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.

Consider $A$ is some singular $N \times N$ real matrix. I learned that range space of $A$ (denote as $R(A)$) shrinks as one takes its powers (I learned a proof based on vector spaces for that also). So $R(A) \supseteq R(A^2) \supseteq R(A^3)\dots $ and so on. This should imply that $rank(A) \geq rank(A^2) \geq rank(A^3)\dots$. since the dimension is decreasing (non-increasing). I have also learned that rank of a matrix is also same as the number of non-zero eigen values. But then, eigen-values of powers of $A$ are the powers of eigen-values of $A$. So doesn't it imply that $rank(A)=rank(A^2)$. If it is so, then how can range space shrink, for after all, rank is the dimension of range space. I believe I am missing some thing in this chain of arguments I made.

share|improve this question
    
You need to be careful when you say the rank of a matrix is "the number of nonzero eigenvalues". What would you say the rank of the $2 \times 2$ identity matrix is? –  wj32 Oct 22 '12 at 5:21
    
@wj32 Rank of $2 \times 2$ identity matrix is 2 and its non-zero eigen values are also 2!!. I would be really happy if you can give a formal proof to that statement. –  dineshdileep Oct 22 '12 at 5:26
    
@GerryMyerson Corrected!! –  dineshdileep Oct 22 '12 at 5:27
2  
@dineshdileep: Sorry, the example I wanted to use was $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. –  wj32 Oct 22 '12 at 5:29
    
Oh,yes!! thanks a lot. That clarifies everything. But rank=non-zero eigen values, does it hold for symmetric(hermitian) matrices? Can you post it as an answer. –  dineshdileep Oct 22 '12 at 5:35

1 Answer 1

up vote 1 down vote accepted

It is not quite true that the rank is the number of non-zero eigenvalues. If the matrix is diagonalizable then the statement holds. In general, the multiplicity of zero as an eigenvalue will always be greater than or equal to the nullity of the matrix (as a consequence of algebraic multiplicity always being larger than geometric multiplicity).

As for when the matrix loses rank as you take powers of it, it is more informative to examine the Jordan blocks of the matrix. If your matrix has a Jordan block of size $k$ corresponding to zero, then it is easy to see that the matrix will lose a rank for each power up to the $k$th power. How many ranks it loses "per power" will be dependent on the number and the sizes of the null Jordan blocks. Notice also, that the example that wj32 gave is precisely a Jordan block of size $2$.

share|improve this answer
    
Does it mean that for diagonalizable matrices, the range do not shrink? –  dineshdileep Oct 22 '12 at 6:05
    
The range does not shrink for diagonalizable matrices. In fact, the range will shrink if and only if the matrix has a non-trivial zero Jordan block. –  EuYu Oct 22 '12 at 6:16

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.