Question about free variables?

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SUMMARY

The matrix A given by the entries |1 1 1 1|, |0 1 0 0|, |0 0 1 0|, |0 0 0 1| is proven to be non-diagonalizable due to the discrepancy between its algebraic and geometric multiplicities. The eigenvalue identified is 1, with an algebraic multiplicity of 3. The reduction of the matrix leads to a system where the free variables are x1, x3, and x4, confirming that the geometric multiplicity is less than the algebraic multiplicity, thus establishing non-diagonalizability.

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Homework Statement


Show that A=
|1 1 1 1|
|0 1 0 0|
|0 0 1 0|
|0 0 0 1|
is not diagonalizable.

Homework Equations


The Attempt at a Solution


I know i have to show the algebraic multiplicity does not equal the geometric mult. of the eigenvalue. So i found an eigenvalue to be 1 which then reduces the matrix to be:
|0 1 1 1|
|0 0 0 0|
|0 0 0 0|
|0 0 0 0|

And now i know that x2=-x3-x4 with the free variables being x1,x3,x4 but i do not know how to write this as a solution. Would it be
x1[1 0 0 0]T + x3[-1 0 1 0]T + x4[-1 0 0 1]T??
 
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so what is the algebraic multiplicity guessing it is 3

try testing the vectors you found on the original matrix, it should be clear x3 & x4 do not have eigenvalue of 1, in fact they do not satisfy the condition you stated, which is correct
 

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