Re: Show this matrix is not diagonalizable-

Carl R. wrote:

>Let A be the matrix [1, 1 ; 0,1] and consider the following
>differential equation:
>
>dy/ dt = A*y
>
>Now here's the problem:
>
>By considering the solution of dy/dt = A*y
>Show the matrix A is not diagonalizable.
>
>
>Ok so far I found the solution is given by y(t)=C_1* e^t + C_2* t*e^(t)
> where C_1, C_2 are constants.
>Then I have no clue how to proceed, any ideas?
>
>
>Thanks in advance.
>
It is all linear algebra! Linear differential equations is just one of its many applications.

I did a little Google search for "derogatory matrix" - I thought that that was the terminology for NxN matrices with fewer than N independent eigenvectors.

I found this at the third Google position: http://www.psi.toronto.edu/matrix/index1.html ;
this leads to http://www.psi.toronto.edu/matrix/ ,

which yields in turn http://www.psi.toronto.edu/matrix/matrix.html - Matrix Reference Manual by Mike Brookes

The correct name is "defective matrix" or better: "defective linear operator".

To prove that [[1, 1], [0, 1]] is defective, launch a direct attack: just try and solve the equations
(?lambda, a, b | a+b = lambda . a, b = lambda . b). You will find that necessarily lambda=1 and b=0. Done!

When starting with the characteristic equation, observe that there is the only =algebraically double= eigenvalue 1, and that the column vector (1, 0) and its multiples are its only eigenvectors.

You should find this stuff and more in any college or university textbook on linear algebra.

Cheers: Johan E. Mebius

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