Symmetric Matrices to Jordan Blocks

In summary, when a matrix A is diagonalizable, it can be written as A= S \Lambda S^{-1} where S is formed from eigenvectors that correspond to the eigenvalues in \Lambda. However, when A is not diagonalizable, it does not have a complete set of eigenvectors. In order to form S, generalized eigenvectors must be used to fill out the matrix. This can be done by finding vectors that satisfy (A-\lambda I)v= <1, -1> and adding them as columns to S.
  • #1
LogicalTime
114
0
I've been working through the Linear Algebra course at MITOCW. Strang doesn't go into the Jordan form much.

When a matrix A is diagonalizable then

[itex]
A= S \Lambda S^{-1}
[/itex]

and the matrix S can be formed from eigenvectors that correspond to the eigenvalues in \Lambda

Question:
how do I form S when A is not diagonalizable?
ie.
[itex]
\left[
\begin{array}{rr}
5&1\\
-1&3\\
\end{array}
\right]
=S \left[
\begin{array}{rr}
4&1\\
0&4\\
\end{array}
\right]
S^{-1}
[/itex]
 
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  • #2
The fact that A is NOT diagonalizable means that it does NOT have a complete set of eigenvectors. You need to form S from eigenvectors as much as possible and use "generalized" eigenvectors to fill out S.
In your example, the eigenvalue equation is
[tex]\left|\begin{array}{cc} 5-\lambda & 1 \\ -1 & 3-\lambda\end{array}\right|[/tex]
[tex]= \lambda^2- 8\lambda+ 16= (\lambda- 4)^2= 0[/itex]
so 4 is a double eigenvalue (which you knew).

An eigenvector corresponding to eigenvalue 4 must satisfy
[tex]\left[\begin{array}{cc}5 & 1 \\ -1 & 3\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]= \left[\begin{array}{c}4x \\ 4y\end{array}\right][/tex]
and so must satisfy 5x+ y= 4x and -x+ 3y= 4y, both of which reduce to y= -x. Any vector of the form <x, -x>= x<1, -1> is an eigenvector. But that's only one eigenvector which is why we cannot diagonalize this matrix.

Now, every matrix satisfies it own characteristic equation: (A- 4I)^2= 0 which means (A- 4I)^2v= 0 for every vector v. Obviously, if v is an eigenvector with eigenvalue 4, (A- 4I)v= 0 so (A- 4I)(A- 4I)v= (A- 4I)0= 0. But it might also be possible that (A-4I)v is not 0 but (A- 4I)((A-4I)v)= 0. Such a vector is a "generalized" eigenvector In that case, (A-4I)v must be an eigenvector! To find another vector such that (A- 4I)^2v= 0, we must find a vector such that (A-4I)v= <1, -1>. That gives the equation
[tex]\left[\begin{array}{cc}5- 4 & 1 \\ -1 & 3- 5\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]= \left[\begin{array}{cc}1 & 1 \\ -1 & -2\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]= \left[\begin{array}{c}1 \\ -1\end{array}\right][/tex]
which gives the two equations x+ y= 1 and -x- 2y= -1. Adding the two equations, -y= 0 so y= 0. Then x= 1. A "generalized" eigenvector is <1, 0>
Take
[tex]S= \left[\begin{array}{cc} 1 & 1 \\ -1 & 0\end{array}\right][/tex]
taking the eigenvector and "generalized" eigenvector as columns. Then
[tex]S^{-1}= \left[\begin{array}{cc} 0 & -1 \\ 1 & 1\end{array}\right][/tex]
and
[tex]S^{-1}AS= \left[\begin{array}{cc} 0 & -1 \\ 1 & 1\end{array}\right]\left[\begin{array}{cc}5 & 1 \\ -1 & 3\end{array}\right]\left[\begin{array}{cc} 1 & 1 \\ -1 & 0\end{array}\right][/tex]
[tex]= \left[\begin{array}{cc}1 & -3 \\ 4 & 4\end{array}\right]\left[\begin{array}{cc} 1 & 1 \\ -1 & 0\end{array}\right]= \left[\begin{array}{cc} 4 & 1 \\ 0 & 4\end{array}\right][/tex]

If your matrix, A, had [itex]\lambda[/itex] as a triple eigenvalue but only one eigenvector corresponding to that eigenvalue, say, [itex]v_1[/itex], then you would look for a vector [itex]v_2[/itex] such that [itex](A-\lambda I)v_2= v_1[/itex] and a vector [itex]v_3[/itex] such that [itex](A- \lambda I)v_3= v_2[/itex].

By the way, your title "Symmetric Matrices to Jordan Blocks" puzzled me. All symmetric (real) matrices are diagonalizable so the question of Jordan Blocks doesn't arize with them. And, of course, your example is not a symmetric matrix.
 
Last edited by a moderator:
  • #3
Sry meant to write "Similar matrices and Jordan Blocks", thought one thing and wrote another. Is there a way to change the name of the thread?

Thanks for answering my question, it makes sense. I'll have to go look up generalized eigenvectors now and see what they are all about.
 
Last edited:
  • #4
Hey! I recognize the matrix [itex]
\left[
\begin{array}{rr}
4&1\\
0&4\\
\end{array}
\right]
[/itex] is a Jordan Block. :biggrin:
 

Related to Symmetric Matrices to Jordan Blocks

1. What is a symmetric matrix?

A symmetric matrix is a square matrix that is equal to its own transpose. In other words, the elements on either side of the main diagonal are equal to each other. For example, A is symmetric if Aij = Aji for all values of i and j.

2. What are Jordan blocks?

Jordan blocks are square matrices that have a specific form, often used in linear algebra and matrix analysis. They are defined as a square matrix with a non-zero diagonal element, and all other elements on the first subdiagonal are equal to 1. The remaining elements are all 0. For example, a 3x3 Jordan block looks like this:

B = [a 1 0
        0 a 1
        0 0 a]

3. How do you convert a symmetric matrix to Jordan blocks?

The process of converting a symmetric matrix to Jordan blocks is known as diagonalization. It involves finding the eigenvalues and eigenvectors of the matrix and then using them to construct a diagonal matrix. The diagonal matrix will have the same eigenvalues as the original matrix, but the eigenvectors will be used to create the Jordan blocks. The resulting diagonal matrix will be similar to the original symmetric matrix, but with the Jordan blocks instead of the original eigenvalues on the diagonal.

4. What is the significance of converting a symmetric matrix to Jordan blocks?

Converting a symmetric matrix to Jordan blocks can be useful in various applications, such as in solving systems of linear equations, analyzing the behavior of dynamical systems, and studying the properties of quadratic forms. Jordan blocks provide a simpler form to work with compared to the original matrix, making it easier to perform calculations and gain insights into the behavior of the system.

5. Are there any limitations to converting a symmetric matrix to Jordan blocks?

Yes, there are some limitations to converting a symmetric matrix to Jordan blocks. First, the symmetric matrix must have distinct eigenvalues. If the matrix has repeated eigenvalues, the Jordan blocks will not be unique. Additionally, not all symmetric matrices can be converted to Jordan blocks. For example, a symmetric matrix with complex eigenvalues cannot be diagonalized using real Jordan blocks. In such cases, complex Jordan blocks must be used instead.

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