Se
A = ⎡ ⎣ ⎢ ⎢ ⎢ 2 0 0 0 0 6 0 0 0 0 2 0 0 0 0 4 ⎤ ⎦ ⎥ ⎥ ⎥ A = [ 2 0 0 0 0 6 0 0 0 0 2 0 0 0 0 4 ] A = \begin {bmatrix} 2 & 0 & 0 & 0 \ 0 & 6 & 0 & 0 \ 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 4 \end {bmatrix}
Allora:

Question

Se  
 A = ⎡ ⎣ ⎢ ⎢ ⎢ 2 0 0 0 0 6 0 0 0 0 2 0 0 0 0 4 ⎤ ⎦ ⎥ ⎥ ⎥ A = [ 2 0 0 0 0 6 0 0 0 0 2 0 0 0 0 4 ]  A = \begin {bmatrix} 2 & 0 & 0 & 0 \ 0 & 6 & 0 & 0 \ 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 4 \end {bmatrix}    
 Allora:

Answer

K 2 ( A ) = 2 K 2 ( A ) = 2 K_2(A) = 2 .

Answer

K 2 ( A ) = 4 3 K 2 ( A ) = 4 3 K_2(A) = \frac {4}{3} .

Answer

K 2 ( A ) = 3 K 2 ( A ) = 3 K_2(A) = 3 .

Crowdly

Se A = ⎡ ⎣ ⎢ ⎢ ⎢ 2 0 0 0 0 6 0 0 0 0 2 0 0 0 0 4 ⎤ ⎦ ⎥ ⎥ ⎥ A = [ 2 0 0 0 0 6 ...

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