Check whether the matrices A and B are diagonalisable?Diagonalise those matrices which are diagonalisable. (i) #A={[-2,-5,-1],[3,6,1],[-2,-3,1]}# (ii) #B={[-1,-3,0],[2,4,0],[-1,-1,2]}#.

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Feb 15, 2018

Answer:

#A# is not diagonalizable.
#B# is diagonalizable and we have #B = SDS^{-1}# where one diagonalizing matrix is # S = ((1,0,3),(-1,0,-2),(0,1,1))# and the diagonal matrix is # D = ((2,0,0),(0,2,0),(0,0,1))#

Explanation:

For the matrix #A#, the characteristic equation is
#|[-2-lambda,-5,-1],[3,6-lambda,1],[-2,-3,1-lambda]| = 0 #
After some algebra this can be recast in the form
# 4-8 lambda + 5 lambda^2-lambda^3=0#
or #(1-lambda)(2-lambda)^2 = 0#
Thus the eigenvalues are 1,2 and 2.

Since the eigenvalue 2 is repeated (it is algebraically degenerate) there is a possibility that there may not be enough independent eigenvectors to form a diagonalizing matrix. To check that, let us try to find the independent eigenvector(s) corresponding to #lambda = 2# :

#([-4,-5,-1],[3,4,1],[-2,-3,-1])([x],[y],[z]) = ([0],[0],[0]) #

By elementary row operations (or just by appropriately adding or subtracting the equations) we get #x = -y=z# so that the only independent eigenvector corresponding to #lambda =2# is # (1,-1,1)^T#.
Since #A# has only two independent eigenvectors (one for #lambda =1# and one for #lambda=2#) it is not diagonalizable.

It is easy to see that #B# has the same characteristic equation as #A#. Once again, let us examine the case for #lambda=2# closely.
Here the eigenvector(s) satisfy

#([-3,-3,0],[2,2,0],[1,1,0])([x],[y],[z]) = ([0],[0],[0]) #

It is obvious that in this case the only constraint on the eigenvectors is #x+y=0#. This means that we can write down two independent eigenvectors, say #(1,-1,0)^T# and #(0,0,1)^T#

To complete the diagonalization, we need to find the eigenvector corresponding to #lambda=1#. It is easy to check that one such eigenvector is #(3,-2,1)^T#

So, a diagonalizing matrix is
# S = ((1,0,3),(-1,0,-2),(0,1,1))#

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