Rank of S-K*identity: What is the rank of S-K*identity matrix?

In summary, to prove that the rank of S-K*identity is 3, we can show that there can be at most one zero row in the reduced matrix of S-K*identity, and that if one row is zero, the other three must be non-zero. This is because if k is an eigenvalue of S, then the rank of S-K*identity is < 4. Therefore, the rank of S-K*identity is 3.
  • #1
hellokitten
12
1
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?
 
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  • #2
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
Oh Sorry! X can be any number. a,b,c are non zero numbers.
 
  • #4
hellokitten said:
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?

The determinant of ##S## has the form
[tex] \det(S) = x^4 + f_3 x^3 + f_2 x^2 + f_1 x + k^4, [/tex]
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
Ray Vickson said:
The determinant of ##S## has the form
[tex] \det(S) = x^4 + f_3 x^3 + f_2 x^2 + f_1 x + k^4, [/tex]
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##?

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
hellokitten said:
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.

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
Ray Vickson said:
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.

Why does having k as an eigenvalue of S automatically mean that the rank of S-KI < 4?
 
  • #8
Nevermind. I got it! Thanks!
 

1. What is the rank of S-K*Identity Matrix?

The rank of S-K*Identity Matrix is equal to the number of linearly independent rows or columns in the matrix. This can range from 0 to the number of rows/columns in the matrix.

2. How do you calculate the rank of a matrix?

To calculate the rank of a matrix, you can use several methods such as Gaussian elimination, finding the determinant, or using row reduction. The method used will depend on the size and complexity of the matrix.

3. What does it mean if the rank of a matrix is 0?

If the rank of a matrix is 0, it means that all the rows and columns in the matrix are linearly dependent. In other words, the matrix can be reduced to a matrix with all zeros. This also means that the matrix has no independent rows or columns.

4. Can the rank of a matrix be greater than its number of rows or columns?

No, the rank of a matrix cannot be greater than its number of rows or columns. The maximum possible rank of a matrix is equal to the smaller of the two values.

5. How does the identity matrix affect the rank of a matrix?

The identity matrix does not affect the rank of a matrix. Multiplying a matrix by the identity matrix does not change the number of linearly independent rows or columns in the original matrix.

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