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Symmetric Matrices to Jordan Blocks 
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#1
Feb1209, 01:53 PM

P: 115

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] 


#2
Feb1209, 03:01 PM

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Thanks
PF Gold
P: 39,552

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 (A4I)v is not 0 but (A 4I)((A4I)v)= 0. Such a vector is a "generalized" eigenvector In that case, (A4I)v must be an eigenvector! To find another vector such that (A 4I)^2v= 0, we must find a vector such that (A4I)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. 


#3
Feb1209, 03:24 PM

P: 115

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. 


#4
Feb1309, 10:11 PM

P: 333

Symmetric Matrices to Jordan Blocks
Hey! I recognize the matrix [itex]
\left[ \begin{array}{rr} 4&1\\ 0&4\\ \end{array} \right] [/itex] is a Jordan Block. 


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