# Check if $\det(I + S) = 1 + \operatorname{trace}(S)$ holds ?

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

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

This statement is false : take $S = I$ for example. It would give us $2^n = 1 + n$.

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).

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.

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 $1+\operatorname{Tr}(S)=2+a$

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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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.

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 $|S+I|=-3 \neq 4= 1+ \mbox {trace} S$.