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For each invertible $2$ x $2$ matrix $A$, does there exist an invertible $2$ x $2$ matrix $B$ such that the following conditions hold?

(1) $A + B$ is invertible

(2) det($A+B$) = det($A$) + det($B$)

I know that for $2$ x $2$ matrices det($A+B$) = det($A$) + det($B$) + tr($A$)tr($B$) - tr($AB$). So this means tr($A$)tr($B$) = tr($AB$). Right now I am having trouble proving that there exists a $B$ that satisfies this equation as well as condition (1).

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    $\begingroup$ Are you working in $\mathbb{R}$ or $\mathbb{C}$ or are you looking for a solution that does not depend on the field we are considering ? $\endgroup$ – charmd Nov 14 '18 at 7:27
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    $\begingroup$ Note that in $\mathcal{M}_2(\mathbb{Z}/2\mathbb{Z})$, the result is obviously false, since all invertible matrices have determinant $1$ $\endgroup$ – charmd Nov 14 '18 at 7:36
  • $\begingroup$ Last remark: if you are working in $\mathbb{C}$, this is easy : triangularize $A = PTP^{-1}$ (see en.wikipedia.org/wiki/Triangular_matrix#Triangularisability), define $B = P\mbox{Diag}(\alpha_2,-\alpha_1)P^{-1}$ where $(\alpha_1,\alpha_2)$ is the diagonal of $T$ $\endgroup$ – charmd Nov 14 '18 at 7:44
  • $\begingroup$ @CharlesMadeline Thank you. What would I do for $\mathbb{R}$? $\endgroup$ – Ryan Greyling Nov 14 '18 at 7:51
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    $\begingroup$ I have a solution in the real case provided that $\mbox{Tr}(A)\neq -1$. Take $B = \lambda I_2-A$ for a proper $\lambda$ (use the triangularized form to see why it works) $\endgroup$ – charmd Nov 14 '18 at 10:02
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Since we aretalking of $2\times2$ matrices, its slightly easier to write down explicitly.

So $\det(A+B)=\det(A)+\det(B)$ happens when $(a_{11}+b_{11})(a_{22}+b_{22})-(a_{12}+b_{12})(a_{21}+b_{21})=(a_{11}a_{22}-a_{12}a_{21})+(b_{11}b_{22}-b_{12}b_{21})$

$\implies a_{11}b_{22}+b_{11}a_{22}-a_{12}b_{21}-b_{12}a_{21}=0=\det\begin{bmatrix}a_{11}\,a_{12}\\b_{21}\,b_{22}\end{bmatrix}+\det\begin{bmatrix}b_{11}\,b_{12}\\a_{21}\,a_{22}\end{bmatrix}$

Choosing $b_{11} = -a_{21}, b_{12} = -a_{22}, b_{21} = a_{11}$ and $b_{22} = a_{12}$ we get our matrix B.

The key point here is to choose $b_{ij}$ such that the two determinants are zero. Part 1 follows by considering matrix B such that $\det(A+B)\neq0$

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Example for $\det(A+B)\neq0$ consider the matrix $A=\begin{bmatrix}4\: 5 \\7\: 9\end{bmatrix}$. We can choose matrix B as $\begin{bmatrix}-7 -9 \\4 \:\: 5\end{bmatrix}$.

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  • $\begingroup$ This answer cannot be correct because it works in $\mathcal{M}_2(\mathbb{Z}/2\mathbb{Z})$, which is impossible. More precisely, with your definitions, $A+B = \begin{pmatrix} a_{11}+a_{21} & a_{12}+a_{22}\\ a_{21}+a_{11} & a_{22}+a_{12}\end{pmatrix}$. So it is imposible to "choose $B$ such that $\mbox{det}(A+B)\neq 0$" $\endgroup$ – charmd Nov 14 '18 at 9:53
  • $\begingroup$ Choose $b_{11} = -a_{21}, b_{12} =- a_{22}, b_{21} = a_{11}$ and $b_{22} = a_{12}$. $\endgroup$ – Yadati Kiran Nov 14 '18 at 9:56
  • $\begingroup$ Determinant will still be zero since the rows (or columns) are identical upto scalar multiplication. The sum of determinants will still be equal to determinant of sum as shown in example. $\endgroup$ – Yadati Kiran Nov 14 '18 at 11:11
  • $\begingroup$ You can make the edit @Charles Madeline. But just would like to know what was your argument regarding $\mathcal{M}_2(\mathbb{Z}/2\mathbb{Z})$. $\endgroup$ – Yadati Kiran Nov 14 '18 at 11:35

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