# diagonal

• Nov 23rd 2010, 12:39 AM
alexandrabel90
diagonal
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?
• Nov 23rd 2010, 01:07 AM
DrSteve
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.
• Nov 23rd 2010, 01:17 AM
alexandrabel90
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?
• Nov 23rd 2010, 04:37 AM
DrSteve
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

0 -1
1 0
• Nov 23rd 2010, 05:20 AM
alexandrabel90
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.
• Nov 23rd 2010, 05:25 AM
alexandrabel90
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.
• Nov 23rd 2010, 06:05 AM
FernandoRevilla
There is a nice generalization:

$T:\mathbb{R}^{n\times n}\rightarrow \mathbb{R}^{n\times n},\quad T(A)=A+A^t$

(i) $T(A)=0 \Leftrightarrow A+A^t=0 \Leftrightarrow A^t=-A \Leftrightarrow{\;A\textrm{\;is\;skew\;symmetric}}$

So, $\lambda_1=0$ is an eigenvalue of $f$ and $\ker (T-0I)=F_1$ (subspace of skew symmetric matrices).

(ii) $B$ symmetric implies $B=(1/2)(B+B^t)$ , as a consequence $T(B)=2B$. Reciprocally,

$T(B)=2B\Rightarrow B+B^t=2B \Rightarrow B^t=B\Rightarrow B\textrm{\;is\;symmetric}$

That is, $\lambda_2=2$ is an eigenvalue of $T$ and $\ker (T-2I)=F_2$ (subspace of symmetric matrices). But $\mathbb{R}^{n\times n}=F_1\oplus{F_2}$. This implies:

(a) $T$ is diagonalizable on $\mathbb{R}$.

(b) $D=\textrm{diag}(0,\ldots,0,2,\ldots,2)$ with $0$, $n(n-1)/2$ times and $2$, $n(n+1)/2$ times.

As a particular case for $n=2$:

$V_0=\left\{{\begin{bmatrix}{0}&{-1}\\{1}&{0}\end{bmatrix}}\right\},\quad V_2=\left\{{ \begin{bmatrix}{1}&{0}\\{0}&{0}\end{bmatrix},\begi n{bmatrix}{0}&{1}\\{1}&{0}\end{bmatrix},\begin{bma trix}{0}&{0}\\{0}&{1}\end{bmatrix} }\right\}$

and $D=\textrm{diag}(0,2,2,2)$

Regards.

Fernando Revilla
• Nov 23rd 2010, 06:16 AM
alexandrabel90
But i haven learnt abt symmetry. Is there any other mtd?
• Nov 23rd 2010, 07:56 AM
FernandoRevilla
Quote:

Originally Posted by alexandrabel90
But i haven learnt abt symmetry. Is there any other mtd?

My last post was only a generalization, for your particular case follow DrSteve's outline. The canonical basis for $\mathbb{R}^{2\times 2}$ is:

$B=\left\{{e_1=\begin{bmatrix}{1}&{0}\\{0}&{0}\end{ bmatrix},e_2=\begin{bmatrix}{0}&{1}\\{0}&{0}\end{b matrix},e_3=\begin{bmatrix}{0}&{0}\\{1}&{0}\end{bm atrix},e_4=\begin{bmatrix}{0}&{0}\\{0}&{1}\end{bma trix}}\right\}$

and:

$\begin{Bmatrix}T(e_1)=2e_1\\T(e_2)=e_2+e_3\\T(e_3) =e_2+e_3\\T(e_4)=2e_4\end{matrix}$

So, the matrix of $T$ in respect to $B$ is:

$A=\begin{bmatrix}{2}&{0}&{0}&{0}\\{0}&{1}&{1}&{0}\ \{0}&{1}&{1}&{0}\\{0}&{0}&{0}&{2}\end{bmatrix}$

as DrSteve pointed out. You'll find as eigenvalues $\lambda_1=0$ (simple) and $\lambda_2=2$ (triple). And now, if you obtain for example $(0,-1,1,0)$ as eigenvector for $\lambda_1=0$, these are coordinates on $B$, so this vector is exactly:

$0e_1-1e_2+1e_3+0e_4=\begin{bmatrix}{0}&{-1}\\{1}&{0}\end{bmatrix}$

Regards.

Fernando Revilla