I would say the book's answer is correct. You have three degrees of freedom (three arbitrary parameters), and so you really have a 3-dimensional space. You could add your two matrices that are multiplied by b in order to obtain the book's matrix.
We know by definition that If a vector space has a basis consisting of vectors, then the number is called dimension of denoted by
Let be the subspace of all symmetric matrices in . What is the dimension of ?
I wanted to find the basis. So I did it like this:
A symmetric matrix is:
so
So the set
spans Moreover, can be shown to be linearly independent, and conclude that dimension of is
The book did it like this.
A symmetric matrix is:
so
So the set
spans Moreover, can be shown to be linearly independent, and conclude that dimension of is
Are these both correct? or am I wrong? Why?
with a symmetric matrix, you have two kinds of entries: diagonal entries, and off-diagonal entries.
by the definition of a symmetric matrix, the i,j-th entry is equal to the j,i-th entry. for diagonal entries, these are the same,
so the elementary matrices Eii can serve as the start of a basis for the symmetric matrices.
if we have a 1 in the i,j-th position, symmetry forces us to also have a one in the j,i-th position, so although Eij (i ≠ j) is not
a symmetric matrix, Eij + Eji is. in general, for nxn matrices, a basis for the symmetric matrices of size n,
is {Eii: 1 ≤ i ≤ n} U {Eij+Eji : 1 ≤ i,j ≤ n, i ≠ j}.
there are n matrices in the first set, and n(n-1)/2 matrices in the second set, giving a total dimension of:
n + n(n-1)/2 = (2n + n^2 - n)/2 = n(n+1)/2, for any n.
when n = 2, this is 2(3)/2 = 3, as you have seen above.
alternatively, once can write any matrix A as (1/2)(A+Aᵀ) + (1/2)(A-Aᵀ) (a symmetric part, and an anti-symmetric part).
if A is symmetric to begin with, A = (1/2)(A+Aᵀ).
writing A = a11E11 + a12E12 +.......+ annEnn = (1/2)[(a11E11 + a12E12 +.......+ annEnn) + (a11E11 + a12E12 +.......+ annEnn)ᵀ]
and collecting like terms, we get:
A = (a11E11 + a22E22 +...+ annEnn) + a12(E12 + E21) + a13(E13 + E31) +......+ a(n-1,n)(E(1-n,n) + E(n,n-1))
so it is clear the above set {Eii: 1 ≤ i ≤ n} U {Eij+Eji : 1 ≤ i,j ≤ n, i ≠ j} spans the set of symmetric matrices.
a litle reflection should convince you it is linearly independent as well, and thus a basis.
I thought I understood this but I'm having confusion. So I'm reopening this thread. Sorry about that.
Then Let's say we want to find the dimension of each subspace of
and are real numbers}
By writing the representative vector as
In this case why and are not basis when all of them are in ?
Why the answer to the above question the basis is and ?
Why do I always have to have same number of vectors in basis equal to the number of variables? Is this a norm?
really you need to write W as {c(0,1,1) + d(1,-1,0): c,d in R}, and it is clear that a proper basis is {(0,1,1), (1,-1,0)}.
if two variables are independent of each other, it makes sense that a vector space based just on linear combinations
of those two variables, would have dimension....2.