Invertible matrix implies linear independent columns

[jamesb1]

Is the title statement true?

Was doing some studying today and this caught my eye, haven't looked into linear algebra in quite a while so I'm not sure how it is true :/

Internet couldn't provide any decisive conclusions neither

Many thanks

 

[HallsofIvy]

As a very simple example, note that
$$\begin{pmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}= \begin{pmatrix}a_{11} \\ a_{21} \\ a_{31}\end{pmatrix}$$
and similarly for $\begin{pmatrix}0 \\ 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}$

That is, applying the linear transformation to the standard basis vectors gives the three columns. The linear transformation is invertible if and only if it maps R³ to all of R³. That is true if and only if those three vectors, the three columns, are a basis for R³ which is, in turn, true if and only if the three vectors are independent.

Generalize that to Rⁿ.

 

[jamesb1]

But that means you CAN'T have linearly dependent and invertible linear transformations .. no?

 

[HallsofIvy]

Yes, If those n vectors, the columns of the n by n matrix, are linearly dependent, they span only a subset of Rⁿ and so the linear transformation is NOT invertible.