Can we say that $\det(A+B) = \det(A) + \det(B) +\operatorname{tr}(A) \operatorname{tr}(B) - \operatorname{tr}(AB)$?

Question

Let $A,B \in M_n$. Is this formula true?

$$\det(A+B) = \det(A) + \det(B) + \operatorname{tr}(A) \operatorname{tr}(B) - \operatorname{tr}(AB).$$

Answer 1

In the cases of $2 \times 2$ matrices it always holds. This appears to be a calculation, for transparency introduce a variable $\lambda$

$$\text{det}(\lambda A + B)-\lambda^2 \text{det}(A) - \text{det}(B) - \lambda \text{Tr}(A)\text{Tr}(B) + \lambda \text{Tr}(AB)$$

is a polynomial. To get the quadratic part of $\text{det}(\lambda A + B)$, write

$$A=\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22} \end{pmatrix}\qquad B=\begin{pmatrix}b_{11}&b_{12}\\b_{21}&b_{22} \end{pmatrix}$$

then

$$\text{det}(\lambda A+B)=(\lambda a_{11}+b_{11})(\lambda a_{22}+b_{22})-(\lambda a_{21}+b_{21})(\lambda a_{12}+b_{12})$$

After distributing the products, the quadratic part is obviously $\text{det}(A)$. The part without any $\lambda$ is the determinant of $B$.

The linear terms that are left are just

$$\lambda (a_{11}b_{22}+a_{22}b_{11}-a_{21}b_{12}-a_{12}b_{21}) $$

We almost see $\lambda \text{Tr}(A)\text{Tr}(B)=\lambda(a_{11}+a_{22})(b_{11}+b_{22})$ here, so after subtracting that what is left is

$$-\lambda(a_{11}b_{11}+a_{21}b_{12})-\lambda(a_{22}b_{22}+a_{12}b_{21})$$

But this expression is just $-\lambda\text{Tr}(AB)$

I wonder if there is an analogous formula for dimension 3, or an extension to arbitrary dimension.

Answer 2

$$\det(A+B) = \det(A*(1+A^{-1}*B)) = \det(A)\det(1+A^{-1}B).$$

Now applying your formula to $\det(1+A^{-1}B)$ leads to:

$$\det(A+B)=\det(A)(\det(1)+\det(A^{-1}B)+\mathrm{tr}(1)\mathrm{tr}(A^{-1}B)-\mathrm{tr}(A^{-1}B)) = \det(A)+\det(B)+(\dim(V)-1)\det(A)\mathrm{tr}(A^{-1}B),$$

with vector space dimension $dim(V)$.

But it is false that:

$$\det(A)(\dim(V)-1)\mathrm{tr}(A^{-1}B)=\mathrm{tr}(A)\mathrm{tr}(B)-\mathrm{tr}(AB).$$

Simply take $A=B=1$ and then you will see contradiction.

Answer 3

If $A$ is a diagonal matrix with entries $a_1,..,a_n$ and $B$ is a diagonal matrix with entries $b_1,..,b_n$ your formula becomes $$(a_1+b_1)...(a_n+b_n)=a_1...a_n +b_1...b_n-(a_1+...+a_n)(b_1+..+b_n)-a_1b_1-a_2b_2-..-a_nb_n$$

It is easy to come up with counterexamples.

Answer 4

Here's an attempt to make the $2\times 2$ case seem just slightly horrible, while practicing to master the align* environment.

Let $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ and $B = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}$.

Then the left-hand side expands to eight terms:

\begin{align*} \det (A+B) ={}& (a_{11}+b_{11})(a_{22}+b_{22}) - (a_{12}+b_{12})(a_{21}+b_{21}) \\ ={}& \hphantom{(}a_{11}a_{22} - a_{12}a_{21} \\ &+ b_{11}b_{22} - b_{12}b_{21} \\ &+ a_{11}b_{22} - a_{21}b_{12} \\ &+ a_{22}b_{11} - a_{12}b_{21} \end{align*}

The first two lines of the latter expression are respectively $\det A = a_{11}a_{22} - a_{12}a_{21}$ and $\det B=b_{11}b_{22} - b_{12}b_{21}$.

That leaves only the expansion and cancellation for the trace expressions:

\begin{align*} (\operatorname{tr} A)(\operatorname{tr} B) - \operatorname{tr}(AB) ={}& (a_{11}+a_{22})(b_{11}+b_{22}) \\ &- (a_{11}b_{11} + a_{12}+b_{21} + a_{21}+b_{12} + a_{22}+b_{22}) \\ ={}& a_{11}b_{22} - a_{21}b_{12} \\ &+ a_{22}b_{11} - a_{12}b_{21} \end{align*}

Thus in the $2\times 2$ cases the identity is valid:

$$ \det A+B = \det A + \det B + (\operatorname{tr} A)(\operatorname{tr} B) - \operatorname{tr}(AB) $$

