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Linear Algebra - Sebastian Pancratz

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A A <br />

det A = 0<br />

A A <br />

det A det A = 0<br />

A 0 A<br />

A 0 <br />

det A = 0 <br />

A, B ∈ Mn(F) det(AB) = det(A) det(B)<br />

A dA : (B (1) , . . . , B (n) ) ↦→ det(AB) B = (B (1) , . . . , B (n) ) <br />

F n det(AB) = dA(AB (1) , . . . , AB (n) ) dA <br />

<br />

<br />

det(AB) = dA(B (1) , . . . , B (n) )<br />

= det BdA(e1, . . . , en)<br />

= det(B) det(A),<br />

<br />

⎛<br />

det(AB) = det ⎝ <br />

bj11A (j1)<br />

<br />

, . . . , bjnnA (jn)<br />

⎞<br />

⎠<br />

σ∈Sn<br />

j1<br />

= <br />

⎛ ⎞<br />

n<br />

⎝ b ⎠<br />

σ(j)j det(A σ(1) , . . . , A σ(n) )<br />

σ∈Sn<br />

j=1<br />

jn<br />

= <br />

⎛ ⎞<br />

n<br />

⎝ b ⎠<br />

σ(j)j ε(σ) det A<br />

j=1<br />

= det(A) det(B).<br />

B AB det B = 0 = det(AB) B <br />

B = E1 · · · Ek <br />

<br />

det(AB) = det(AE1 · · · Ek)<br />

= det A det E1 · · · det Ek<br />

= det A det B.<br />

A det A −1 = (det A) −1 <br />

A AA −1 = I (det A)(det A −1 ) = det I = 1 <br />

det A −1 = (det A) −1 <br />

n × n

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