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I got stuck on part (c) of the following question:

Suppose T : V → W is a rank k linear transformation between finite dimensional vector spaces.

(a) Show that T has k linearly indepenent columns,

(b) Show that T has k linearly independent rows,

(c) Show that T has a k × k submatrix which is invertible

Part (a) and (b) are easy but for part (c) I had the following counter example:

T = \begin{bmatrix}1&0&0\\0&0&1\\0&0&0\end{bmatrix}

where k = 2. It is a valid linear transformation (it satisfies T(a+b) = T(a) + T(b) and T(ca) = cT(a)), it has 2 linearly independent rows and 2 linearly independent columns (satisfying (a) & (b)) but all 2 x 2 submatrices have determinant 0 and hence are not invertible. Is this counter-example valid? If not then why is it invalid?

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  • $\begingroup$ You don’t have to take consecutive columns to form the submatrix. If you take the top 2 rows and the 1st and 3rd columns you get the invertible submatrix. $\endgroup$ – Paul Jun 4 at 10:52
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No, it is not a valid counterexample. Your matrix is$$\begin{bmatrix}\color{red}1&0&\color{red}0\\ \color{red}0&0&\color{red}1\\0&0&0\end{bmatrix},$$whose submatrix $\left[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right]$ has non-zero determinant.

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