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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!

- 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!

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JasonRox

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

- #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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