2
$\begingroup$

In the linear algebra book I'm working through, we've already proved two related theorems, namely:

  1. $A: V\mapsto{W}$ is an isomorphism iff $Av_k = w_k$ where $v_1,...,v_n$ and $w_1,...,w_n$ are bases in $V$ and $W$ respectively.
  2. Let $A: V\mapsto{W}$ be a linear transformation. $A$ is invertible iff for any right side $b\in{Y}$ the equation $Ax = b$ has a unique solution $x\in{X}$.

The corollary: an $m$ by $n$ matrix is invertible iff its columns form a basis in $\mathbb F^n$ is then given without proof.

How can we show that the corollary is true from the two theorems above? I'm assuming it should be fairly intuitive since the proof is omitted, but I'm struggling to put the pieces together in a rigorous way.

$\endgroup$
0
$\begingroup$

There are many ways to think about it. Since the given theorems verse about linear transformations while your corollary about matrices, you should first keep in mind that any linear transformation $T$ has an associated matrix $A_T$ such that $[T (x)]_B = A_T [x]_B$ for a given basis $B $, from where it follows that the matrix $A_T $ is invertible iff $T$ is (or equivalently, iff $T $ is an isomorphism.) Also, isomorphisms can only be established between spaces of equal dimensions, so necessarily $m = n$ in your statement.

Consider the second theorem. If the columns of matrix $A $ do not form a basis of $\mathbb {R}^n $, then the columns of $A $:

1) are not linearly independent; and/or

2) do not span all $\mathbb{R}^n$

If 1) is the case, the solutions for $ A x = b $ are not unique; if 2) is the case, there are vectors $b $ for which the equation has no solutions. In both cases, A is not invertible (cause its associated linear transformation is not invertible.)

$\endgroup$
1
$\begingroup$

The main issue is "can every $b$ be written as a linear combination of the columns of $A$." The answer is "yes" by your second theorem. Which linear combination is it? The unique solution $x$.

$\endgroup$
0
$\begingroup$

The set $B = \{e_i\}_{1 \leq i \leq n}$ with $(e_i)_j = \delta_{ij}$ is a basis for $\mathbb{F}^n$. That is, the vector $e_i$ has zeros in every coordinate but the $i$-th one, where it has a one. By $(1)$, $A$ will be an isomorphism iff $Ae_i$ forms a basis of $\mathbb{F}^n$. Note (or prove) that these are precisely the columns of $A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.