Answer 5

• $n=2$, $\newcommand{\tr}{\operatorname{tr}}$

  $$\det(A+B) = \det(A) + \det(B) + \tr(A) \tr(B) - \tr(AB).$$

• $n=3$, letting $c(X) = (\tr(X)^2 – \tr(X^2)) / 2$,

  \begin{align*} \det(A + B) ={}& \det(A) + \det(B) – \tr(AB)\tr(A) – \tr(AB)\tr(B) +{} \\ &{}+ c(A)\tr(B) + \tr(A)c(B) + \tr(AAB) + \tr(ABB) \end{align*}

• $n>3$, a formula with $2^n$ terms should be obtainable from the 1987 Reutenauer and Schützenberger's "A formula for the determinant of a sum of matrices"

Answer 6

For a $2\times 2$ matrix $A$ with eigenvalues $\lambda_1$, $\lambda_2$ we have $$\det A = \lambda_1 \lambda_2 \\ \operatorname{tr}A = \lambda_1 + \lambda_2 \\ \operatorname{tr} A^2= \lambda_1^2 + \lambda_2^2$$ Therefore we have $$\det(A) = \frac{1}{2}\left( (\operatorname{tr} A)^2 - \operatorname{tr}A^2\right)$$

Now, apply this for $A+B$ and get

$$\det(A+B) = \frac{1}{2}\left(\, (\,\operatorname{tr}(A+B)\,)^2 - \operatorname{tr}\,(A+B)^2\,\right)$$

Now expand on RHS and get $$\det(A+B) =\det(A) + \det(B) + \operatorname{tr}A \operatorname{tr} B - \operatorname{tr}(A B)$$

There is a simple mnemonic to remember this formula. $\det (A)$ is not additive and $\operatorname{tr} $ is not multiplicative. So the error terms $$\det(A+B)-( \det A + \det B) \\ \operatorname{tr}(A B) - \operatorname{tr}A \cdot \operatorname{tr} B$$ are not $0$, but their sum is.

In general, for $A$ an $n\times n$ matrix we can express $\det A$ in terms of $\operatorname{tr} A^k$, $1\le k \le n$. This is because we can express $\lambda_1 \cdots \lambda_n$ in terms of the power sums $\sum \lambda_i^k$. In fact: for every square matrix $A$ and $t$ small ( a parameter) we have

$$I + t A = \exp (\log (I + t A) ) = \exp (\sum_{k\ge 1} \frac{(-1)^{k-1}t^kA^k}{k})$$ Now take the $\det$ of both sides and use the fact that $\det (\exp (M)) = \exp \operatorname{tr}(M)$. We get $$\det(I + t A) = \exp\left(\sum_{k\ge 1} \frac{(-1)^{k-1} \operatorname{tr}(A^k) t^k}{k} \right)$$

Say $A$ is $n\times n$. We look at the coefficient of $t^n$, which from LHS is $\det A$. On RHS we can ignore the factors with $k> n$. Therefore

$$\det A= [t^n] \exp\left(\sum_{k=1 1}^n \frac{(-1)^{k-1} \operatorname{tr}(A^k) t^k}{k} \right)$$

Let's apply this for $n=3$. We get on RHS, for $\operatorname{tr}(A^k) = p_k$,

$$[t^3]\exp\left( p_1 t - \frac{p_2 t^2}{2} + \frac{p_3 t^3}{3}\right)=\frac{1}{6}( p_1^3 - 3 p_1 p_2 + 2 p_3)$$ and so

$$\det A = \frac{1}{6}(\, (\operatorname{tr} A)^3 - 3 \operatorname{tr} A \cdot \operatorname{tr} (A^2) + 2 \operatorname{tr} (A^3)\,)$$

We can try to get the formula in @rych:'s answer using the above.

Answer 7

As long as we're necrobumping this thread, here's another way to derive the identity in the case where $A$ and $B$ are $2\times 2$ matrices:

It can be shown that $$\det(A+B)=\sum_{p+q=2}\mathop{\mathrm{tr}}(\textstyle\bigwedge^p A\mathop{\square}\bigwedge^q B)=\mathop{\mathrm{tr}}(\bigwedge^2A)+\mathop{\mathrm{tr}}(A\mathop{\square}B)+\mathop{\mathrm{tr}}(\bigwedge^2 B)$$ where $\square$ is the box product (generalizaton of the exterior power to more than one matrix). We know that $\mathop{\mathrm{tr}}(\bigwedge^2A)=\det(A)$ and $\mathop{\mathrm{tr}}(\bigwedge^2 B)=\det(B)$. By definition of the box product, $$\begin{align*} \mathop{\mathrm{tr}}(A\mathop{\square} B)&=\sum_{\sigma,\tau\in S_2}(-1)^{\sigma\tau}A_{\sigma(1)\tau(1)}B_{\sigma(2)\tau(2)}\\ &=A_{11}B_{22}-A_{12}B_{21}-A_{21}B_{12}+A_{22}B_{11}\\ &=\mathop{\mathrm{tr}}(A)\mathop{\mathrm{tr}}(B)-\mathop{\mathrm{tr}}(AB) \end{align*}$$ so the result follows.