i just started on diagonalizable matrix and i am trying to prove that f: 2x2 matrix--> 2x2 matric where f(A)= A + A^t. find a basis for the 2x2 matrix which consist of eigenvectors for f.
im wondering, the basis would be in the form of a 2x2 matrix yet eigenvectors are column matrices..so how do i go about doing it when the sizes are different?
A basis is a set of vectors.These vectors must be linearly independent and must span the given space.
So you are looking for 2 eigenvectors that form a basis - that is you are looking for 2 column vectors. You put these two vectors together to form the 2x2 matrix P.
that makes sense.
so to find the eigenvetors, we need to know A such that f(v)=Av where A has to be a 2x2 matrix right?
since the funct is defined by f(A)= A+A^t, what i tried to do what such A to be 2x2 elementary matrices and obtain that the image is of the form
(1 0 ),( 0 1), (0 0)
(0 0),(1 0), (0 1)
[ read it as 3 2x2 matrices]
how do i find A then so that i can find its eigenvectors?
Apologies - I misread the question. I didn't realize there was a linear transformation involved. The question actually doesn't quite make sense as written. You should check the wording and maybe repost it. Did you mean to prove that f is a linear transformation? If so, then this is straightforward. f is essentially a map from R^4 to R^4, so its eigenvectors have four components - not 2. Which means the P matrix would be 4x4. I think we can get that as follows. It's been a while since I've done a problem like this, so if someone could confirm that this is correct I'd appreciate it.
First, identify each 2x2 matrix with a vector in R^4, and think of the linear transformation f as a mapping from R^4 to R^4 (in other words write the matrix with entries a, b, c, d as a column vector with entries a, b, c, d).
Next, find the matrix of the transformation by computing f of each basis vector. The matrix of the transformation is the following:
2 0 0 0
0 1 1 0
0 1 1 0
0 0 0 2
Now find the eigenvalues and eigenvectors of this matrix in the usual way. You should get eigenvalues of 0 and 2 (multiplicity 3)
The corresponding eigenvectors are
0 1 0 0
-1 0 1 0
1 0 1 0
0 0 0 1
Rewrite these eigenvectors as matrices. For example the eigenvector with eigenvalue 0 is
The exact qns is
Let f : 2x2 matrix -> 2x2 matrix be the linear map defined by f(x)=A+A^t
. Prove that f is diagonalizable by finding a basis for 2x2 matrix consisting of eigenvectors for f.
How do you know when its from r^4 to r^ 4 and not a 2x1 basis column matrix? I thought your first explAnation made sense. Thanks for your help.
There is a nice generalization:
So, is an eigenvalue of and (subspace of skew symmetric matrices).
(ii) symmetric implies , as a consequence . Reciprocally,
That is, is an eigenvalue of and (subspace of symmetric matrices). But . This implies:
(a) is diagonalizable on .
(b) with , times and , times.
As a particular case for :
But i haven learnt abt symmetry. Is there any other mtd?