- #1

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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

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- Thread starter jamesb1
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- #1

- 22

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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

- #2

HallsofIvy

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[tex]\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}[/tex]

and similarly for [itex]\begin{pmatrix}0 \\ 1 \\ 0 \end{pmatrix}[/itex] and [itex]\begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}[/itex]

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 R

Generalize that to R

- #3

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

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HallsofIvy

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