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My lecturer showed me something like a mxn matrix postmultiplied with a x-vector to get R^n, thus n = no. of cols. Makes sense when he was explaining but when i stepped out i realised that i didnt get it. Any help pls? Thanks!

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- Thread starter nyxynyx
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- #1

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My lecturer showed me something like a mxn matrix postmultiplied with a x-vector to get R^n, thus n = no. of cols. Makes sense when he was explaining but when i stepped out i realised that i didnt get it. Any help pls? Thanks!

- #2

JasonRox

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It's pretty quite intuitive after you get further into Linear Algebra and have become more comfortable with Nullspaces and such. I'll find a link with a proof.

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JasonRox

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I can't one what doesn't use linear transformations!

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Hurkyl

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e.g. you may have done an exercise solving Ax=b, where A is 3x3, and got a two-dimensional space of solutions. The rank-nullity theorem says that the rank of A is one -- so confirm that by computing the rank of A!

- #6

JasonRox

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foundationof your intuition for linear algebra. If you're looking forunderstanding, your best bet is probably to review previous exercises where you actually solved systems of equations and apply the rank-nullity theorem to describe the system.

e.g. you may have done an exercise solving Ax=b, where A is 3x3, and got a two-dimensional space of solutions. The rank-nullity theorem says that the rank of A is one -- so confirm that by computing the rank of A!

Exactly. It comes around. If it hasn't yet, continue solving systems.

- #7

radou

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Once you learn something about linear operators and their matrix representation, it should become formally clear.

Edit: actually, you can investigate this fact by going into Gaussian elimination.

Edit: actually, you can investigate this fact by going into Gaussian elimination.

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