{K}^n$ also $m < n$. Now let's take $I_n$, hence $I_n * e_i = e_i$ and $rk(I_n) = n = dim(im(I_n))$. Still $V$ "is" only m-dimensional. $\endgroup$ – Patrick Abraham Jul 7 '16 at 7:52
  • $\begingroup$ @PatrickAbraham You are confusing the identity on $\Bbb K^n$ with the identity on $V$. They are two different maps. In your example, the identity $I_V$ on $V$ has rank $m$ not rank $n$. $\endgroup$ – Brian Fitzpatrick Jul 7 '16 at 8:02
  • $\begingroup$ You couldn't multiply $I_m$ with the basis since $e_i \in \mathbb{K}^n$, you might have $$I_v = \begin{bmatrix} I_m \\ 0_{n-m} \end{bmatrix}$$ and I never said that $I_n$ is the identity of $V$, but that there exists a linear-map in $\mathbb{K}^{n,n}$ such that a sub-vectorspace $V$ gets transformed. But the map $I_n$ is actually an left identity of $V$ the right would be $I_m$. $\endgroup$ – Patrick Abraham Jul 7 '16 at 8:11

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