# Rank of a matrix

1. Dec 2, 2014

### hellokitten

• Originally posted in a technical math section, so missing the template
Let S = [ x x x x ; a x x x ; 0 b x x ; 0 0 c x]
Find the rank of S-K*identity.

Attempt I basically did straight forward row reduction and got

[ (x-k) x x x ; 0 ((x-k)^2-ax)/(x-k) x-ax/(1-k) a- ax/(1-k); 0 0 (x-k)-b[x(x-k)-ax]/ [(x-k)^2-ax] (x-k)-b[x(x-k)-ax]/[(x-k)^2-ax]; 0 0 0 [(x-k)^2-ax(x-k)-c]/[(x-k)^2 -ax] ]

It is given that the rank is 3. I am not sure how to prove this. Shouldn't I have gotten a zero row when row reducing?

Are there any alternatives to find the rank of a matrix like finding the upper bound of the rank is 3 and the lower bound for the rank is 3?

Last edited: Dec 2, 2014
2. Dec 2, 2014

### HallsofIvy

Staff Emeritus
I don't understand your notation. I assume that "a", "b", and "c" are numbers but what about "x" is it simply another number?

3. Dec 2, 2014

### hellokitten

Oh Sorry! X can be any number. a,b,c are non zero numbers.

4. Dec 2, 2014

### Ray Vickson

The determinant of $S$ has the form
$$\det(S) = x^4 + f_3 x^3 + f_2 x^2 + f_1 x + k^4,$$
where the $f_i$ are polynomials in $a,b,c,k$. For any given value of $k,a,b,c$ the determinant cannot be identically zero for all $x$, but it may equal zero for at most 4 real values of $x$ (and maybe for no real values of $x$ at all). What does this tell you about the rank in general?

I am not really sure how to interpret your question, where you say "it is given that the rank is 3". Does that mean that the person posing the question thinks the rank is always 3, or does it mean you are told the rank is 3 and are then required to say something about $a,b,c,k,x$?

5. Dec 2, 2014

### hellokitten

K,a,b,c is not identically zero then det(S) =/= 0 => S is full rank.

The question stated to verify that the rank of S-I*lamda = 3. This is needed to prove the geometric multiplicity. of lamda =1.

6. Dec 2, 2014

### Ray Vickson

OK: the question is mis-stated, but that can be fixed. A proper statement is that "If k is an eigenvalue of S, then S - kI has rank 3". (Of course, if k is an eigenvalue of S then the rank of S-kI is < 4, but you want it to = 3 exactly). I have not checked your row-reduction in your first post, but I think what you need to do is to show that there can be at most one zero row of the reduced matrix; that is, if one row = 0 the other three must be non-zero.

7. Dec 2, 2014

### hellokitten

Why does having k as an eigenvalue of S automatically mean that the rank of S-KI < 4?

8. Dec 3, 2014

### hellokitten

Nevermind. I got it! Thanks!