Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $B = A - I$?

share|improve this question

4 Answers 4

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

share|improve this answer

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.

share|improve this answer

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)

share|improve this answer
    
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 $\det(A-I)=P_A(1)$.

share|improve this answer
    
Merry Christmas. –  user1551 Dec 12 '12 at 9:43
    
@user1551: and Happy New Year! –  robjohn Dec 12 '12 at 10:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.