# how does addition of identity matrix to a square matrix changes determinant?

Suppose there is $n \times n$ matrix $A$. If we form matrix $B = A+I$ where $I$ is $n \times n$ identity matrix, how does $|B|$ - determinant of $B$ - change compared to $|A|$? And what about the case where $B = A - I$?

In case you are interested, there is a result which expresses $\det(A+B)$ in terms of $\det(A)$ and $\det (B)$, it is given by following $$\det (A+B)=\det(A)+\det(B)+\sum_{i=1}^{n-1}\Gamma_n^i\det(A/B^i)$$ Where $\Gamma_n^i\det(A/B^i)$ is defined as a sum of the combination of determinants, in which the $i$ rows of $A$ are substituted by the corresponding rows of $B$. You can find the proof in this IEEE article: http://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=262036&userType=inst

• Note: This formula is just a consequence of applying the multilinearity of the determinant once to each row (consecutively, so you end up with a sum of $2^n$ addends). – darij grinberg Mar 12 '18 at 8:20
• Do you have an answer to this?: math.stackexchange.com/questions/3184171/… – Susan_Math123 Apr 11 at 19:53

There is no simple answer. $P(\lambda) = \det(A-\lambda I)$ is the characteristic polynomial of $A$. This is a polynomial of degree $n$: the coefficient of $\lambda^n$ is $(-1)^n$ and the coefficient of $\lambda^0$ is $\det(A)$. The coefficients of other powers of $\lambda$ are various functions of the entries of $A$. $\det(A+I)$ and $\det(A-I)$ are $P(-1)$ and $P(1)$; that's about all there is to say about them.

As others already have pointed out, there is no simple relation. Here is one answer more for the intuition. Consider the (restricting) codition, that $A_{n \times n}$ is diagonalizable, then $$\det(A) = \lambda_0 \cdot \lambda_1 \cdot \lambda_2 \cdot \cdots \lambda _{n-1}$$ Now consider you add the identity matrix. The determinant changes to $$\det(B) = (\lambda_0+1) \cdot (\lambda_1+1) \cdot (\lambda_2+1) \cdot \cdots (\lambda _{n-1} +1)$$ I think it is obvious how irregular the result depends on the given eigenvalues of A. If some $\lambda_k=0$ then $\det(A)=0$ but that zero-factor changes to $(\lambda_k+1)$ and det(B) need not be zero. Other way round - if some factor $\lambda_k=-1$ then the addition by I makes that factor $\lambda_k+1=0$ and the determinant $\det(B)$ becomes zero. If some $0 \gt \lambda_k \gt -1$ then the determinant may change its sign...
So there is no hope to make one single statement about the behave of B related to A - except that where @pritam linked to, or except you would accept a statement like $$\det(A)=e_n(\Lambda_n) \to \det(B)= \sum_{j=0}^n e_j(\Lambda)$$ where $\Lambda = \{\lambda_k\}_{k=0..n-1}$ and $e_k(\Lambda)$ denotes k'th elementary symmetric polynomial over $\Lambda$... (And this is only for diagonalizable matrices)

• You don't need diagonalizable in your arguments. You can always work with the Jordan form, or with the Schur decomposition, and reason exactly like you did. – Martin Argerami Dec 12 '12 at 13:07

The characteristic polynomial $P_A(\lambda)$ of a matrix $A$ is defined as $$P_A(\lambda)=\det(A-I\lambda)$$ Therefore, $\det(A)=P_A(0)$, while $\det(A+I)=P_A(-1)$ and $\det(A-I)=P_A(1)$.

• Merry Christmas. – user1551 Dec 12 '12 at 9:43
• @user1551: and Happy New Year! – robjohn Dec 12 '12 at 10:02

I would like to contribute the following special case which I encountered when I was looking for an answer and stumbled upon this thread.

If $$A=uv^\intercal$$, then $$\det(I+A) = 1+u^Tv$$

• P.S Someone please help with formatting this to correct transposes. Didn't manage. – mexmex Dec 4 '18 at 9:54

For I and A $N\times N$ matrices we have:

$$det(I-A)=1-\sum_{j}^{N}A_{jj}+\sum_{n\geq 2}^{N}(-1)^{n}\sum_{1\leq j_{1}<...<j_{n}\leq N}det((A_{j_{a},j_{b}})^{n}).$$