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Let $A$ and $B$ be two $2 \times 2$ matrices with integer entries. Prove that $\det(A+xB)$ is an integer polynomial of the form $$P(x) = \det(A+xB) = \det(B)x^2+mx+\det(A).$$

I tried expanding the determinant of $\det(A+xB)$ for two arbitrary matrices, but it got computational. Is there another way?

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  • $\begingroup$ Why not the expansion ? It is not so hardworking and gives also the vaue of $m$. $\endgroup$ – Emilio Novati Mar 16 '17 at 17:37
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    $\begingroup$ @Widawensen I also wanted to generalize and not just for $2 \times 2$. $\endgroup$ – user19405892 Mar 16 '17 at 17:38
  • $\begingroup$ @user19405892 Ok. I see. $\endgroup$ – Widawensen Mar 16 '17 at 17:39
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The constant term $\mathrm{det}(A)$ comes from setting $x = 0.$ The coefficient $\mathrm{det}(B)$ is the constant term of $x^2 P(1/x) = \mathrm{det}(xA + B),$ again setting $x = 0.$

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  • $\begingroup$ Can you explain why $x^2P(1/x) = \det(xA+B)$? $\endgroup$ – user19405892 Mar 16 '17 at 17:32
  • $\begingroup$ By homogeneity $\mathrm{det}(xA+B) = x^2 \mathrm{det}(A + B/x)$ $\endgroup$ – user399601 Mar 16 '17 at 17:45
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$$\begin{pmatrix}a_{11}&&a_{12}\\a_{21}&&a_{22}\end{pmatrix}+x\begin{pmatrix}b_{11}&&b_{12}\\b_{21}&&b_{22}\end{pmatrix}=\begin{pmatrix}a_{11}+xb_{11} &&a_{12}+xb_{12}\\a_{21}+xb_{21}&&a_{22}+xb_{22}\end{pmatrix}=C$$ $$\det(C)= a_{11}a_{22}+a_{11}xb_{22}+a_{22}xb_{11}+x^2b_{11}b_{22}- a_{21}a_{12}-a_{21}xb_{12}-a_{12}xb_{21}-x^2b_{21}b_{12} $$ $$=\det(A)+x\left[\det\begin{pmatrix}a_{11}&&a_{12}\\b_{21}&&b_{22}\end{pmatrix}+\det\begin{pmatrix}b_{11}&&b_{12}\\a_{21}&&a_{22}\end{pmatrix}\right]+x^2\det(B)$$

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Another approach. All matrices are complex valued. This argument can be extended to $n\times n$ matrices: see the end of the post. (EDIT sorry, I have found another error, the general case is incomplete).

First step: Assume that $A=I$ and let $\lambda_1, \lambda_2$ denote the eigenvalues of $B$. (They might not be distinct, this does not affect the argument). Then $$\tag{1}\det(I+xB)=(1+x\lambda_1)(1+x\lambda_2)=1 + x\,\mathrm{trace} (B) + x^2 \det B.$$ Here we use the fact that the eigenvalues of $I+xB$ are $1+x\lambda_1, 1+x\lambda_2$ and that the determinant equals the product of the eigenvalues.

Second step: Now assume that $A$ is invertible. Write $$\tag{2}\det(A+xB)=\det(A)\det(I+xA^{-1}B) = \det(A) + x\,\mathrm{trace}(\det(A)A^{-1}B) + x^2\det(B).$$ Here we used the first step together with Binet's theorem (i.e., $\det(MN)=\det(M)\det(N)$).

Final step Now we remove the invertibility assumption on $A$ with a continuity argument. Note that, if $A=\begin{bmatrix} a & b \\ c & d\end{bmatrix}$ is invertible, one has $$\det(A)A^{-1}=\begin{bmatrix} d & -b \\ -c & a\end{bmatrix}.$$ If one rewrites formula (2) as follows $$\tag{3} \det(A+xB)=\det(A) + x\, \mathrm{trace}\left( \begin{bmatrix} d & -b \\ -c & a\end{bmatrix}B\right) + x^2 \det(B), $$ one sees at once that it holds true for all matrices, not only for invertible ones. More precisely, the formula (3) is true for all invertible $A$ and invertible matrices are a dense subset of the space of all matrices, so the formula extends by continuity.

When all matrices are integer valued, the term $$m=\mathrm{trace}\left( \begin{bmatrix} d & -b \\ -c & a\end{bmatrix}B\right)$$ is integer and this terminates the proof.

The $n\times n$ case. In the general case, one obtains in the $A=I$ case the formula \begin{equation}\begin{split}\det(I+xB)&=1+\mathrm{trace}(B)x + \sum_{i<j} \lambda_i\lambda_j x^2+\ldots +\sum_{i_1<\ldots<i_{n-1}} \lambda_{i_1}\ldots\lambda_{i_{n-1}}x^{n-1} + \det(B)x^n \\ &= 1+\mathrm{trace}(B)x+p_2(B)x^2+\ldots + p_{n-1}(B)x^{n-1}+\det B x^n.\end{split}\end{equation}

The coefficients $p_2(B)\ldots p_{n-1}(B)$ are the sums of all principal minors of $B$ of order $2,\ldots n-1$ respectively.

Therefore, using the same argument as before, we obtain in the case of invertible $A$ the formula $$ \det(A+xB)=\det(A) + \mathrm{trace}(\det A A^{-1}B) x+ \det A p_2( A^{-1}B)x^2+\ldots+ \det A p_{n-1}( A^{-1}B)x^{n-1} + \det (B) x^n.$$

To remove the invertibility assumption on $A$, the formula one should use is the following: $$\det(A)A^{-1}=\mathrm{adj}(A), $$ where $\mathrm{adj}(A)$ denotes the adjugate matrix of $A$.

The final result is (warning incomplete) $$\tag{4} \det(A+xB)=\det(A) + \mathrm{trace}(\mathrm{adj}(A)B) x+ \det A p_2(A^{-1}B)x^2+\ldots +\det A p_{n-1}(A^{-1}B)x^{n-1} + \det (B) x^n.$$

TO DO Find formulas for $\det A\cdot p_{2}(A^{-1}B)m \ldots \det A\cdot p_{n-1}(A^{-1}B)$ which do not rely on invertibility of $A$.

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  • $\begingroup$ So the polynomial is always a quadratic for any $n$? Shouldn't it be $n$? $\endgroup$ – user19405892 Mar 16 '17 at 20:27
  • $\begingroup$ @user19405892: I have added the $n\times n$ case, I hope it is correct and that it can be useful to you. $\endgroup$ – Giuseppe Negro Mar 19 '17 at 16:31
  • $\begingroup$ @user19405892: I am sorry, there was an error in the general formula. The coefficients of $x$ and $x^n$ are correct. The others are correct for an invertible $A$, I need to find a way to express them for general $A$. $\endgroup$ – Giuseppe Negro Mar 20 '17 at 11:47
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    $\begingroup$ Your "TO DO" implies finding a general formula for $\det(A+B)$ for two matrices $A$ and $B$. (You have an $x$ there as well, but obviously you can let it get swallowed by the $B$.) I know of only two such formulas. One is what you get by expanding the Leibniz expansion and then combining terms (spelled out in all the painful details in Theorem 5.157 of my Notes on the combinatorial fundamentals of algebra, version of 15 February 2017. It is sometimes useful, but completely impractical in general. ... $\endgroup$ – darij grinberg Mar 23 '17 at 19:50
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