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I have read the proof. I do not get why the entires that are not on the diagonal are equal to zero, or why the following part of the proof is true:

If $k\neq \ell$

$$(A\cdot\hat A)_{k\ell}=\sum_{i=1}^n (-1)^{i+\ell}a_{ki} \det A(\ell\mid i)=0$$

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    $\begingroup$ It's the determinant of the matrix you get when in $A$ you replace the $\ell$-th row with the $k$-th. The resulting matrix has two identical rows, hence the determinant is $0$. $\endgroup$ Commented Aug 20, 2015 at 18:43
  • $\begingroup$ @DanielFischer so there is an action of replacing a row? It do not derive from the formula? $\endgroup$
    – gbox
    Commented Aug 20, 2015 at 18:45
  • $\begingroup$ I'm not sure what the question is. Interpret the formula bearing Laplace expansion in mind. $\endgroup$ Commented Aug 20, 2015 at 18:53
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    $\begingroup$ Laplace expansion by the $\ell$-th row gives you $$\det B = \sum_{i = 1}^n (-1)^{\ell + i} b_{\ell i}\det B(\ell \mid i).$$ Now, if $B$ is the matrix you obtain from $A$ by replacing the $\ell$-th row with the $k$-th, then $B(\ell \mid i) = A(\ell \mid i)$, and also $b_{\ell i} = a_{k i}$. $\endgroup$ Commented Aug 20, 2015 at 19:23
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    $\begingroup$ Hypothetically. As a Gedankenexperiment. We don't actually do that. But that way we see that, and why, we have $A\cdot \operatorname{Adj} A = (\det A)\cdot I$. $\endgroup$ Commented Aug 20, 2015 at 19:32

1 Answer 1

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Consider the $3\times 3$ case $$ \underbrace{\left[\matrix{\color{red}{a_{11}} & \color{red}{a_{12}} & \color{red}{a_{13}}\\ \color{blue}{a_{21}} & \color{blue}{a_{22}} & \color{blue}{a_{23}}\\ a_{31} & a_{32} & a_{33}}\right]}_{A}\cdot \underbrace{\left[\matrix{A_{11} & A_{21} & A_{31}\\ A_{12} & A_{22} & A_{32}\\ A_{13} & A_{23} & A_{33}}\right]}_{\text{adj}(A)}. $$ If you multiply the first row with the first column you get exactly the determinant expansion along the first row $$ \color{red}{a_{11}}A_{11}+\color{red}{a_{12}}A_{12}+\color{red}{a_{13}}A_{13}=\left|\matrix{\color{red}{a_{11}} & \color{red}{a_{12}} & \color{red}{a_{13}}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}}\right|=\det(A). $$ Now if you multiply the second row with the first column you get $$ \color{blue}{a_{21}}A_{11}+\color{blue}{a_{22}}A_{12}+\color{blue}{a_{23}}A_{13} $$ which in the same way can be interpreted as the determinant expansion along the first row, but for what matrix? Well, the cofactors are built of the elements of the second and third rows of $A$, so those rows remain unchanged, and the first row must be made of the blue elements, since it is them who replace the red elements in the formula above, hence $$ \color{blue}{a_{21}}A_{11}+\color{blue}{a_{22}}A_{12}+\color{blue}{a_{23}}A_{13}= \left|\matrix{\color{blue}{a_{21}} & \color{blue}{a_{22}} & \color{blue}{a_{23}}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}}\right|. $$ But this matrix has first two rows being equal, which means that the determinant is zero.

P.S. For other rows and columns the idea is exactly the same. The second row times the second column and the third row times the third column will be the determinant expansion along the second and the third rows, respectively, and give $\det(A)$. All other combinations give zero since there are two equal rows in the corresponding determinant.

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  • $\begingroup$ Thanks a lot, the only point that I do not get is why "first row must be made of the blue elements" $\endgroup$
    – gbox
    Commented Aug 20, 2015 at 19:37
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    $\begingroup$ This is how the expansion works. Compare with the red elements above. The blue elements replace the red ones. $\endgroup$
    – A.Γ.
    Commented Aug 20, 2015 at 19:40
  • $\begingroup$ this is what we get: $$ \color{blue}{a_{21}} \left|\matrix\\ a_{22} & a_{23}\\ a_{32} & a_{33}}\right|+\color{blue}{a_{22}}\left|\matrix{\color{blue}\\ a_{21} & & a_{23}\\ a_{31} & & a_{33}}\right|+\color{blue}{a_{23}}\left|\matrix{\color{blue}\\ a_{21} & & a_{22}\\ a_{31} & & a_{32}}\right|$$ $\endgroup$
    – gbox
    Commented Aug 20, 2015 at 19:58
  • $\begingroup$ Minus in the middle term. $\endgroup$
    – A.Γ.
    Commented Aug 20, 2015 at 20:02
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    $\begingroup$ Right. Now the last step is to see the same principle for the $n\times n$ case. (This is what @DanielFischer was saying in the comments). $\endgroup$
    – A.Γ.
    Commented Aug 20, 2015 at 20:17

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