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I saw the following statement in my homework and we are asked to make use of the statement:

If $S$ is a symmetric matrix then

$$\det(I + S ) = 1 + \operatorname{trace}(S).$$

However, I am not sure if this really holds. Please give me a hand here please

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    $\begingroup$ ...was it a true/false question? $\endgroup$ – anon Mar 26 '13 at 14:21
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    $\begingroup$ YES. Because the teacher says "make use of this statement for blablabla" but I dont think the statement even holds. $\endgroup$ – user1769197 Mar 26 '13 at 14:24
  • $\begingroup$ trace of S + n? where $n$ is the dimension of the matrix? $\endgroup$ – Lost1 Mar 26 '13 at 14:25
  • $\begingroup$ A related exercise on Hoffman-Kunze is $\det (I+S)=1+\det (S)$ iff $\mathrm{trace}(S)=0$ for a $2\times 2$ matrix $S$. $\endgroup$ – StubbornAtom Jan 6 '17 at 15:15
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This statement is false : take $S = I$ for example. It would give us $2^n = 1 + n$.

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What is true is that the expansion of the characteristic polynomials is given by traces of powers of the matrix $A$; explicitly, the characteristic polynomial $\chi_A(T)=\det(T\cdot{\rm Id}-A)$ is given by

$$T^n-{\rm tr}(A)T^{n-1}+\frac{{\rm tr}(A)^2-{\rm tr}(A^2)}{2}T^{n-2}-\frac{{\rm tr}(A)^3-3{\rm tr}(A){\rm tr}(A^2)+2{\rm tr}(A^3)}{6}T^{n-3}+\cdots $$

with the final term being $(-1)^n\det A$. This does not depend on $A$ being symmetric or anything else and so $\det(1+A)=(-1)^n\chi_A(-1)\approx (-1)^n[(-1)^n-{\rm tr}(A)(-1)^{n-1}]=1+{\rm tr}(A)$ is in some sense a valid first-order approximation (we set $T=-1$ and truncated our expansion at just two terms).

This can also be seen more simply as in Alex's answer: if $\{\lambda_i\}$ are the eigenvalues of $A$ with multiplicity, then $I+A$ has eigenvalues $\{\lambda_i+1\}$ with multiplicity, and $$\det(1+A)=\prod_{i=1}^n(1+\lambda_i)=1+\left(\sum_{i=1}^n\lambda_i\right)+{\rm higher~order~terms}\approx 1+{\rm tr}(A).$$

I have been assume $A$ is an $n\times n$ matrix throughout, but the underlying field is arbitrary.

The $\chi_A(T)$ expansion is possible thanks to the fundamental theorem of symmetric polynomials, and in particular may be recursively computed by hand using the NG formulas (see this section).

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This is false. By the (finite-dimensional) spectral theorem, every symmetric matrix is diagonalizable, so we can write $S=QDQ^{-1}$. Thus we have $$\det(I+S)=\det(Q^{-1}(I+S)Q)=\det(I+D)=\prod\limits_{i=1}^n (1+d_i)\\\ne 1+\sum\limits_{i=1}^n d_i=1+\mathrm{trace}(D)=1+\mathrm{trace}(S).$$ Perhaps the formula $\prod\limits_{i=1}^n (1+d_i)$ may be of use, however.

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If the matrix is also real, it suffice to check it over diagonal matrices. But it doesn't hold in dimension $\geqslant 2$, because if $S:=\pmatrix{1&0\\0& a}$, $\det(I+S)=2(a+1)$ and $1+\operatorname{Tr}(S)=2+a$

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Holds if $S$ is of rank $1$. A simple case for that is when $S=a\cdot b^T$ where $a$ and $b$ are column vectors.

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  • $\begingroup$ Welcome to math.SE: for some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – A.P. May 6 '13 at 20:04
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It is not true. Use diagonal matrices and the fact that the determinant of a diagonal matrix is the product of the elements while the trace is their sum. Only few of these matrices have this property.

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This statement does not hold. Take $S=\begin{pmatrix} 1 &3 \\ 3 &2 \end{pmatrix}$ .Then $S+I=\begin{pmatrix} 2 &3 \\ 3 &3 \end{pmatrix}$.So $|S+I|=-3 \neq 4= 1+ \mbox {trace} S$.

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